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webass
- Posts: **666** Member

2010 AP Calculus AB Form A solutions.

Here are quick answers, let's check compare and check them.

**1.**

a) ∫(0→6) f(t) = 142.xxx cubic ft.

b) f(8) - g(8) = 48.417 - 108 = -59.58 cubic ft/hr

c) forgot already, but i did it

d) (∫(0→9) f(t)) - (125 + 108*2) = 26.335 cubic ft.

man it's so uncomfortable to write here, will add others later, or u guys can add too...

Here are quick answers, let's check compare and check them.

a) ∫(0→6) f(t) = 142.xxx cubic ft.

b) f(8) - g(8) = 48.417 - 108 = -59.58 cubic ft/hr

c) forgot already, but i did it

d) (∫(0→9) f(t)) - (125 + 108*2) = 26.335 cubic ft.

man it's so uncomfortable to write here, will add others later, or u guys can add too...

Post edited by webass on

This discussion has been closed.

## Replies to: 2010 AP Calculus AB Form A solutions (Official)

68Junior Member242Junior Member666Memberber1023, my bad. but the answer is still the same.Action52, on calculator part, where the intervals are given, you just can write the answer.2,429Senior Member****. I better get a 5 on this, or I'll be really mad. I was getting easy (and I mean easy) 5s on my practice tests, and in less time than allotted too.

666Member2.a) (E(7) - E(5)) / 7-5 = (21-13)/2 = 4 hund. of entries / hr.

b) ∫(0→8) E(t) = (2*4) + (3*13) + (2*21) + 23 = 112 hund. of entries

1/8(∫(0→8) E(t)) = 112/8 = 14 hund. of entries

1/8(∫(0→8) E(t)) represents the average number of entries, in hundreds, visitors deposited between noon - 8 PM.

c) 112 - (∫(8→12) P(t)) = 96 hund. of entries

d) at t=9.183; the max occurs

666Member2,429Senior Member68Junior Memberand b) should be trapezoidal rule

1,327Senior Member666MemberFallenAngel9, I'm not 100% sure about (C), but 99% sure about others.666Member666Member68Junior Member68Junior Memberam i retarded?

666Memberi really hope E(t) should be the rate.