Answer:
I think it would be B.
I took 528÷ 264=2 this tell me how many MB one picture equals. Then I took 2×35 which gave me 70. Have a Lovely Day.
In order for the inverse to exist, the matrix cannot be singular, so we need to first examine the conditions for existence of the inverse.
Compute the determinant. The easiest way might be a cofactor expansion along either the first row or third column; I'll do the first.
![\begin{vmatrix}x&0&0\\0&1&0\\w&0&x\end{vmatrix}=x\begin{vmatrix}1&0\\0&x\end{vmatrix}=x^2](https://tex.z-dn.net/?f=%5Cbegin%7Bvmatrix%7Dx%260%260%5C%5C0%261%260%5C%5Cw%260%26x%5Cend%7Bvmatrix%7D%3Dx%5Cbegin%7Bvmatrix%7D1%260%5C%5C0%26x%5Cend%7Bvmatrix%7D%3Dx%5E2)
The matrix is then singular whenever
![x\neq0](https://tex.z-dn.net/?f=x%5Cneq0)
.
With this in mind, compute the inverse.
Answer:
6 ft longer (total length of 15ft)
Step-by-step explanation:
the initial dimensions are:
length: ![l=9ft](https://tex.z-dn.net/?f=l%3D9ft)
width: ![w=8ft](https://tex.z-dn.net/?f=w%3D8ft)
the area area of the garden is given by:
![a=l*w](https://tex.z-dn.net/?f=a%3Dl%2Aw)
so the original area is:
![a=9ft*8ft=72ft^2](https://tex.z-dn.net/?f=a%3D9ft%2A8ft%3D72ft%5E2)
Since we need the area to be
, and we can only change the length, the width will still be 8ft.
We substitute the value of the new area and the width to the equation for the area:
![a=l*w\\120ft^2=l*8ft](https://tex.z-dn.net/?f=a%3Dl%2Aw%5C%5C120ft%5E2%3Dl%2A8ft)
and we clear for the new length:
![l=\frac{120ft^2}{8ft}\\ \\l=15](https://tex.z-dn.net/?f=l%3D%5Cfrac%7B120ft%5E2%7D%7B8ft%7D%5C%5C%20%5C%5Cl%3D15)
The length of the garden for the area to be
, must be 15ft.
This means that if originally the length was 9 ft, now it has to be 6 ft longer.
the answer is negative two over zero