For more information, please see full course syllabus of AP Calculus AB

For more information, please see full course syllabus of AP Calculus AB

## Discussion

## Download Lecture Slides

## Table of Contents

## Transcription

### AP Practice Exam: Section 1, Part A No Calculator

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro 0:00
- Exam Link 0:10
- Problem #1 1:26
- Problem #2 2:52
- Problem #3 4:42
- Problem #4 7:03
- Problem #5 10:01
- Problem #6 13:49
- Problem #7 15:16
- Problem #8 19:06
- Problem #9 23:10
- Problem #10 28:10
- Problem #11 31:30
- Problem #12 33:53
- Problem #13 37:45
- Problem #14 41:17

### AP Calculus AB Online Prep Course

### Transcription: AP Practice Exam: Section 1, Part A No Calculator

*Hello, welcome back to www.educator.com, and welcome back to AP Calculus.*0000

*Today, we are going to start on the AP practice exam.*0004

*What I would like to do, the practice exam that we are going to use is going to be a 2006 version.*0009

*You can find it on, I will write the link here.*0015

*The link is also going to be down at the bottom of the page in the quick notes.*0019

*www.online.math.uh.edu/apcalculus/exams.*0024

*When you enter that or when you click on the link down below,*0044

*the page will come up and it is going to give you a bunch of different versions.*0048

*The version that we are going to use is going to be version 5.*0051

*Just click on version 5 for part A, the section on part A.*0054

*We are also going to be doing when we get to, it is going to be version 5 for the part B.*0068

*For the written part, the free response questions, we are going to go with version 2.*0073

*You can either pull it up, head on your screen for you.*0079

*If you want, you can print it out, whichever you prefer.*0082

*Let us jump right on in.*0086

*Question number 1, we have got a couple of functions here.*0088

*We have f is equal to -2x + 1 and our g(x), I will write it as g = -x/ x² + 1.*0096

*In this particular case, we are to find f of g(1).*0109

*Nice and straightforward composition of functions.*0120

*First thing we do is we find the g(1).*0123

*g(1) is equal to -1/ 1² + 1 = -½.*0125

*f of g(1) is equal to f(-½).*0138

*We have -2 × -1/2 + 1.*0145

*This is 1 + 1 is equal to 2.*0152

*In this particular case, our answer is c.*0156

*Question number 2, I think I will do this on the next page.*0164

*You know what, I think I’m going to work in blue.*0170

*Question number 2, we have f(x) is equal to -x² – 4√x and we want f’ at 4.*0173

*Take the derivative, plug in 4, and see what it is that you actually get, the slope of the tangent line.*0194

*The slope of the tangent line is the value of the derivative at that point.*0205

*f(x) is that, f’(x), what do we get for f’(x)?*0215

*The derivative of this is -2x and this is going to be -4 × 1/2 × x⁻¹/2 = -2x - 2/ √x.*0222

*f’ at 4 is -2 × 4 - 2/ √4.*0245

*This is equal to -8 – 1, we have -9.*0255

*Therefore, in this particular case, the answer is going to be c.*0261

*Let us see what we have got, let me move this a little bit over here.*0270

*Question number 3, question number 3 is asking us to evaluate,*0278

*they want the limit as x goes to infinity of 2x³ + 4x/ -2x⁵ + x² – 2.*0286

*To evaluate this particular limit, a couple ways that we can do it.*0309

*One of the systematic way, whenever you are dealing with a rational function is to divide the top and bottom,*0314

*the numerator and the denominator by the highest degree in the denominator.*0322

*Basically, divide everything on the top by x⁵ and divide everything in the bottom by x⁵.*0326

*And then, take the limit as x goes to infinity.*0331

*Let us go ahead and do that.*0334

*Let us go ahead and bring this over here.*0336

*This is going to be dealing with a functional.*0338

*I do not want to keep writing limit symbol over and over again.*0340

*When I divide the top, it is going to be 2x³/ x⁵ - 4x/ x⁵.*0343

*On the bottom, I’m going to get -2x⁵/ x⁵ + x²/ x⁵ - 2/ x⁵.*0353

*This becomes 2/ x² + 4/ x⁴/ -2 + 1/ x³, I think -2/ x⁵.*0365

*Now we take the limit as x goes to infinity of this function.*0385

*As x goes to infinity, this goes to 0 because this becomes really big, therefore, this becomes really small, it goes to 0.*0394

