 So we're going to do a solution key for the January 2008 exam, Form B. That's this one here. I'll try and talk you through how I'm doing the questions. I've never tried doing a verbal video solution key before, so it may not quite go as smoothly as I hope, but we'll see how well this works. So we're on the non-calculator section. Number one, determine the amplitude of the function. Amplitude, right there. Amplitude is always positive. Amplitude is four. Remember, originally the amplitude of cosine and of sine is one. The amplitude of tangent is undefined because tangent has a range of all reels. And secant, cosecant, and cotangent, you can figure out by looking at their reciprocal sine, cosine, and tangent. Number two, the terminal arm of angle theta in standard position intersects the unit circle at point m comma n. Hey, that's x. That's y. And which expression represents tan theta? Well, tangent is y over x, which is going to be n over m. A. Determine the exact value of cosecant of 7 pi by 4. Exact value is one of our trigger phrases. That means special triangles. Let's see. Here they told me the angle, so I'll do a quick sketch. There's 4 pi by 4, 6 pi by 4, 7 pi by 4, right there. And if I use the cast rule, I can already tell cosecant, cosecant, cosecant goes with sine. Sine is negative, so cosecant will be negative in this quadrant. That means those two answers are wrong. My reference angle is pi by 4 because all the way around is 2 pi or 8 pi by 4. Which triangle has the pi by 4 in it? That's the 1, 1, root 2 triangle. And that has a cosecant, cosecant goes with sine. Cosecant is hypotenuse over opposite, which is going to be root 2 over 1. Oh, but we're in this quadrant, negative. Number 4, in a circle, an arc, so this is an arc length question. Remember the arc length equation is, it looks like the word arc, it's that. They want us to find the radius. The radius is going to be a divided by theta. The key here is, though, that theta has to be in radians. So the arc length is 30, 120 degrees in radians. What's that? Well, you can either go 120 degrees times pi radians over 180, and yes, that's fair game on the non-calp. All you can realize that 120 degrees is 60 degrees times 2. And 60 degrees, you may remember from your special triangle, is pi by 3. This is 2 pi by 3 radians, how to divide by a fraction, flip the reciprocal, multiply by the reciprocal. So this is going to be 30 times 3 over 2 pi, which is 90 over 2 pi, or 45 over pi. The answer to number 4 is B. Number 5, determine an equivalent expression for that thing. Well, hopefully you have your formulas sheet handy. I'm looking on my formulas sheet, and I'm noticing that this looks an awful lot like sine something, cos plus cos something, sine. This looks an awful lot like sine alpha, cos beta, plus cos alpha, sine beta, which is actually, from my formula sheet, the sine of alpha plus beta, sine cos plus cos sine. Look at what's sitting where the alpha is. This is going to be the sine of 3x, what's sitting where the beta is, x. So I get the sine of 4x. Is that one to pick from there? No. That must mean they want me to do a little more. I don't think this is the right answer. I think they're trying to trick me with that, so uh-uh. And I'm looking at the rest of these. The rest of these have a sine cos, I think this is a double angle identity in disguise. By the way, this is a fairly tough question in case you're wondering. I think this is actually the sine of 2 times a, where a equals 2x. 2 times 2x is 4x. But this helps me visualize a little better. 2a, that's 2, sine a cos a, that's our double angle identity, if I make a temporary substitution to visualize better, which means this is 2, what's a, sine 2x cos 2x. So I used two different identities. The answer to number 5 is d. Number 6, which of the following is the graph of one period of the function y equals secant of 2x? Well, let's see. Secant goes with cos, cos x would look like this. Starts up here, ends here, there's pi, there's 2pi, cos would look something like that. So secant would be the reciprocal. It would have an asymptote there, have an asymptote there, it would touch one high, it would touch negative one high. I'm doing a real quick rough sketch here. If secant is the reciprocal, this is secant of 2x. So this is undergoing a horizontal compression by a half. This point right here is 0, 1. If I compress that by a half, it's still 0, 1. So I'm looking for a graph that goes through 0, 1, which means it's either c or d. But not only that, these values would all get compressed by a half. This would be instead of pi, pi by 2. This instead of pi by 2 would be pi by 4. So I'm looking for a graph that has an asymptote at pi by 4. Let's see. Here, this one. There's pi by 2, halfway between 0 and pi by 2 is pi by 4. Here's my vertical asymptote, and it also goes through 0, 1. D. Tough question. Given the graph of that thing, find the value of B. Remember, B is 2 pi over the period. So how long is the period here? This is a bit of a tough one. They haven't given me nice easy points to work with. Let's see. Normally I try and go from a top point to a top point, but that's kind of halfway between. I'll have another one. Or I'd go from bottom to bottom. That's kind of halfway between. I'm going to go from middle and on my way up to middle and on my way up. That's one full wave, right there. How long is that from there to there? Let's see. 1, 2, 3, 4, 5, 6, 7, 1, 2, 3, 4, 5, 6, 7. Each square is pi by 7, and this is 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 squares long. So the period is 10 pi by 7. This is going to be 2 pi divided by 10 pi by 7. How do I divide by fraction? Multiply 2 pi times 7 over 10 pi. The pi's cancel. I get 14 over 10, and I'm pretty sure that 14 over 10 reduces to 7 over 5. Number 8. Solve this. This is a quadratic trig, which means I'm expecting four answers. I'm pretty sure A and C are wrong. We would get the cos squared by itself. This would become cos squared x equals 3 over 4. How do I get rid of a squared? Square root. So it's going to be cos x equals root 3 over 2, because the square root of 3 is root 3. Square root of 4 is 2. I can do that evenly. Ah, but I have to remember when I square root what, what, what, yes, plus or minus. That's where the four answers are going to come from, by the way. I'm already going to nuke those two. So cosine can be positive or negative. It's going to be here or it's going to be here. What's the reference angle? Well, I have a triangle with a root 3 and a 2 in it. That's the 1, 2, root 3 triangle. Which of these is a cosine of root 3 over 2? The bottom angle or the top angle? You know what? I've