The formula for the surface area of a cube is: SA= 6((length of side)^2).
So if the surface area is 96, first divide by 6. And you get 16. Next you find the square-root of 16 and get 4. So the length of each side is 4 inches.
The formula for volume of a cube is V=(length of side)^3.
So 4 raised to the third power is: 4*4*4.
The volume of the cube is 64 inches cubed.
Answer:
Domain ![=(-\infty,\ \infty)](https://tex.z-dn.net/?f=%3D%28-%5Cinfty%2C%5C%20%5Cinfty%29)
Range ![=(-\infty,\ \infty)](https://tex.z-dn.net/?f=%3D%28-%5Cinfty%2C%5C%20%5Cinfty%29)
Step-by-step explanation:
Domain : Domain of a function
is the set of all possible values of
for which
exists.
Range : range of a function
is the set of all possible values of
.
Here ![f(x)=36-3x](https://tex.z-dn.net/?f=f%28x%29%3D36-3x)
can be any value from
to
.
![\forall\ x=a\ there\ exists\ f(x)\ such\ that\ f(a)=36-3a](https://tex.z-dn.net/?f=%5Cforall%5C%20x%3Da%5C%20there%5C%20exists%5C%20f%28x%29%5C%20such%5C%20that%5C%20f%28a%29%3D36-3a)
hence possible value of
can be any value between
and ![\infty](https://tex.z-dn.net/?f=%5Cinfty)
![domain =(-\infty,\ \infty)](https://tex.z-dn.net/?f=domain%20%3D%28-%5Cinfty%2C%5C%20%5Cinfty%29)
let ![y=-f(x)](https://tex.z-dn.net/?f=y%3D-f%28x%29)
![y=36-3x\\3x=36-y\\\\\\x=\frac{36-y}{3}\\](https://tex.z-dn.net/?f=y%3D36-3x%5C%5C3x%3D36-y%5C%5C%5C%5C%5C%5Cx%3D%5Cfrac%7B36-y%7D%7B3%7D%5C%5C)
so
.
hence
can have any value between
.
Range ![=(-\infty,\ \infty)](https://tex.z-dn.net/?f=%3D%28-%5Cinfty%2C%5C%20%5Cinfty%29)
Your answer would most likely be C. 2^1/3 i'm sorry if I got it wrong
Answer:
As shown in picture, the area is divided into 4 parts: 3 triangles and 1 rectangles.
Total area:
A = smallest triangle + medium triangle + largest triangle + rectangle
= 2 x 2 x 1/2 + 2 x 6 x 1/2 + 4 x (2 + 2 + 6) x 1/2 + 2 x 2
= 2 + 6 + 20 + 4
= 32
Hope this helps!
:)
Answer:
2xy+10z-5yz-4x
Step-by-step explanation:
Just put the positives and negatives together. You can't add or subtract any of them because they aren't like terms.