Answer:
3.6 x 10^3
Step-by-step explanation:
Answer:
The 2 pound bag of gummy bears
Step-by-step explanation:
You do 1.5/4.35 which = 2.9 the 2.9 is $2.90 per pound
Then, you do 5.3/2 which = 2.6 this 2.6 is $2.60 per pound
$2.60 < $2.90 there for the 2 pound bag of gummy bears is the better deal
Answer:
They're the exact same, except for the variables.
Step-by-step explanation:
Since the variable can be any letter, it is the exact same equation but with different letters as x. Although the equation of G(y) has a y instead of x, you know y is the x and g(y) is the y because the y inside of the parentheses tells us that y is the independent variable.
Answer:
About the x axis
![V = 4\pi[ \frac{x^5}{5}] \Big|_0^2 =4\pi *\frac{32}{5}= \frac{128 \pi}{5}](https://tex.z-dn.net/?f=%20V%20%3D%204%5Cpi%5B%20%5Cfrac%7Bx%5E5%7D%7B5%7D%5D%20%5CBig%7C_0%5E2%20%3D4%5Cpi%20%2A%5Cfrac%7B32%7D%7B5%7D%3D%20%5Cfrac%7B128%20%5Cpi%7D%7B5%7D)
About the y axis
![V = \pi [4y -y^2 +\frac{y^3}{12}] \Big|_0^8 =\pi *\frac{32}{3}= \frac{32 \pi}{3}](https://tex.z-dn.net/?f=%20V%20%3D%20%5Cpi%20%5B4y%20-y%5E2%20%2B%5Cfrac%7By%5E3%7D%7B12%7D%5D%20%5CBig%7C_0%5E8%20%3D%5Cpi%20%2A%5Cfrac%7B32%7D%7B3%7D%3D%20%5Cfrac%7B32%20%5Cpi%7D%7B3%7D)
About the line y=8
![V = \pi [64x -\frac{32}{3}x^3 +\frac{4}{5}x^5] \Big|_0^2 =\pi *(128-\frac{256}{3} +\frac{128}{5})= \frac{1024 \pi}{5}](https://tex.z-dn.net/?f=%20V%20%3D%20%5Cpi%20%5B64x%20-%5Cfrac%7B32%7D%7B3%7Dx%5E3%20%2B%5Cfrac%7B4%7D%7B5%7Dx%5E5%5D%20%5CBig%7C_0%5E2%20%3D%5Cpi%20%2A%28128-%5Cfrac%7B256%7D%7B3%7D%20%2B%5Cfrac%7B128%7D%7B5%7D%29%3D%20%5Cfrac%7B1024%20%5Cpi%7D%7B5%7D)
About the line x=2
![V = \frac{\pi}{2} [\frac{y^2}{2}] \Big|_0^8 =\frac{\pi}{4} *(64)= 16\pi](https://tex.z-dn.net/?f=%20V%20%3D%20%5Cfrac%7B%5Cpi%7D%7B2%7D%20%5B%5Cfrac%7By%5E2%7D%7B2%7D%5D%20%5CBig%7C_0%5E8%20%3D%5Cfrac%7B%5Cpi%7D%7B4%7D%20%2A%2864%29%3D%2016%5Cpi)
Step-by-step explanation:
For this case we have the following functions:

About the x axis
Our zone of interest is on the figure attached, we see that the limit son x are from 0 to 2 and on y from 0 to 8.
We can find the area like this:

And we can find the volume with this formula:


![V = 4\pi [\frac{x^5}{5}] \Big|_0^2 =4\pi *\frac{32}{5}= \frac{128 \pi}{5}](https://tex.z-dn.net/?f=%20V%20%3D%204%5Cpi%20%5B%5Cfrac%7Bx%5E5%7D%7B5%7D%5D%20%5CBig%7C_0%5E2%20%3D4%5Cpi%20%2A%5Cfrac%7B32%7D%7B5%7D%3D%20%5Cfrac%7B128%20%5Cpi%7D%7B5%7D)
About the y axis
For this case we need to find the function in terms of x like this:

but on this case we are just interested on the + part
as we can see on the second figure attached.
We can find the area like this:

And we can find the volume with this formula:


![V = \pi [4y -y^2 +\frac{y^3}{12}] \Big|_0^8 =\pi *\frac{32}{3}= \frac{32 \pi}{3}](https://tex.z-dn.net/?f=%20V%20%3D%20%5Cpi%20%5B4y%20-y%5E2%20%2B%5Cfrac%7By%5E3%7D%7B12%7D%5D%20%5CBig%7C_0%5E8%20%3D%5Cpi%20%2A%5Cfrac%7B32%7D%7B3%7D%3D%20%5Cfrac%7B32%20%5Cpi%7D%7B3%7D)
About the line y=8
The figure 3 attached show the radius. We can find the area like this:

And we can find the volume with this formula:


![V = \pi [64x -\frac{32}{3}x^3 +\frac{4}{5}x^5] \Big|_0^2 =\pi *(128-\frac{256}{3} +\frac{128}{5})= \frac{1024 \pi}{5}](https://tex.z-dn.net/?f=%20V%20%3D%20%5Cpi%20%5B64x%20-%5Cfrac%7B32%7D%7B3%7Dx%5E3%20%2B%5Cfrac%7B4%7D%7B5%7Dx%5E5%5D%20%5CBig%7C_0%5E2%20%3D%5Cpi%20%2A%28128-%5Cfrac%7B256%7D%7B3%7D%20%2B%5Cfrac%7B128%7D%7B5%7D%29%3D%20%5Cfrac%7B1024%20%5Cpi%7D%7B5%7D)
About the line x=2
The figure 4 attached show the radius. We can find the area like this:

And we can find the volume with this formula:


![V = \frac{\pi}{2} [\frac{y^2}{2}] \Big|_0^8 =\frac{\pi}{4} *(64)= 16\pi](https://tex.z-dn.net/?f=%20V%20%3D%20%5Cfrac%7B%5Cpi%7D%7B2%7D%20%5B%5Cfrac%7By%5E2%7D%7B2%7D%5D%20%5CBig%7C_0%5E8%20%3D%5Cfrac%7B%5Cpi%7D%7B4%7D%20%2A%2864%29%3D%2016%5Cpi)
Area of the cage : A = L * W
A = 3 * 5
A = 15 sq ft....so ur cage is 15 square ft...and they need at least 12 sq ft...so yes, it is big enough