always told you guys to draw this very elongated, so it's clear which is the smaller one and which is the bigger one. So why don't I model that? 1, 2, root 3. This is the one that has a cosine of root 3 over 2 and it's a reference angle of pi by 6. You know what? That means pi by 6, pi pi by 6, 7 pi by 6, 11 pi by 6 are my answers. None of those ones. I'm not even going to waste my time on the multiple choice finding all four answers. I know it's that one. It's going to look out my reference angle. And yes, you're allowed to do that. They have no way of knowing on the multiple choice. Be clever. Final equation for the asymptote of this exponential graph. Maybe an exponential graph has the asymptote, the x-axis, y equals 0, except this graph has been moved four up. Solve. I have to write this as equivalent exponents. I have to equate the exponents. I can't though. My bases are not the same yet. So the first thing that I want to do is write this as identical bases. This is going to be 3 squared to the x equals 3 cubed to the x minus 3. Now I can use my power to a power rule. This is going to be 3 to the 2x equals 3 to the 3x minus 9. Since my bases are the same, then my exponents are the same. 2x equals 3x minus 9, minus 3x from both sides, and you get negative x equals negative 9. I'm pretty sure that x equals 9. Am I right? There it is. We're blazing. Number 11. Solve. By the way, this also occasionally appears as a written question, although usually there'll be a multiplication, a plus sign, and it'll end up being a quadratic. Here, my bases are the same, so I can write this as 1 log. Subtracting outside the log is the same as dividing inside a log. Don't forget my base equals 2. Now if I had 1 log equals 1 log, then I could take the anti-log of both sides and cancel the logs. Here, I have to write this as an exponent. This to this equals what's inside the log. 5 squared equals what's inside the log. 5 squared is 25. How do I solve this? Yes, cross multiply. Hopefully, I don't need to put the 1 there. You'll see it, but I'll put it there anyways. You're going to get 3x equals 25x minus 75, minus 25x from both sides. You're going to get negative 22x equals negative 75. I think x is going to be 75 over 22. I think the answer to number 11 is C. Can they give us big numbers like 75s and 20s? Yes. I'll expect you to do some basic arithmetic without a calculator. Live with it. Evaluate this. Well, let's see. I'm probably going to write this as 1 big exponent. This is log base 5 of 5 cubed to the 1 half, because square root is the same as 1 half exponent. Oh, and I can use my power to a power rule. This is going to be the log base 5 of 5 to the 3 over 2, and I can move the exponent to the front. I'll get 3 over 2, log base 5 of 5, and what is log base 5 of 5? Just 1. The answer is 3 over 2, 13. These ones are tricky. These are those if-then questions. What I'm going to do is take this expression, and I'm going to try and find these in there. Let's see. My base is 2, my base is 2, my base is 2, so I'm not using the base change rule. I think I'm going to try and break this up. How will I break this up? Well, let's see. I notice that 15 is 5 times 3, and I notice there's a 5 there and a 3 there, so that's a good start. I think I'm going to rewrite this as log base 2 of 5 plus log base 2 of 3, because 15 is 3 times 5, or 5 times 3, and that's adding outside of log, the same as multiplying inside of log, and dividing inside of log is the same as subtracting outside of the log. I'm done. Watch. According to this question, the log base 2 of 5 can be replaced with an x, and the log base 2 of 3 can be replaced with a y, and log base 2 of 2 is just a 1. Solve this. Oh, boy. These are tough. The trick is to start in the middle and work your way out, okay? So let's... Sorry, start on the outside and work your way in. I said that totally backwards, and since it's a live recording, I can't fix that. Everyone now knows I made a mistake. Huh? I never make mistakes. I'm going to write this as 2 to the 1 equals what's inside the log, 2 to the 1 equals what's inside the log, and by the way, what is 2 to the 1? Hey, that's just plain old 2. I'm going to do the same thing now. This to that equals what's inside the log. X squared equals what's inside the log. See how I got my equation? I know. A little tricky. But this is a quadratic. How do I know? It's got a squared. That's equal to 0. Factor, you'll get x minus 3, x plus 2 equals 0. You'll get roots of 3 and negative 2, but if I put a negative 2 in for the base right there with the little x's, that's extraneous because bases cannot be negative. The answer to number 14 is A. Sorry, B. I circled the right one. Evaluate this thing here, okay? I think what they want you to do is realize that this is 6 times 5 times 4 times 3 times 2 times 1 all over 3 times 2 times 1. Yes, I could have just written 3 factorial, 3 factorial, and oh, times 2. Oh, you want me to go further? Okay. 6 times 5, that's 30 times 4, 120 divided by 260. Or you could have gone 4 divided by 2 and said, hey, that's a 2. I got 30 times 260. All right, 16. If my original graph is that, find the zeros of that. Well, this one has zeros of negative 4 and positive 2. Those are the roots. And this here, oh, I should write those as coordinates. In other words, this one has an x-intercept of negative 4, 0 and an x-intercept of 2, 0. This here is a horizontal compression by a half. Compress that by a half, compress that by a half. These are vertical, so they're not going to change at all. It looks like my new zeros are negative 2 and positive 1. The answer to number 16 is C. Now we move on to the calculator section, although a lot of these you can do without a calculator. So when you're writing the provincial, don't stop, move on. Which equation represents a graph of y equals f of x after it's expanded vertically by a factor of 5, that one there. Not there. Next to the x is horizontal, so those are definitely wrong. And I told you everything's backwards, so if you put this next to the y be a one-fifth, let's move it over here. B. So here's the only question on that page. Okay. The graph of that is shown below. Which of these represents that there? Okay. Remember the correct order for transformations is expansions, compressions, stretches, reflections, flips, slides, translations. I don't see any stretches, no coefficients. Oh, negative. There's a reflection. We're going to have a vertical reflection and then three left and one up. Vertical reflection. So this point is going to end up negative three down. Three left, one, two, three, one up. This point goes through negative six, negative two. Is that