I believe that’s it .... i’m so so sorry if it’s not
A) 2x +3
Plug in the x
2(4.1) + 3
8.2+ 3= 11.2
A is 11.2
B) 16-5y
Plug in the y
16 - 5(2.3)
16- 11.5= 4.5
B is 4.5
C) x+y
Plug in x and y
4.1 + 2.3= 6.4
C is 6.4
Answer:
The scale factor is 2.
Step-by-step explanation:
Take the vertical sides of the 2 triangles.
Their length is 3 and 6 so the scale factor is 6/3 = 2.
Also we see than the horizontal lines are in the same ratio 4 : 2 = 2:1.
Given that the octagon rotates 360 degrees, it is a given that for every regular division of the polygon (360/8 = 45 degrees), the image would coincide with the pre-image during the rotation. Therefore, the octagon would coincide a total of 8 times as it rotates 360 degrees about its center.
Answer:
The volume of the solid is 
Step-by-step explanation:
In this case, the washer method seems to be easier and thus, it is the one I will use.
Since the rotation is around the y-axis we need to change de dependency of our variables to have 
. Thus, our functions with 
as independent variable are:
For the washer method, we need to find the area function, which is given by:
![A=\pi\cdot [(\rm{outer\ radius)^2 -(\rm{inner\ radius)^2 ]](https://tex.z-dn.net/?f=A%3D%5Cpi%5Ccdot%20%5B%28%5Crm%7Bouter%5C%20radius%29%5E2%20-%28%5Crm%7Binner%5C%20radius%29%5E2%20%5D)
By taking a look at the plot I attached, one can easily see that for a rotation around the y-axis the outer radius is given by the function 
and the inner one by 
. Thus, the area function is:
![A(y)=\pi\cdot [(\sqrt{y} )^2-(y^2)^2]\\A(y)=\pi\cdot (y-y^4)](https://tex.z-dn.net/?f=A%28y%29%3D%5Cpi%5Ccdot%20%5B%28%5Csqrt%7By%7D%20%29%5E2-%28y%5E2%29%5E2%5D%5C%5CA%28y%29%3D%5Cpi%5Ccdot%20%28y-y%5E4%29)
Now we just need to integrate. The integration limits are easy to find by just solving the equation 
, which has two solutions 
and 
. These are then, our integration limits.
