For # four Draw something like this <
[email protected] (the @ being a filled circle) starting at positive 5.
I'm assuming the function is f(x) = 100(0.7)^x. This is the same as y = 100(0.7)^x because y = f(x).
Plug in x = 0 to get
y = 100(0.7)^x
y = 100(0.7)^0
y = 100(1)
y = 100
So (x,y) = (0,100) is one point on this function curve
Plug in x = 2 to get
y = 100(0.7)^x
y = 100(0.7)^2
y = 100(0.49)
y = 49
So (x,y) = (2,49) is another point on this curve
In summary, the two points on this function curve are (0,100) and (2,49)
Answer:

Step-by-step explanation:
First of all, let's draw the two curves. First one (red) is a classic parabola, second (blue) is a similar curve with a tigher form. Imagine is made with paint so it's not high quality. The intersection between the two curves are easily found by solving a system of equations and are (0;0) and (1;1). At this point, the volume comes from the rotation of the shaded area, and can be obtained as difference from the rotation of the blue curve minus the red curve.
Remembering that the volume of the solid of revolution bound between
is given by
, in our case the volume is given by the expression
![\pi [\int_0^1 (\sqrt[4]{x})^2-(x^2)^2 dx] = \pi[\int_0^1 \sqrt{x}-x^4 dx] = \pi[\frac23x^\frac32 -\frac15 x^5 ] _0^1 = \pi (\frac23 -\frac15) = \frac7{15}\pi](https://tex.z-dn.net/?f=%5Cpi%20%5B%5Cint_0%5E1%20%28%5Csqrt%5B4%5D%7Bx%7D%29%5E2-%28x%5E2%29%5E2%20dx%5D%20%3D%20%5Cpi%5B%5Cint_0%5E1%20%5Csqrt%7Bx%7D-x%5E4%20dx%5D%20%20%3D%20%5Cpi%5B%5Cfrac23x%5E%5Cfrac32%20-%5Cfrac15%20x%5E5%20%5D%20_0%5E1%20%3D%20%5Cpi%20%28%5Cfrac23%20-%5Cfrac15%29%20%3D%20%5Cfrac7%7B15%7D%5Cpi)
Nine wholes with five twelves
9 5/12
The greatest number is 5. 3 cans of regular and 4 cans of diet. The most 15 can be divided by similarly to 20 is by 5.