enough? You know what? I think that's all I need to do. There's not a single other point that goes through negative six, negative two. I'm quitting. That's not lazy. That's not cheating. That's a good test right here. Find the inverse of that. How do I find the inverse? Switch the x and y around. Actually, there is no y. This is one of the few times I don't like function notation. I'll put a y there and now I'll switch the x and y around. We're going to have x equals y cubed minus two. Oh, we've got to get the y by itself. Y cubed equals x plus two. I've plus the two over to this side. How do I get rid of the cube? Cube root. Do you have to plus or minus the cube root? No. Only an even root, like a square root or a fourth root. Et cetera, et cetera. The answer to 19 is A. Number 20. Graph of blah, blah, blah is shown below. Find the equation on the right. Well, definitely there's a vertical reflection because it used to open up and now it opens down. This is too wide. This is too wide. No horizontal stretch. This is too high. This is one high. So there's also vertical compression by a half. I wonder if that's enough. You know what? Every single one of these has a vertical compression by a half and a vertical reflection. That didn't help much. I'm going to try and follow this point right here. If you vertically compressed it, it would end up instead of two high, one high. And then if you vertically reflected it, it would end up negative one high. And it looks like now it's ended up right there. So I think it's been moved two up. I'm kind of leaning towards this one. Let's try it here. Vertical compression instead of four high, two high. Vertical reflection instead of negative, a positive two, negative two. Two up would end up there and that matches that. You know what? I'm looking at B. 21. So here's my original six comma negative 12. This here needs to be factored. I've taught my students that when you see it like this and there's a coefficient in front of the X and it's not factored. We need to factor. I need to rewrite this as F of negative one-third bracket X minus what, yeah, 18. Because negative one-third times negative 18 gives you positive six. So this is actually going to be horizontal compression by a third. Horizontal reflection and then 18 to the right. Horizontal compression by a third. Horizontal reflection, 18 to the right. Hopefully I have 16 comma negative 12 here. I do. Uh-oh, I'm wrong. What did I do here? Sloppy math, Mr. Deweyck? Oh, it's not a horizontal compression. Good gosh, now you're all laughing at me, shush. I could redo the video, I'm not going to bother. Not a horizontal compression, Mr. Deweyck. So horizontal expansion by three. So this would become an 18. This would become a negative 18, horizontal reflection. And then 18 right, that would become a zero. There's my correct answer. Good gosh. Well, if it makes you feel any better, even your math teachers get these wrong sometimes. Maybe I did that on purpose just to make you feel better. Yeah, if you believe that or whatever. Number 22. Ooh, Trig equation, decimal answers, graphing calculator. Don't waste your time trying to solve this algebraically. This is meant to be solved on a graphing calculator. Graph left side, five times, to type sine squared at sine x, sine x is the quickest way to type that. Y2 is cos x. I better make sure I'm in radians, which I'm not. And my window, I'm going to go from zero to two pi. And this is an amplitude of five, so I'm going to go, oh, scale one. Actually scale, I usually chose pi by six, didn't I? Amplitude, I'm going to go from negative six to six. I think that'll work better. Scale, I don't know, to graph. If I went too fast, pause the video, eh? Okay, I got two answers, two solutions. You know, I'm going to be clever. These ones all look really close. I'm going to find the right hand one, because those are the ones that look like they have the biggest difference between them. In fact, I'm really leaning towards 5.84, because that's two pi. Let's find out. Second function, calculate intersection. First curve, second curve, guess 5.84. You know what? Since that's the only one that has the 5.84, I'm not going to bother finding the first one. 23. The point 5, negative 6 is on the terminal arm of standard position angle theta. Determine the smallest positive measure of theta in radians. I think what they're telling me is that x is 5, that y is negative 6. I think what they're really telling me is that tan theta is negative 6 over 5. x is 5, and y is negative 6. I know I'm in this quadrant here. I have to be, which means I'm between 3 pi by 2 and 2 pi. Those answers there are wrong. I'm going to find the reference angle. The reference angle, I'm going to ignore the negative. I'm just going to go inverse tangent of 6 over 5. I get 0.876. That's the reference angle, 0.876. To find the actual angle, it's going to be 2 pi minus 0.876. 2 pi minus the answer. Let's try that again, Mr. Dewey. Minus the answer, 5.41. The answer number 23 is C. Determine the restrictions for trig equations or trig equations. Restrictions for trig equations or trig functions can't divide by 0. First of all, in the denominator, sin x can't be negative a half, because if sin x works out to negative a half, I'm going to have a 0 in the denominator. I just found the root of that. What about in the numerator? Well, C can't is 1 over cosine. What we're also saying is that cos x can't be 0. Please tell me that's right there. Cos x can't be 0, sin x can't be a half. The answer number 24 is C. 25, ah, word problem. So let's see. I always say that when I'm doing a word problem, I do a sketch. And for my sketch, I want to find highest, lowest, and middle. Highest 16 when at midnight, so right at time zero. I'll even put a big dot there. Lowest 4 when 5.8 hours later. So we have a graph that looks like that. That's half a wave. There's an entire wave. So how long is one wave? It looks like 5.8 times 2, which is 10 plus 1.6. I think the period is 11.6. This graph starts up high, so it's going to be a positive cosine graph. And instead of y, I'll use h. What's the amplitude? Well, the total distance here is 12, so the amplitude is 6. Positive cosine starts up high. What's the period? 