*This goes to 0, this stays -2, goes to 0, goes to 0.*0403

*What you are left with is 0/ -2 which is equal to 0.*0409

*In this particular case, the answer will be a.*0414

*That is it, nice and straightforward limit.*0418

*Nothing particularly difficult about it.*0421

*Question number 4, let us see what question number 4 is asking us.*0425

*We have a particular function and it is bounded by certain numbers, certain functions.*0434

*It wants you to express the integral that gives you the area of the particular region that is bounded.*0443

*We have f(x), in this particular case is equal to x³.*0454

*Let us go ahead and we know what x³ looks like.*0462

*x³ looks like that, let us see, from -1 to 0.*0470

*They say that region 1, x goes from -1 to 0.*0480

*From -1 to 0, we are bounded by,*0490

*We have got, this is 1, this is -1.*0508

*We have this region and we have this region.*0518

*From -1 to 0, we call this our region 1.*0523

*From 0 to 1, this is our region 2.*0527

*Our total area is going to be the area of region 1 + the area of region 2.*0534

*This is fully symmetric here about the origin.*0542

*I can basically just take the integral of one of them and multiply it by 2.*0545

*I can just say it is 2 × the area of region 2.*0549

*It is equal to 2 × the integral from 0 to 1 of the upper function - the lower function.*0555

*It is going to be 1 - x³ dx.*0565

*As far as our choices are concerned, 0 to 1,2, just pull the constant in there, 1 - x³ dx.*0573

*It looks like our answer is c.*0584

*They give you a particular graph, they tell you the bounds, and you just have to find a particular area.*0589

*Let us see what number 5 says.*0600

*Question number 5 on this particular version.*0604

*We are given a particular function, we want you to determine the change in y with respect to x.*0606

*They want us to find the derivative dy dx.*0613

*We have 3x³ - 4xy – 4y² is equal to 1.*0619

*We see that in this particular case, our function is given implicitly.*0632

*We are going to use implicit differentiation.*0636

*Let us go ahead and do that.*0639

*The derivative of this with respect to x.*0641

*We are looking for dy dx.*0643

*The question asks, determine the change in y with respect to x.*0645

*Dy dx is what we are looking for.*0651

*The derivative of this is going to be 9x² – 4.*0653

*I just tend to pull my constants.*0661

*It is going to be 4 ×, this is a function of x and this is a function of x.*0664

*It is going to be this × the derivative of that.*0669

*x × y’ which is dy dx, just a shortened version, + y × the derivative of that which is 1.*0671

*- the derivative of this is -8yy’.*0681

*The derivative of 1 is 0.*0686

*We have got 9x² - 4xy’ - 4y – 8yy’ is equal to 0.*0690

*I have got 9x² - 4y, I’m going to put the y’ on one side = 4xy’ + 8yy’.*0703

*Factor out the y’, I got 9x² – 4y = y’ × 4x + 8y.*0720

*I’m left with y’ is equal to 9x² – 4y/ 4x + 8y.*0733

*As far as our choices are concerned, that is equivalent to 9x² - 4y/ - and -4 x is -8y.*0756

*The answer that you get, depending on how you did it, which you move to the left or the right,*0776

*maybe slightly different on the choices that you have.*0781

*What is probably going to be the most tricky part of this is,*0784

*just because you got something that does not look like any of your choices, it does not mean you are wrong.*0788

*Just see if you can change what you got and if it is equivalent to one of your choices.*0792

*In this case, it is equal to one of the choices.*0798

*It is going to be d.*0801

*Do not freak out, if what you got is going to be different.*0805

*Clearly, you figured out by now, having gone through calculus that three of you can do the problem and all get it right.*0808

*And three of you have different answers, that is not a problem because there is a lot of symbolism going on.*0815

*The symbolism is going to take different shapes, depending on the particular approach that was used to solve the problem.*0820

*That was number 5, let us take a look at number 6.*0828

*We have got f(x) = 4 × sec(x) - 3 × csc(x).*0837

*We are asked to find the derivative of this.*0848

*Very straightforward, as long as you know your derivative formulas, = 4.*0851

*The derivative of sec x is sec tan.*0858

*4 sec x tan x -, let us do it this way,*0861

*Let me put -3.*0881

*The derivative of csc x is –csc cot.*0883

*-csc x cot x, you end up with 4 sec x tan(x) + 3 csc x cot x.*0888

*In this particular case, our answer is b.*0905

*Just a straight application, just have to watch the sign, that is about it.*0908

