<p>@tmiddlet
5d-that is 100% correct. i remember c=ln1100 so u can just bring it down</p>
<p>5a and 5b are also right</p>
<p>not sure bout 5c dont remember</p>
<p>I think i got what u got for A but i didnt actual work it out, u know what i mean? I didnt answer b. I got 1/625 for c. I Got W= 300 + 44e^(-1/25t) i think</p>
<p>@tmiddlet </p>
<p>yes i got all the same answers 
!!!</p>
<p>for question 1. it was sin(pix) and 8x^3 right?</p>
<p>oh yeah, bioboy, tell us your secretsss! :D</p>
<p><a href=“Supporting Students from Day One to Exam Day – AP Central | College Board”>Supporting Students from Day One to Exam Day – AP Central | College Board;
<p>Yayy, the FRQs are up. I wanna see some answers, people :P</p>
<p>a) just a tangent line approximation, f(0) + f’(0)(1/4)</p>
<p>b)implicit differentiation, you’ll find that it’s positive or concave up so the tangent line will be an underestimate</p>
<p>c)separation of variables, you’ll get 1/(W-300)dw=1/25dt and when you integrate you’ll get an ln(W-300)=1/25t + C. take an e to each side to cancel out ln and solve for C with the given initial condition. you’ll get W=1100e^(t/25)+300</p>
<p>that’s how I did it. kinda ugly but very doable.</p>
<p>5 that was posted isn’t entirely correct.
A. 1411
B. Underestimate
C. If w’=(w-300)/25, then w’‘=w’/25 => (w-300)/(25^2)
D. It’s w=1100e^(t/25) +300. Not -t.</p>
<ol>
<li> </li>
</ol>
<p>a. decreasing – a(t) is negative at t = 5.5
b. -.947 – 2.6829 / 6 – 0 = .289
c. Integral (2sin(e^.25t) + 1) from 0 to 6 = 11.696
d. V(t) = (2sin(e^.25t) + 1) = 0 at 5.195
Integral((2sin(e^.25t) + 1) from 0 to 5.195 + 2 = 14.134</p>
<p>let me know of any disagreements</p>
<p>W=1100e^(t/25)+300</p>
<p>@tm1211
You’re right, and that’s actually what I got. I don’t know why I posted the - sign.</p>
<p>2)
a)-8/3
b)52.95 degrees C average temperature on the interval
c) -23 degrees C temperature change from start to end
d) 28.81 degrees cooler</p>
<ol>
<li> .
a. (H(5) – H(2) / 3) = (52 – 60 / 3) = -8/3 degrees per minute
b. This is the average temperature of the tea throughout time = 0 and time = 10
Area 1 = (2/2)(66+60)
Area 2 = (3/2)(60+52)
Area 3 = (4/2)(52+44)
Area 4 = (1/2)(44+43)
Add them all up and divide by 10 yields 52.95 degrees celcius avg temp
c. 43-66 = -23 degrees. This is the temperature change from t = 0 to t = 10
d. Integral of B’ = 80(e^-0.173t) + C
When t = 0, B = 100, so C = 20
B(10) = 80(e^-1.73) + 20 = 34.18 degrees celcius so it is 8.81 degrees cooler</li>
</ol>
<p>for #1 a) i got decreasing at 5.5 because v(5.5)= -0.453
b) i got average velocity = 1.949
c) i got total distance traveled= 14.573
d) i didnt get to do it (time restraints)
did anyone else get these answers too? i hope so. lol</p>
<p>mattgtodude had 2d) right i forgot about t=0 and solving for c
how many points will i get for getting 2 a,b,c right and showing the integral of d
7 or 8 points?</p>
<ol>
<li> .
a. F’(x) = 24x^2
F’(.5) = 6
y-1 = 6(x-.5)
y = 6x+2
b. Integral from 0 to .5 (sin(pi<em>x) – 8(x^3)) = 0.193
c. V = ((sin(pi</em>x)-1)^2) - ((8x^3)-1)^2)) dX</li>
</ol>
<p>@mattgtodude
I got 1.9494 for 1.b
Also got 12.573 for 1.c because you have to take the absolute value.
I agree with a and d</p>
<p>for #3 a) i got y-1=6(x-1/2)
b) i got A= 1/pi * 1/8
c) i got V= pi integral from 0 to 1/2 of ((1-sin(pix))^2-(1-8x^3)^2)dx
anyone get these answers?</p>
<p>I got 1/pi - 1/8 for 3.b</p>
<p>@BadGirlsClub
I believe you have the terms switched for 3.c. You will get a negative answer.</p>