11.6. What's b? 2 pi over the period. I'll just write that in, I'm not fussy. T minus actually no phase shift, because I'm starting up high right at time zero, so T. And then my vertical displacement is going to be, well, the amplitude, the maximum minus the amplitude, or the minimum plus the amplitude. The vertical displacement is 10. There's my equation. Now I go to my calculator. Type in the equation, whatever I have here. So it's going to be 6 cosine of 2 pi x instead of T divided by 11.6, close bracket on the trig function, plus 10. Window, 0 to, well, one whole period is 11.6. I'll go from 0 to 12 hours. Good scale, I don't know, up by 2s. Y min, I like to see the ground is 0. My maximum height is 16, I'll go to 20. Scale, 2, sure. Now, this is asking me to find a time value. They've given me another x value, so what I do is I graph y2 is 8. There's my tide goes down, tide goes up. There's my second line, 8 meters high. What am I looking for? Where those two lines cross. So I use the intersection method. It's nice, it's visual. Guess close to there, enter. And I get 3.53 hours C. Occasionally, one of these will appear on the written section as well instead of a multiple choice question. Hey, we've just changed topics. Which of the following is a geometric sequence? I'm looking for a sequence where I'm multiplying by the same number, times by 4, times by now, times by, you know what? Well, let's see. Times by 2, times by 2, no. Times by, oh, this is looking better. Because to find r, it's any term divided by the 1 in front of it. So we have 12 divided by 18, that's r. And 8 divided by 12, that's r. Hey, that's the one that has a common ratio, 26d. Determine the number of terms in this. So they're asking me to find n. They told me a is 3. And I can figure out that r is 2, any term divided by the 1 in front of it. Not only that, they've also told me the n-th term. The n-th term is 49152. So I can use t sub n equals a r to the n minus 1. 49152 equals 3 times 2 to the n minus 1. Divide by 3, 49152 divided by 3. 16384, 16384 equals 2 to the n minus 1. Now, we could solve this algebraically by taking the log of both sides. But I'm going to tell you right now, it's going to be way faster just to plug in these answers on our calculator, because we're on the calculator section. So I'm going to go 2 to the power of 16. No. 2 to the power of 15. No. 2 to the power of 14. No. Oh. OK. This needs to work out to 14, which means I'm pretty. Oh, Mr. Doe, nice job. If that's supposed to work out to 14, I'm pretty sure that n needs to be 15, because 15 take away 1 is 14. Determine the sum of this infinite series here. So for an infinite series, it's a over 1 minus r. Let's see, a is 16. r is any term divided by the 1 in front of it, negative 12 over 16, which is in lowest terms, divide by 4, divide by 4, negative 0.75. So it's going to be 16 divided by 1 minus negative 0.75. It's going to be 16 divided by bracket 1 minus negative 0.75. Yes, I know that's 1.75. I'm getting all of that. Math, enter, enter. 64 over 7 to make it a fraction. Evaluate. OK. I got a 3 here. I got a 12 here. 12 take away 3 is 9 plus 1. There are 10 terms. The first one is 32 times bracket negative 1 half to the power of 3. The first one is negative 4. The second one is this same thing to the power of 4. The second one is 2. That's enough for me. I think I can figure out the rest of this. It looks like a is negative 4, and it looks like r is any term divided by the 1 in front of it, negative 1 half, which I kind of see in the original question. Because now I can use my s sub n equals a bracket 1 minus r to the n all over 1 minus r. So this is actually going to end up being the 10th sum. It's going to be negative 4 bracket 1 minus negative 0.5, negative 1 half to the 10th all over 1.5. 1 minus negative 0.5 is 1.5. We're on the calculator section, so I can use this. Let's just make that a little bit neater so you can actually read what the heck is going on. Negative 4 bracket 1 minus bracket negative 0.5 closed bracket to the 10th closed bracket. I think that's the top. My brackets match. Yes, divided by 1.5. And I get an answer of negative 2.66, which is right there. 29d. Number 30, they gave me the fourth term and the seventh term. There's all sorts of ways to do this. You can do this algebraically. I've suggested just kind of draw a little picture. Fourth term, fifth term, sixth term, seventh term, fourth term, 250, seventh term, negative 16. I don't know what I multiplied by to get to there, but I called it r, and we multiplied by r, and we multiplied by r. In fact, if you start with a 250 and you multiply by r cubed, you get negative 16. r cubed is going to be negative 16 over 250. r is going to be the cube root of negative 16 over 250. Negative 16 divided by 250, math, cube root, cube root of that. And I get r is negative 0.4. I could now go backwards and figure out what a is. If this is term 7, term 8, term 9, term 10. 1, 2, 3, r, r, r. I think if I go negative 16 times negative 0.4 cubed, that's going to give me the 10th term. 1.024. Oh, they wanted a fraction fine. Math integer. 128 over 125. Positive? Yep. Number 30, D. I need to tie this up a bit so that when I print this, it doesn't look so hideous. So there's the cube root and there's a letter r. For what values of x will the following series have a finite sum? Well, if it's an infinite series, what we're really saying is that r has to be between negative 1 and 1. What's r here? 3x squared over x. r is 3x. What that means is that negative 1 has to be less than 3x has to be less than 1. I think what that's really saying is if I divide everything by 3, negative 1 third has to be less than x, has to be less than positive 1 third. I think x has to be between negative 1 third and positive 1 third b. Yeah, that's a tough question. You might not have seen that one before because the textbook, the workbook doesn't have that in there. I think it's in the review package that I gave you, but nobody asked me about that one. Ah, okay, change to exponential form. You know this is always on there. 2 to the fifth equals 3x. 2 to the fifth equals 3x, 8. Number 33. Population grows continuously according to the formula p equals p0e to the k, ooh, base e. Where p is the final population, t is the time, p0 is the initial population, k is the annual growth rate. What will the population be at the end of 10 years? Okay, so they want me to find final population. That means they gave me everything else. Ooh, initial population is 5,000 times e. The growth rate is 3%, so 0.03 and 10 years. So I'm gonna go like this, 5,000 times e. There's my e to the x button right there, second function ln to the power of 0.03 times 10. 