No, you need to divide by $\displaystyle 12^3$...essentially a 12 for each of the 3 spatial dimensions.

$\displaystyle 7128\text{ in}^3\cdot\left(\frac{1\text{ ft}}{12\text{ in}} \right)^3=\frac{7128}{12^3}\,\text{ft}^3=156.75 \text{ ft}^3$

Printable View

- Nov 19th 2012, 08:07 PMMarkFLRe: Calc Word Problem. Please Help!!!
No, you need to divide by $\displaystyle 12^3$...essentially a 12 for each of the 3 spatial dimensions.

$\displaystyle 7128\text{ in}^3\cdot\left(\frac{1\text{ ft}}{12\text{ in}} \right)^3=\frac{7128}{12^3}\,\text{ft}^3=156.75 \text{ ft}^3$ - Nov 19th 2012, 08:14 PMophischaussRe: Calc Word Problem. Please Help!!!
Oh, ok that makes since. So for part 5 would I use the formula 1/2*b*h so 1/2*(99)*(104) = 5148 in^2

- Nov 19th 2012, 08:22 PMMarkFLRe: Calc Word Problem. Please Help!!!
No, you have a trapezoid, not a triangle. You want to use:

$\displaystyle A=\frac{h}{2}(B+b)$

where:

$\displaystyle h=99$

$\displaystyle B=104$

$\displaystyle b=32$ - Nov 19th 2012, 08:27 PMophischaussRe: Calc Word Problem. Please Help!!!
O.k., sorry for not getting this right away, thanks for your patience lol.

So the answer would then be 6732 - Nov 19th 2012, 08:48 PMMarkFLRe: Calc Word Problem. Please Help!!!
An impatient person has no business trying to help on a forum. You are doing fine.

Yes, that is the correct answer. Now, in order to find this area using integration, I recommend orienting your coordinate axes such that the origin is at the bottom corner under the lowest point of the ceiling. We will then want to write the upper slanting edge of the wall as a linear function. What would the*y*-intercept and the slope be? - Nov 19th 2012, 08:54 PMophischaussRe: Calc Word Problem. Please Help!!!
Thank-you, I appreciate your help.

Would the slope be 8 and the y-intercept 32? - Nov 19th 2012, 08:58 PMMarkFLRe: Calc Word Problem. Please Help!!!
You have the correct intercept, but for the slope, think of the rise over run of an individual step.

- Nov 19th 2012, 09:00 PMophischaussRe: Calc Word Problem. Please Help!!!
ok so the slope would be 8/11?

- Nov 19th 2012, 09:18 PMMarkFLRe: Calc Word Problem. Please Help!!!
Yes, good work! So, what would the linear function representing the slanted edge be?

- Nov 19th 2012, 09:20 PMophischaussRe: Calc Word Problem. Please Help!!!
(8/11)x+32?

- Nov 19th 2012, 09:26 PMMarkFLRe: Calc Word Problem. Please Help!!!
Exactly! Now, over what interval do you want to integrate?

- Nov 19th 2012, 09:27 PMophischaussRe: Calc Word Problem. Please Help!!!
Would it be from 5 to 13?

- Nov 19th 2012, 09:35 PMophischaussRe: Calc Word Problem. Please Help!!!
and if that is the interval would it be:

(4/11)x^2+32x evaluated from 5 to 13= 308.364 - Nov 19th 2012, 09:36 PMMarkFLRe: Calc Word Problem. Please Help!!!
No, recall where the origin is and how many inches wide the back wall is.

- Nov 19th 2012, 09:41 PMophischaussRe: Calc Word Problem. Please Help!!!
well the back wall is 99 inches wide, and the origin is at the lowest point of the ceiling, so would it be from 11 to 88