*That was number 6, let us go ahead and see what number 7 has to offer.*0913

*We are asked to compute an integral here.*0922

*We have the integral from 0 to 1/4 of 16/ 1 + 16t² dt.*0926

*You remember some of your integral formulas, there was an integral formula*0941

*where the integral of 1/ 1 + x² dx was equal to the inv tan(x) + c.*0945

*This thing, then I'm going to rewrite, I’m going to pull the 16 out.*0961

*16 × the integral from 0 to 1/4 of 1/,*0967

*I’m going to write this as 1 + 4t².*0984

*I just change the 16t², I wrote it as (4t)² dt.*0994

*I’m just going to do a little bit of u substitution here.*0999

*I’m going to call u for t, therefore du = 4 dt.*1003

*Therefore, dt is equal to du/ 4.*1011

*What we get is 16.*1019

*This integral with the u substitution, we get 16, 0 to ¼, 1/ 1 + u² × dt is du/ 4.*1023

*I go ahead and pull the 4 out.*1041

*I can write this as 16/ 4 × the integral from 0 to 1/ 4 of + 1/ 1 + u² du,*1043

*which is equal to 4 × the inv tan of u which is equal to 4 × the inv tan or u is 4t.*1057

*From 0 to 1/4 which is equal to, I will go to the next page, not a problem.*1077

*Which is equal to 4 × the inv tan of 1 – inv tan of 0.*1088

*We get = 4 × the inv tan of 1 is π/4 -,*1110

*The inv tan of 0 is 0 = π.*1121

*Our answer is d, that is it.*1130

*A little bit of a recognition of what the integral formula is for 1/ 1 + x².*1132

*In this case, 1 + 1/ u², slight manipulation.*1138

*And then, use the u substitution to go ahead and take care of that.*1141

*Let us move on to number 8.*1147

*What is question number 8 asking us to do.*1156

*We have to determine the derivative of this particular expression.*1159

*ddx, 2x⁴ – 2x/ 2x⁴ + 2x.*1168

*What is the best way to approach this?*1184

*What is the best way to do this?*1197

*Okay, that is not a problem.*1204

*Before we actually start with the quotient rule, more than likely we are going to be using the quotient rule here.*1205

*Before we do that, let us see if there is something that we can actually do this function to make it a little easier on us.*1210

*Especially, when you take a look at the choices.*1219

*The choices have 2x³ + 2 in the denominator, except one of them which is 1 + x³.*1222

*Before we actually start taking the derivative, let us see if there is something we can do here.*1230

*This 2x⁴ - 2x/ 2x⁴ + 2x.*1234

*Let me factor out a 2x and I end up with x³ - 1, if I’m not mistaken.*1247

*I got a 2x here and I have got an x³ + 1 over here.*1254

*2x actually vanishes.*1259

*What I'm left with is x³ - 1/ x³ + 1.*1261

*Our f is the x³ - 1 and our g is our x³ + 1.*1269

*Now f’(x), the quotient rule is gf’ – fg’/ g².*1274

*We just have to work it out.*1285

*This is going to equal x³ + 1 × 3x² – fg’ - x³ - 1 × 3x²/ x³ + 1².*1288

*That is going to equal 3x⁵ + 3x² – 3x⁵ + 3x²/ x³ + 1².*1310

*3x⁵ and 3x⁵ goes away, that leaves us with a final answer of 6x²/ x³ + 1².*1327

*This happens to coercive with answer d.*1339

*That is about it, pretty straightforward.*1342

*If you just gone ahead and started doing that,*1345

*you are going to end up with a pretty complicated expression to have to simplify algebraically.*1347

*Would you have come up with the same answer, honestly,*1354

*I did not actually carry out that particular algebraic manipulation.*1356

*I took a look at the choices and I notice the x³.*1362

*I looked back at the function and thought can I factor it.*1366

*It turned out that I can.*1369

*Again, this is calculus we are dealing with.*1372

*There is a million ways to do something.*1377

*That is question number 8.*1380

*Let us go ahead and move on to question number 9 here.*1382

*How long is 9, that is not a problem.*1389

*Question number 9, in this particular case, we are given a function and*1392

*we are asked to find the equation of the normal line to the graph at a given point.*1399