6,749 b. 34. In 1872, Washington State experienced an earthquake of magnitude 6.8 on the Richter scale. Find an earthquake that is half as intense as the Washington State. So what we've said is 10 to the bigger minus smaller equals how many times as intense. So we're gonna go 10 to the bigger one that's 6.8 minus x, the smaller one we don't know, but we know that it was 0.5 or half as intense. How do I solve this? Take the log of both sides. Log of 10 to the power of 6.8 minus x equals the log of 0.5. This exponent can move to the front. You'll get 6.8 minus x times the log of 10 equals the log of 0.5. Log of 10 is just one. So really what you have is 6.8 minus x equals the log of 0.5. X ends up being 6.8 minus the log of 0.5. Minus that to that side plus that to that side. It's either gonna be b or c. 6.8 minus the log of 0.5. Oh, wait a minute. No, the log of 0.5 is negative. Silly me. D, 7.1. Is that correct? 34 wants c. You know why this is wrong, Mr. Dewick? Let me pause for a second and think about this. All right, we're back. Mr. Dewick thought about this for a while and he figured out his mistake. It's about 10.30 at night. I'm pretty tired, so my brain's not quite what it wants to be. I call this the bigger earthquake. This is the smaller earthquake. If I want the bigger earthquake to be Washington state, if the smaller earthquake is half as intense, that means the bigger earthquake is twice as intense. If the smaller one is half as big, the bigger one is twice as big. What I want here is not a 0.5. Because I'm calling the bigger earthquake and I'm using that, the Washington state one is my frame of reference, I don't want a 0.5 there. I don't want a 0.5 there. I want a two there. I want a two there. I want a two there. I want a two there. And now, if you go 6.8 minus log two, you get 6.5, which is C. Sorry about that. That's a pretty obscure mistake, but I shouldn't have made it anyways. Let's see if I can redeem myself on 35. We want to choose eight players, 10 males, 12 females. It doesn't look like order matters. Bucket. Males, females. 10, 12. We want at least three of each gender on the team. How many teams are possible? Well, let's see. So I can have three males, or four males, or five males. I can't have six males because then that would give me only two girls and I've got to have at least three girls. Does that make sense? Okay. I can also go with, can I go with more girls? Let's see. I can go with three girls. That would be five guys. I've got that. I can go four girls. That would be four guys. I can go five girls. That would be three guys. I can't go six girls because that would only be two guys and they want at least three of each gender. Okay, here we go. Or means add. For the first one, three males, that must mean five females. 10, choose three. 12, choose five. Or four and four. 10, choose four. 12, choose four. Or five and three. 10, choose five. 12, choose three. What's the answer gonna be? I'm gonna do this one group at a time. So 10, choose three times. 12, choose five. 95,040. Plus four and four. 103,950. Plus five and three. 55,440. So plus 103,950. Plus 95,040. I get 254,430. 35, see. Number six. How many even four digit whole numbers are there? For example, 12, 20 is acceptable but 0, 6, 7, 8 is not. The problem here is that zero is part of both restrictions. You can get that a zero, two, four, six, or eight in the last digit but you can't have a zero in the first digit. So I think the easiest way to do this is to divide this into two cases. Ends with a zero. Ends in a two, four, six, or eight. We have four digits, one, two, three, four. And we know for this it has to end in a zero. There's only one way to pick a zero. How many choices do I have for the first? Now you might think nine but remember zero has been picked, yeah, nine. Then eight, then seven. Or you can end in a two, four, six, or eight. That gives you four choices for the last digit. How many choices do you have for the first digit? Not nine because zero is still in your grab bag so you only have eight to pick from. Let's say you picked a seven. How many choices now? Not seven because now you can pick a zero and that is still in the grab bag. There's eight to pick from, then seven to pick from. The total number is gonna be this plus that. Nine times eight times seven and eight times eight times seven times four. Total number of even numbers, four digit even numbers is 2,296 and I'm completely wrong. What have I done wrong, Mr. Dewick? That type of thing wrong? Yeah, favors. Oh no, nine times eight times seven. Or eight times eight times seven times four. Getting later, Mr. Dewick, come on, brain. How many even four digit whole numbers? So if you end in a zero, you have one way to choose a zero. That leaves nine, then eight, then seven. If you end in a two, four, six or eight, you have four ways to choose a two, four, six or eight. So let's suppose you pick an eight. That means there's nine left in the grab bag. Still a zero though so you can only pick eight of them. Let's suppose you picked a seven. Now there's eight numbers left in the grab bag and all eight are available. You can pick a zero. Now there's seven. Then pause again and think. All right, I figured out what my mistake was. I didn't read the question carefully. I've written here and read, repetitions are allowed. This question doesn't say that I can't do repetitions. And that's how I solved that question. I actually thought, oh, I couldn't use the same number twice. But look, it says I can use the two twice. I can use any number twice. So this is still gonna be fundamental counting principle one, two, three, four digits. Now the last digit can be a zero, a two, a four, a six, or an eight. So I got five choices for that. The first digit can't be a zero. So I only have nine choices. But then there's no repetitions. I have 10 choices, zero through nine, 10 choices, zero through nine. And if I multiply those together, because the fundamental counting principles multiply, I get C. Number 37, an Olympic final race has seven competitors and how many possible ways could the gold, silver, and bronze medals be awarded? Well, order matters, gold, silver, bronze. So this is a permutation. It's gonna be from seven, permutate three, which is seven factorial over four factorial. Or you could go three lines, three blanks, and you could say seven choices for gold, six choices for silver, five choices for bronze. But no matter what, you're gonna get 210. Number 38, a little bit of a twist here. It's telling me the first term that they haven't told me N. Okay, let's see, let's think. Well, if this is the first term and the exponent on the A is a six and the original exponent here was a one, the only way the first term could have an A to the sixth would be if that was to the sixth power. So I can now solve this. I'm gonna use the term K plus one equals N, choose K, A to the N minus K, B to the K, where K equals, well, if I want the fourth term, if I want this to be a four, K is three. N is six, A is actually two, A, and B is actually negative B. It's