If there is 2 digits and 7 numbers the combos would be:
1,1 2,1 3,1 4,1 5,1 6,1 7,1
1,2 2,2 3,2 4,2 5,2 6,2 7,2
1,3 2,3 3,3 4,3 5,3 6,3 7,3
1,4 2,4 3,4 4,4 5,4 6,4 7,4
1,5 2,5 3,5 4,5 5,5 6,5 7,5
1,6 2,6 3,6 4,6 5,6 6,6 7,6
1,7 2,7 3,7 4,7 5,7 6,7 7,7
Therefore there are 49 possible combinations :)
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)
Answer:
y = 5x -12
Step-by-step explanation:
point-slope form:
y - y1 = m(x-x1)
m= slope
m= (y2-y1)/ (x2-x1)
we have (4, 8) and (2,-2)
x1 = 4 y1= 8
x2= 2 y2= -2
m=( -2-8) / (2- 4)
m= -10/ -2
m= 5
so we have:
y - 8 = 5(x-4)
y - 8= 5x -20
y= 5x -20 +8
y = 5x -12
Answer:
1. -18x¹¹
2. 3n⁷
Step-by-step explanation:
For these problems, there are two things you need to worry about: negative signs and exponents.
1. Let's look at the signs first. There is only one value with a negative sign, meaning that the negative sign will stay.
When multiplying with exponents, you have to add up the exponents. Don't forget the numerical coefficients.
-3x² · 3x · 2x³ · x⁵ = -18x¹¹
2. There are two negative signs in this probem, meaning that they will cancel out. Multiply the rest like we did in the first problem.
3n² · -n² · -n³ = 3n⁷
5x + 3 ≥ 10
3 is adding on the left, then it will subtract on the right
5x ≥ 10 -3
5x ≥ 7
5 is multiplying on the left, then it will divide on the right
x ≥ 7/5
x ≥ 1.4
Then, the smallest integer value of x is 2