The correct question is
The raised vegetable garden in Susan's yard is in the shape of a rectangular prism with a volume of 48 cubic feet and a height of 3/4 foot. <span>The base of the rectangular prism is not a square and the width is greater than 2 feet. What is the length and width of the rectangular prism?
let
x-------> the length of the base of the rectangular prism
</span>y-------> the width of the base of the rectangular prism
<span>
we know that
volume of the prism=area of the base*height
volume=48 ft</span>³
<span>height of the prism=3/4 ft
area of the base=volume/height--------> 48/(3/4)---> 48*4/3----> 64 ft</span>²
<span>
area of the base=64 ft</span>²
<span>area of the base=x*y
64=x*y--------> equation 1
y=x+2--------> equation 2
substitute equation 2 in equation 1
64=x*[x+2]-----> 64=x</span>²+2x------> x²+2x-64=0
<span>
using a graph tool----> to resolve the second order equation
see the attached figure
the solution is
x=7.062 ft
y=x+2-----> y=7.062+2-----> y=9.062 ft
the answer is</span>
the length of the base of the rectangular prism is 7.062 ftthe width of the base of the rectangular prism is 9.062 ft<span>
</span>
Answer:
1) 
2) ![\sqrt[3]{-1331}=-11](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B-1331%7D%3D-11)
3) Evaluating
we get 
4) 
5) 
Step-by-step explanation:
1) 
Prime factors of 1225 : 5x5x7x7
Prime factors of 1024: 2x2x2x2x2x2x2x2x2x2


2) ![\sqrt[3]{-1331}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B-1331%7D)
We know that ![\sqrt[n]{-x}=-\sqrt[n]{x} \ ( \ if \ n \ is \ odd)](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7B-x%7D%3D-%5Csqrt%5Bn%5D%7Bx%7D%20%5C%20%28%20%5C%20if%20%5C%20n%20%5C%20is%20%5C%20odd%29)
Applying radical rule:
![\sqrt[3]{-1331}\\=-\sqrt[3]{1331} \\=-\sqrt[3]{11\times\11\times11}\\=-\sqrt[3]{11^3} \\Using \ \sqrt[n]{x^n}=x \\=-11](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B-1331%7D%5C%5C%3D-%5Csqrt%5B3%5D%7B1331%7D%20%5C%5C%3D-%5Csqrt%5B3%5D%7B11%5Ctimes%5C11%5Ctimes11%7D%5C%5C%3D-%5Csqrt%5B3%5D%7B11%5E3%7D%20%5C%5CUsing%20%5C%20%5Csqrt%5Bn%5D%7Bx%5En%7D%3Dx%20%5C%5C%3D-11)
So, ![\sqrt[3]{-1331}=-11](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B-1331%7D%3D-11)
3) 
It can be written as:

Evaluating
we get 
4) 
Put value of x, y and z in equation and solve:

So, 
5) 
We know (-a)^n = (a)^n when n is even and (-a)^n = (-a)^n when n is odd

So, 
Answer:
m = 3/4; b = -3
Step-by-step explanation:
1) First, put the given equation into slope-intercept form. This is so we can identify its slope and y-intercept easily. Isolate y:
2) Lines that are in slope-intercept form are represented by the formula
. The number in place of
, or the coefficient of the x-term, represents the slope. The number in place of
represents the y-intercept. So, the slope of
is
, and its y-intercept is -3.
B and D and C in that order
Area=width*height
area=(11.7)*(15.4)
area=180.18cm^2