gonna be six, choose three, two A to the N minus K, to the third, negative B to the third. Now all they want here is the coefficient, oh, I need to open up my calculator. All they want here is the coefficient. The coefficient is going to be six, choose three and a two to the third there. There's an eight there. It's gonna be six, choose three times two to the third or eight, coefficient's gonna be 160. Oh, wait a minute, wait a minute. I see a negative 160 and a positive 160. I should be a little bit more careful. Ah yeah, there's a negative one hidden right here, negative one to the third. I'm gonna end up with a negative 160, A cubed, B cubed. The correct answer to number 38 is A. I almost fell for that one. I gotta be a little more careful as you saw from my earlier mistake. Of course, if you have me as a teacher, you used to be making mistakes. Each day, number 39, a student chooses one out of three beverages. Over 10 days, he chooses apple juice three times, orange juice three times, and lemonade three times, four times, and how many different orders can this occur? So they're talking about orders. It's definitely a permutation. I think I can represent this as a word. I think this is three apple juices, three orange juices, and four lemonades. This is going to be 10 factorial divided by three factorial, three factorial, four factorial, which is gonna be 10 factorial divided by three factorial is six, three factorial is six, and four factorial is 24. It's sometimes a little easier just to do those small factorials in your head, less typing. 10 factorial divided by, now a common mistake kids might make is they might go six times, six times 24. But wait a minute, all three of these have to be in the denominator. I have to use brackets so my calculator knows that. And I get 4,200. All right, we've entered the probability zone. 22% of households have dogs as pets. 13% have cats. 3% have both. What percent have neither? So they're giving me percentages. They're giving me both. There's two categories. The best approach for this is a Venn diagram. So we have dogs, cats, 3% have both. That means if 22% have dogs, I only want to put a 19 right there because those combined are my 22. And if 13% have cats, I only want to put a 10 there. What percent have neither? Well, all the percentages have to add to 100. So I have 19 plus three is 22 plus 10, 32. If this is 32%, I'm pretty sure this has to be 68%. Number 40, D. 41, a card from a standard deck of 52 cards is, our card is drawn from a standard deck of 52 cards and then replaced. In 20 draws of this type, what's the probability that a red card will be drawn exactly 12 times? Now believe it or not, even though this is a card question, this is not a bucket. I know I've said card questions are buckets, but what they've done here is they're replacing the card each time, which means the odds never change. The odds every single time of a red card is one half. 26 out of 52, but I reduced it in my head, hopefully I didn't lose you there. How many draws are we doing? 20, so definitely not a tree. This is actually a binomial PDF question. This is going to be from 20, choose 12.5, one half. I want 12 successes, 12 reds. The odds of getting a black is 0.5, I want eight of those. Or you could use binomial PDF of 20.5, 12. This is sneaky and I know I've said to you, card questions aren't binomial, they're buckets. This card question though is unique because you're not hanging onto the card each time, you're putting it back in the deck and reshuffling so the odds never change. Fancy word for that of course is the trials are independent. 20, 0.5, 12, 0.120 on the answer, is C. Number 42, two fair six-sided dice are rolled and the face values showing are added. What's the probability that the sum of the face values is at least nine? You know what, for dice questions we do a table. We have the first dice and the second dice. The second dice has one, two, three, four, five, six. The first dice has one, two, three, four, five, six. So we can get one, one, one, two. You know what, I'm gonna save myself a bit of time because this is a test and I'm trying to go as fast as I can. I can tell you there's going to be 36 outcomes. I want the sum of the face values to be at least, what, we okay here? Well, try that again. I want the sum of the face values to be at least nine. I think that's gonna happen right here, right here, right here, right here, and right here, six comma three, six comma four, six comma five, and six comma six. I'm pretty sure there's going to be one, two, three, four, five, six, seven, eight, nine, 10, 10 out of 36, which I'm pretty sure is five out of 18. So you can take shortcuts like that long as you're really confident. Number 43, a rock concert is attended by 1,300 students of whom 760 are girls, 540 are boys. Four of these students are chosen to randomly receive backstage passes. What's the probability that at most three of the chosen students are boys? At most means zero, or one, or two, or three. That's a lot of work, or does mean add, and I could do each of those. I think it's gonna be the same as one minus the probability of four boys. Now, what's four boys? Well, this is going to be a bucket. How do I know this is going to be a bucket? Because once a kid receives the backstage pass, they're not put back into the pot. They can't get two passes in a row. So once one student has the pass, now there's 1,299 students to pick from. We have boys, girls, 540, 760, four boys, no girls. So it's gonna be one minus 540, choose four times 760, choose zero, all over from 1,300, choose four. My only question is will my calculator actually handle this? Let's see, 540 math, choose four. 760, choose zero is just one, so I'm not gonna bother typing it, divided by 1,300, choose four. Hey, I got an answer, one minus that answer, and I get 0.9704. What are the odds that you're gonna have zero, or one, or two, or three guys on this? Pretty good! Last multiple choice question. 10 discs, by the way, last question's usually pretty tough. I'm gonna expect that, not freak out. 10 discs numbered zero through nine are placed in a box. Two discs are randomly removed from the box without replacement. Determine the probability that one disc will have the digit five written on it, and the other disc will have the digit six written on it. Okay, I'm back, sorry about that, I'm doing this at home. It's kind of tough to find time by yourself, and I've just noticed that Chicago has won the Stanley Cup, so congratulations to them. I have it on mute, but good for them. Let's get back to this question. 