*Our f(x) is equal to 3x × √x² + 6 – 3.*1408

*It is going to be at the point 0 – 3.*1423

*What we are going to do is we are going to find the slope of the tangent line which is just a derivative,*1428

*evaluated at a certain point.*1439

*The normal line, we know that the normal line is perpendicular to the tangent line.*1441

*All we are going to do is, the slope that we get for the tangent line, we are going to take the negative reciprocal of it.*1455

*That will give us the slope of the normal line, the slope of the normal line.*1460

*Because it is still passing through 0 and -3, we have the slope, we have a point.*1468

*We do y - y1 = m × x - x1.*1472

*We see the normal line is perpendicular the tangent line.*1477

*We take the negative reciprocal.*1484

*Let us go ahead and find f’(x) first.*1490

*F’(x), we got 3, we got a function of x × a function of x.*1494

*It is going to be a little long but not too big of a deal.*1498

*Again, I tend to take that out.*1501

*It is going to be this × the derivative of that + that × the derivative of this.*1503

*We are going to get x × 1/2 x² + 6⁻¹/2 × 2x + x² + 6 ^½ × 1.*1509

*That is going to equal 3 × x²/ √x² + 6 + √x² + 6.*1528

*We want to find f’ at 0, it is a function of x.*1545

*We are going to be using the 0 value.*1549

*That is going to equal 3 × 0/ √0 + 6 + √6.*1551

*We are going to end up with 3√6.*1563

*This is the slope of the tangent line, we want the negative reciprocal.*1571

*The normal line slope = -1/ 3√6.*1576

*Therefore, our line is equal to y - y1 - 3 = m which is -1/ 3√6 × x – 0.*1587

*Let us see if I can turn the page here.*1605

*I have a little difficulty getting to the next page.*1607

*We end up with y + 3 = -1/ 3√6 × x.*1611

*When you rearrange this, just multiply by 3√6.*1625

*Rearrange it to make it correspond with one of the choices.*1628

*Again, the answer that you got is not going to match one of the choices, but it is the same object.*1631

*You just have to rearrange it.*1637

*Rearranging to match one of the choices.*1639

*We get x – 3√6, y = 9√6 which is choice a.*1641

*It is going to happen quite a lot.*1656

*Your answer is going to be slightly different.*1658

*When you do the rearrangement, make sure you go very slowly.*1660

*In the choices that they give you, the differences are going to be very subtle.*1663

*There are going to be + and -, all the symbols are going to be there.*1667

*9, √6, 3 √6, just be very careful with your choices.*1673

*Make sure you look at all of your choices to make sure that you have an excluded one.*1678

*Just because you think you found the right choice, there may be something else going on.*1682

*Make sure you look at all of your choices.*1685

*Question number 10, let us see what we got for question number 10.*1690

*Here we want to find the concavity of a particular graph.*1696

*In this particular case, our function f(x) is equal to 2 sin(x) + 3 × cos² x.*1702

*We want to find the concavity at a given point.*1716

*Let us find and we know that concavity has to do with the second derivative.*1722

*Let us find f"(x) and evaluate f” at the point π, to see what the concavity of π is.*1728

*F’x, the derivative of this is going to be 2 × cos(x) + cos(x) – 6 sin x cos x.*1747

*That is one of our f’.*1780

*We want to do our f”.*1784

*F”(x), I will just take the derivative of what it is that we just got.*1787

*We are going to end up with -2 × sin(x) - 6 × sin x × -sin x + cos x cos x,*1791

*which is equal to -2 × sin(x) + 6 sin² x - 6 cos² x.*1812

*Now we go ahead and evaluate f” at π.*1826

*When I put π in for this, -2 × sin(π) + 6 × sin² of π - 6 × cos² of π.*1830

*sin(π) is 0, sin(π)² is 0, cos(π) is -1.*1851

*-1² is 1, we get -6.*1859

*The answer is d because -6 is less than 0.*1869

*We are concave down.*1876

*That is it, nice, straight application.*1878

*Second derivative is positive, you are concave up.*1880

*Second derivative is negative, you are concave down.*1883

*Let us go to question number 11.*1889

*Here it looks like we are computing a derivative.*1893

*Number 11, our particular derivative, we are evaluating at an indefinite integral.*1898

