I'll abbreviate the definite integral with the notation,
![I(f(x),a,b)=\int_a^bf(x)\,\mathrm dx](https://tex.z-dn.net/?f=I%28f%28x%29%2Ca%2Cb%29%3D%5Cint_a%5Ebf%28x%29%5C%2C%5Cmathrm%20dx)
We're given
Recall that the definite integral is additive on the interval
, meaning for some
we have
![I(f,a,b)=I(f,a,c)+I(f,c,b)](https://tex.z-dn.net/?f=I%28f%2Ca%2Cb%29%3DI%28f%2Ca%2Cc%29%2BI%28f%2Cc%2Cb%29)
The definite integral is also linear in the sense that
![I(kf+\ell g,a,b)=kIf(a,b)+\ell I(g,a,b)](https://tex.z-dn.net/?f=I%28kf%2B%5Cell%20g%2Ca%2Cb%29%3DkIf%28a%2Cb%29%2B%5Cell%20I%28g%2Ca%2Cb%29)
for some constant scalars
.
Also, if
, then
![I(f,a,b)=-I(f,b,a)](https://tex.z-dn.net/?f=I%28f%2Ca%2Cb%29%3D-I%28f%2Cb%2Ca%29)
a. ![I(2f,1,5)=2I(f,1,5)=2(I(f,1,8)-I(f,5,8))=2(9-4)=\boxed{10}](https://tex.z-dn.net/?f=I%282f%2C1%2C5%29%3D2I%28f%2C1%2C5%29%3D2%28I%28f%2C1%2C8%29-I%28f%2C5%2C8%29%29%3D2%289-4%29%3D%5Cboxed%7B10%7D)
b. ![I(f-g,1,8)=I(f,1,8)-I(g,1,8)=9-5=\boxed{4}](https://tex.z-dn.net/?f=I%28f-g%2C1%2C8%29%3DI%28f%2C1%2C8%29-I%28g%2C1%2C8%29%3D9-5%3D%5Cboxed%7B4%7D)
c. ![I(f-g,1,5)=I(f,1,5)-I(g,1,5)=\dfrac{I(2f,1,5)}2-I(g,1,5)=10-3=\boxed{7}](https://tex.z-dn.net/?f=I%28f-g%2C1%2C5%29%3DI%28f%2C1%2C5%29-I%28g%2C1%2C5%29%3D%5Cdfrac%7BI%282f%2C1%2C5%29%7D2-I%28g%2C1%2C5%29%3D10-3%3D%5Cboxed%7B7%7D)
d. ![I(g-f,5,8)=I(g,5,8)-I(f,5,8)=(I(g,1,8)-I(g,1,5))-I(f,5,8)=(5-3)-4=\boxed{-2}](https://tex.z-dn.net/?f=I%28g-f%2C5%2C8%29%3DI%28g%2C5%2C8%29-I%28f%2C5%2C8%29%3D%28I%28g%2C1%2C8%29-I%28g%2C1%2C5%29%29-I%28f%2C5%2C8%29%3D%285-3%29-4%3D%5Cboxed%7B-2%7D)
e. ![I(7g,5,8)=7I(g,5,8)=7(5-3)=\boxed{14}](https://tex.z-dn.net/?f=I%287g%2C5%2C8%29%3D7I%28g%2C5%2C8%29%3D7%285-3%29%3D%5Cboxed%7B14%7D)
f. ![I(3f,5,1)=3I(f,5,1)=-3I(f,1,5)=-\dfrac32I(2f,1,5)=-\dfrac32(10)=\boxed{-15}](https://tex.z-dn.net/?f=I%283f%2C5%2C1%29%3D3I%28f%2C5%2C1%29%3D-3I%28f%2C1%2C5%29%3D-%5Cdfrac32I%282f%2C1%2C5%29%3D-%5Cdfrac32%2810%29%3D%5Cboxed%7B-15%7D)
Answer:
Here's what we know:
A = Lw (Area is length times width)
L = 2w + 6 (length is twice the width plus 6)
A = 140 (Area is 140 m2)
Plug in the variable values:
140 = w(2w + 6)
Distribute:
140 = 2w2 + 6w
Subtract 140:
2w2 + 6w - 140 = 0
Factor out a 2:
2(w2 + 3w - 70) = 0
Divide both sides by 2:
w2 + 3w - 70 = 0
(w + m)(w - n)
When we factor out the quadratic, we know it's going to be a +/- situation because the c value in the quadratic is negative, and the two numbers are going to be three away, the plus next to the 3 meaning that the larger number is going to be positive:
(w + 10)(w - 7) = 0
w = -10, 7
We can't have a negative length, so we can toss out the -10, leaving us with w = 7 meters.
L = 2 * 7 + 6
L = 14 + 6
L = 20
Check:
140 = 20 * 7
140 = 140
Answer: 4 Ounces.
Step-by-step Explanation: I don't know the full context of the problem, but if Seth used 4 ounces, that means that he used 4 ounces all together.
The answer should be A. If you need work I can show you.