10 discs numbered zero to nine are placed in a box. Two discs are randomly removed from the box without replacement. Determine the probability that one disc will have the digit five written on it, and the other disc will have the digit six written on it. Okay, I think what we're talking about here, I think we can divide this into winners, a five and a six, and losers, everything else. I think we have two winners, and zero through nine, that's 10 discs, eight losers. We want to choose both winners, no losers. I think it's two, choose two, and eight, choose zero, divided by 10, choose two. Let's see if that's right. I'm a little worried about this one. Two choose two is one, eight choose two is one, so this is one divided by 10, choose two, and then turn it into a fraction. I get one over 45, and I just double check my answer key. I am correct, woo-hoo, onto the written. There's the formula sheet, this page intentionally blanks, some rough work for graphing. I don't think I photocopied all that for you guys. I think yours probably starts right about here. So here's f of x. They want me to graph this thing, okay. Well, this negative means there's going to be a vertical reflection. I also see absolute value, and I see two up, not two left. That would be if it was inside the brackets next to the x. What's the correct order? Well, for absolute value, treat it like brackets, do what's inside first. I'm gonna go two up, then I'm gonna go absolute value, and then I'll go vertical reflection. If they wanted me to do the vertical reflection first, they would have written it this way. If they wanted me to do absolute value before two up, they would have written it that way. That means absolute value first, then two up. So treat it sort of like a bracket, like bed mess. So take this point here, two up, absolute value, it's already above zero, so no problem. And then vertical reflection, it ends up back where it started from, invariant. Two up, absolute value back here. Let's try this point. Two up, absolute value becomes positive. Vertical reflection ends up here. I better try this point then, because I'm not sure it's a straight line. Two up, right there. Absolute value stays there. Vertical reflection stays there. So I actually get out of nowhere a little triangular little shape like that. Let's try this point here. Two up, absolute value stays there. Vertical reflection stays there, ends up like that. How about here? Two up, absolute value, vertical reflection ends up back down here. That's kind of a weird one. Two up, absolute value stays there. Vertical reflection goes to negative two. Connect them. Two up, absolute value stays there. Vertical reflection comes right back. Two up, absolute value stays there. Vertical reflection, in fact, I think it does this. That's a weird one. That's about as tough as I've seen for an absolute value with a twist. You'll notice as soon as I saw the weird one, I went point by point by point by point. No shortcuts here. Reciprocal, ah, we love the reciprocal. We look for anywhere one high or negative one high. Those points are invariant. So that's gonna stay where it is. Same with that. Same with that whole little chunk of line right there. Then anywhere zero high becomes a vertical asymptote. And then I said you pretend you're a little bug walking on an invariant point. So if I start right here, if I move to the left, the graph doesn't exist. If I move to the right, the graph is getting bigger, my reciprocal's gonna get smaller, except my graph doesn't shoot off to infinity. My original goes to negative three, so my reciprocal's gonna go to negative one third. About there. Stand on this invariant point as I move to the left, I approach negative three, the reciprocal is going to approach negative one third about there. As I move to the right, I approach zero from below, I'll shoot off to negative infinity. And this section as I move to my left, I approach zero from below, I'll shoot off to negative infinity. It would look kinda sorta something like that. Two marks. Number three, ah, word problem. Population problem. So we're going to use A equals A zero C to the T over P. And a population of moths, 78 moths increase to 1,000 moths in 40 weeks. What's the doubling time? Okay, it looks like my final population is 1,000. My initial is 78. Doubling time, I'm pretty sure they want the growth rate to be two. They want the doubling time, what they're asking for is the growth period, 40 weeks. I would now divide by 78, divide by 78. 1,000 divided by 78 equals 12.82. Is that a nice fraction? 500 over 39. I'll use 12.82. I'd be a little leery of rounding off, you know what, I'm gonna use 12.8205. 12.8205 equals two to the 40 over P. Take the log of both sides. When you do that, you're gonna get the log of 12.8205 equals 40 log two divided by P. Hopefully by now, you recognize that you can just cross multiply. P ends up being 40 log two divided by the log of 12.8205. P is going to be 40 log two divided by the log of P. And since I have that previous answer stored in my calculator, I'll use it. I get 10.87 weeks. 10.87 weeks, I've rounded off properly. They want me to go to two decimal places, so I'm okay. Number four, probability question. And I see given it's a conditional probability question. Says, it is known that 15% of teenagers ride skateboards, three teenagers are chosen at random. Given that at least one of the three teenagers chosen rides a skateboard, determine the probability that all three rides skateboards. Given at least one, fine, all three. Now I've told you, by the way, three teenagers, you could do this with a tree, it'd be a pretty big tree, but you could do this with a tree. I'm looking at the answer key for the provincial, and it does say you get one mark for a tree diagram, and it does use a tree. I'm gonna try this using a binomial probability distribution function as well. This is going to be the probability of both of these, all three and at least one, divided by the probability of the given one. At least one. This is five marks, and it's fairly tough. I'm gonna do this a little bit separately. All three and at least one. Now at least one means one or two or three. That's this part, and means the overlap. Which of these also has all three? Only this one. If you combine all three and at least one, really that just means all three. I know it sounds weird in English, but if you don't believe me, draw the tree out, put a check mark underneath any tree that has at least one, and then circle the one that also has all three. You'll only circle one tree, all right? And that tree is all three. It's going to be point one five, point one five, point one five. The first guy rides, second guy rides, the third guy rides, point one