*My apologies.*1905

*The integral of -3x² × √x³ + 3 dx.*1906

*I think it is just going to be the straight u substitution.*1917

*I noticed x³ and I noticed an x².*1921

*I’m going to try u is equal to x³ + 3, du = 3x² dx.*1924

*Our integral, -3x² × the integral of x³ + 3 dx, is actually going to equal -the integral,*1939

*3x² x is du, this is going to be e ^½.*1955

*We are accustomed to seeing du on the right.*1965

*It is not a big deal, the order does not matter because you are just multiplying two things.*1968

*The multiplication is commutative, it does not matter the order.*1971

*You are used to seeing the du or the dx, or the dy, on the right hand side in the integrand.*1975

*Let us do it that way.*1983

*- the integral of u ^½ du which is equal to -u³/2/ 3/2 + c which is equal to -2/3 u³/2 + c.*1985

*Of course, we plug the u back in, x³ + 3.*2005

*Our final answer is -2/3 × x³ + 3³/2 + c.*2009

*Our final answer, the choice is going to be e.*2020

*I hope that was reasonably straightforward.*2024

*Let us take a look at the problem number 12.*2030

*Here we have a particular function.*2039

*We want to give the value of x where the function has a local extrema.*2042

*In this particular case, a local maximum.*2046

*Local maximum means you take the derivative, you set the derivative equal to 0.*2050

*You draw yourself a number line and you check points to the left of the critical values,*2058

*to the right of the critical values, to see whether the derivative is increasing or decreasing.*2064

*You decide which is local min and local max.*2068

*f(x) is equal to x³ + 6x² + 9x + 4.*2072

*f’(x), very simple.*2084

*We have 3x² + 12x + 9.*2087

*We want to set the derivative equal to 0.*2092

*We can factor out the 3.*2095

*We are going to get the same roots.*2096

*We have got x² + 4x + 3 is equal to 0.*2097

*We can actually factor this one.*2104

*We get x + 3, we get x + 1.*2107

*Therefore, we have x is equal to -3 and we have x is equal to -1.*2111

*Go ahead and draw myself a little number line here.*2119

*I have got -3, I will put it over here.*2122

*I have got -1, I will put it over here.*2125

*I’m going to check a point here, check a point here, and check a point here, in those intervals.*2127

*I’m going to put them into the derivative.*2132

*To see if the derivative is less than 0, decreasing, or greater than 0, increasing.*2134

*That is all I'm doing.*2140

*Let me go ahead and rewrite f’(x), I’m going to use this version right here.*2143

*I should actually use the original version.*2155

*I should have actually written this as 3 × that.*2157

*This is 3 × x + 3 × x + 1.*2162

*To the left of -3, when I check something in this interval, over here, let us try -4.*2171

*When I plug in -4, I'm going to get, for x, I'm going to get 3 × -4 + 3 is a negative number.*2177

*-4 + 1 is a negative number.*2185

*3 × a negative × a negative is definitely a positive number.*2189

*Here the slope is increasing.*2194

*Or if you want I can put a positive sign.*2197

*Some of you use positive, some of you use increasing/decreasing with an arrow.*2199

*It is that way.*2203

*Let us try a point in between here.*2205

*Let us try -2.*2207

*When I plug -2 in to the derivative, I get 3, -2 + 3 is a positive number.*2210

*-2 + 1 is a negative number.*2218

*3 × a positive × a negative gives me a number that is a negative.*2220

*It is less than 0, it is decreasing there or negative slope.*2226

*There you go, that pretty much takes care of it.*2229

*Because it is increasing, the graph is increasing.*2232

*Then, the graph is decreasing, that is a local max.*2235

*To the left it is increasing, to the right it is decreasing.*2242

*That means it is hitting a maximum point.*2245

*There is a local max at x = -3.*2251

*I think the particular choice was choice e.*2260

*Nothing too crazy so far.*2266

*Here we have number 13 and let us see what number 13 is asking us.*2269

*The slope of the tangent line of the graph that will give the value of c.*2277

*We are give a particular function.*2283

*We have 4x² + cx + 2, e ⁺y is equal to 2.*2284

*They are telling us that the slope of the tangent line of this graph, at x = 0 is 4.*2298