five cubed. Or you could do it as a binomial probability distribution function, binomial PDF of three skate borders. There's the odds of one of them. All three, regardless you're gonna get point one five cubed. I'll leave it like this just so you can see how the bottom works for a binomial PDF. The bottom is at least one, which means one or two or three, now or means add. I could do each one separately. I think that's the same as one minus the probability of none riding. Cause that's the only other option in this group here. This is going to be, well let's figure out binomial PDF of three comma point one five comma three. I don't wanna write it out all the time. Three comma point one five comma three, which is the same as going point one five cubed. Point zero zero three three seven five. Point zero zero three three seven five. All over one minus binomial PDF. So oops, wait a minute Mr. Duke. PDF of three comma point one five comma zero. Equals binomial PDF of three comma point one five comma zero. One minus that answer. This is going to be point zero zero three three seven five divided by three eight five eight seven five. Point three eight five eight seven five. And I'm not gonna round off, it's easy enough to carry that. Point zero zero three three seven five divided by that answer there and I get that. As a fraction, three out of three hundred forty three, that's a nicer answer. Three out of three hundred and forty three four. They said answer accurate to at least four decimal places. Three divided by three hundred and forty three was what, point zero zero eight seven. Zero point zero zero eight seven. So you can also do that with a tree. Number five. Solve algebraically giving exact values. Okay. It's a quadratic trig equation. Exact values means special triangles. Ooh, GCF, greatest common factor. Sine x bracket two sine x plus root three equals zero. What are my roots? Here I'll get sine x equals zero. Here I'll get sine x equals negative root three over two. Sine x equals zero. This is a unit circle question. Where is sine x equal to zero? That's when your y coordinate is zero on the unit circle. Your y coordinate is zero right there and right there. X one is zero radians. X two is pi radians. Here, sine is negative. That means I'm here or here. It does say special triangles and I do have a, oh boy I'm having a hard time drawing a straight line. Good job. I do have a triangle that has a root three and a two and it's the one, two, root three triangle. Where this angle here has a sine of root three over two and that angle is pi by three. So that means that this angle is pi by three and this angle is pi by three. So I'll have x three, well, pi is three pi by three. Four pi by three is your first answer. And x four, two pi is six pi by three where pi by three less than five pi by three. Piece of cake. Number six, this is a bit of an interesting one. This is doing a period change. How can I tell so fast that there's a period change? Hey there's something in front of the x and the trig function. Tells me this. The smallest positive solution of this is 0.18. Find the general solution. The general solution is to add the period. If this was just tangent, you would go x equals 0.18 plus pi n because the period of tangent is pi. Oh if it was sine or cosine, it'd be two pi n. Except the period is not pi. The period is always pi over b or if it's sine or cosine, two pi over b. The period is pi over three. The period is pi over three. Oh and then two years ago they added this requirement. You also have to mention that n is an integer. Here you go. Oh, do I need to write more out between zero and two pi? Tangents, the nice one. You just have to write one and that generates the other one between zero and two pi. Cosine and sine, you usually have to write out both of them. Last question, the identity. Oh there's a cube, hang on. I'm not gonna let it freak me out. Now I am noticing both sides are ugly. I'm pretty sure I'm gonna end up working on both sides. I'm seeing binomial denominators, maybe conjugate, I don't know. I'm gonna tackle the right hand side first because sine two theta is one of those, I'll say no-brainer substitutions. What I mean by that is you always want to substitute sine of two theta. This is two sine theta, cos theta. Right from the formula sheet on the right hand side, about one-third of the way down. On the middle about one-third of the way down. Secant, secant, secant goes with which trig function? It goes with cosine. So this is gonna be times one over cos cubed theta. So I can tidy this up a little bit. I got one cos on the top, I got three on the bottom, now there's two. Can I cancel? It is factored. So this simplifies to two sine theta all over cos squared theta, and I can't do much more with that. Let's go tackle the left hand side. Well, how many fractions do I have on the right hand side? One. How many do I have on the left hand side? Two. Maybe I wanna write it as one fraction with a single common denominator. Now a single common denominator would be one minus sine theta, one plus sine theta. So I would multiply the first fraction by one plus sine theta all over one plus sine theta. And when I do that, I'll get sine theta, one plus sine theta, I'll multiply the brackets out on the next line. The plus sign would drop down like a domino. Here I have a one plus sine theta. I would multiply top and bottom by one minus sine theta, and I'll get plus sine theta, one minus sine theta. Let's keep going. Let's multiply out the bottom. One minus sine, one plus sine. I think I'm gonna get one minus sine squared theta. By the way, one minus sine squared theta. Ooh, I think, ooh, I think that's cos squared. I just got my nerdy adrenaline rush. Here, I'll get sine theta plus sine squared theta plus sine theta minus sine squared theta. I just saw the whole thing. The sine squared's canceled on the top. Sine squared minus sine squared is zero. And sine plus sine, why I'm pretty sure that that's two sine theta, which happens to be my numerator on the right-hand side. And one minus sine squared is cos squared. And I write Q-E-D, question easily done, or Ko Ertas demonstrandum. I get my nerdy adrenaline rush. I kick butt on the very last question, and I am proud because I have just got 100% on a provincial exam. No, you didn't, Mr. Newick. You made a couple of mistakes. Yeah, but I caught myself. Moral of the story is if you're riding the provincial, stay the full three hours. If you finish early, put your head down, let your mind wander. Often a mistake will, or an answer, will pop into your head. Our brains work like that. Don't leave early, even if it's nice sunny weather. Uh-uh.