*They are asking us to find c.*2320

*The slope of the tangent line, for this graph, at x = 0 is equal to 4.*2326

*They want us to find the value of c.*2331

*Let us see what we can do.*2338

*Let us go ahead and take the derivative.*2342

*In this particular case, let us go ahead and take the derivative with respect to,*2344

*8x + c +, this is the function of y, we are going to do implicit differentiation here.*2348

*This is going to be 2e ⁺yy’ is equal to 0.*2356

*When I rearrange this, I'm going to get y’ is equal to -8x - c/ 2 × e ⁺y.*2363

*I will get 2 × e ⁺y.*2377

*2 × e ⁺y, if I move these two over to the right, 2 × e ⁺y from the original function is equal to 2 - 4x² – cx.*2386

*Therefore, if I plug in this into here, I get y’ is equal to -8x - c/ 2 – 4x² – c ⁺x.*2401

*They are telling me that y’ at 0 is equal to 4.*2421

*y' at 0 is equal to -8.*2426

*y’ at 0 is equal to 0 - c/ 2 - 0 – 0.*2431

*They are telling me that this is actually equal to 4.*2440

*I get - c/ 2 is equal to 4 which implies that c is equal to 8.*2445

*Our choice is e.*2457

*Just differentiate, in this particular case, you are going to get something which is 2e ⁺y.*2460

*You notice the 2e ⁺y is actually separate here.*2464

*You can move these over, plug in for x, and then solve for c.*2467

*I hope that made sense.*2472

*Let us try number 14, see how we are doing here.*2477

*Number 14, what is number 14 asking us.*2483

*It looks like it is asking us to evaluate this particular integral.*2486

*Evaluate the integral of 7 ⁺x - 4e⁷ ln x dx.*2492

*Let us see what we can do with this.*2509

*I’m going to separate this out.*2511

*The integral is linear, the linear of the integral of the sum is the sum of the integrals.*2512

*I’m going to do this as, this is the integral of 7 ⁺x dx - 4 × the integral of e⁷ ln x dx.*2518

*Let us go ahead and deal with the first integral.*2534

*This one right here.*2536

*The first integral, the integral of 7 ⁺x dx.*2538

*I’m going to do a u substitution here.*2548

*I’m going to let u equal to 7 ⁺x.*2550

*Du is going to be 7 ⁺x ln 7 dx, du/ ln 7 = 7 ⁺x dx.*2554

*Therefore, this integral actually = the integral of du/ ln 7 = 1/ ln 7 × the integral of du = 1/ ln 7 u + c.*2572

*I plug u back in which is 7 ⁺x.*2601

*I get 7 ⁺x/ ln 7 + c, that is our first integral.*2604

*Let me see here, where am I?*2614

*The second integral is -4, it is that one, × the integral of e ⁺ln x dx = -4 × the integral of e ⁺ln(x)⁷ dx*2619

*= 4 × the integral of e ⁺ln is just x⁷ dx = -4x⁸/ 8 + some constant c.*2648

*The c and c are not the same.*2664

*If you want you can just c1 and c2, it does not really matter.*2666

*This is this one, which is –x⁸/ 2 + c.*2672

*Now we just put them back together.*2691

*We have the integral that we wanted, it is equal to 7 ⁺x/ ln 7 – x⁸/ 2 + c.*2694

*It looks like that is option a.*2709

*That takes care of the first of 14 of the problems.*2714

*For the next lesson, we are just going to continue on with this particular section and go on with the practice exam.*2718

*Thank you so much for joining us here at www.educator.com.*2725

*We will see you next time, bye.*2728

0 answers

Post by R K on April 23, 2017

For number 7, since it is u substitution, shouldn't we change the bounds?

1 answer

Last reply by: Professor Hovasapian

Mon Jul 25, 2016 7:19 PM

Post by Peter Ke on July 23, 2016

For Problem 5, shouldn't the derivative of-4y^2 be -8y and not -8yy' because you bring down the 2 and multiply it by -4 which is -8 and subtract the exponent by 1?

2 answers

Last reply by: R K

Sun Apr 23, 2017 11:41 PM

Post by nathan lau on May 5, 2016

for question number 9 I keep getting letter b do you know what I could be doing wrong?