Answer:
Step-by-step explanation:
The picture of the question in the attached figure
step 1
we know that
----> by complete circle
substitute the given values
solve for x
step 2
Find the measure of arc JM
substitute the value of x
Which statement best describes g (x) = RootIndex 3 StartRoot x + 6 EndRoot minus 6 and the parent function f (x) = RootIndex 3 S
2 answers:
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Two figures are similar if one is the scaled version of the other.
This is always the case for circles, because their geometry is fixed, and you can't modify it in anyway, otherwise it wouldn't be a circle anymore.
To be more precise, you only need two steps to prove that every two circles are similar:
- Translate one of the two circles so that they have the same center
- Scale the inner circle (for example) unit it has the same radius of the outer one. You can obviously shrink the outer one as well
Now the two circles have the same center and the same radius, and thus they are the same. We just proved that any two circles can be reduced to be the same circle using only translations and scaling, which generate similar shapes.
Recapping, we have:
- Start with circle X and radius r
- Translate it so that it has the same center as circle Y. This new circle, say X', is similar to the first one, because you only translated it.
- Scale the radius of circle X' until it becomes
. This new circle, say X'', is similar to X' because you only scaled it
So, we passed from X to X' to X'', and they are all similar to each other, and in the end we have X''=Y, which ends the proof.
First find the tangent line
dy/dx=2x
at x=6, the slope is 2(6)=12
so
use point slope form
y-y1=m(x-x1)
point is (6,36)
so
y-36=12(x-6)
y-36=12x-72
y=12x-36
alright, so we know they intersect at x=6
and y=12x-36 is below y=x^2
so we do
![[\frac{x^3}{3}-6x^2+36x]\limits^6_0=(\frac{6^3}{3}-6(6)^2+36(6))-(0)=\frac{216}{3}-216+216=](https://tex.z-dn.net/?f=%5B%5Cfrac%7Bx%5E3%7D%7B3%7D-6x%5E2%2B36x%5D%5Climits%5E6_0%3D%28%5Cfrac%7B6%5E3%7D%7B3%7D-6%286%29%5E2%2B36%286%29%29-%280%29%3D%5Cfrac%7B216%7D%7B3%7D-216%2B216%3D)
the area under the curve bounded by the lines and the x axis is 72 square units
dy/dx=2x
at x=6, the slope is 2(6)=12
so
use point slope form
y-y1=m(x-x1)
point is (6,36)
so
y-36=12(x-6)
y-36=12x-72
y=12x-36
alright, so we know they intersect at x=6
and y=12x-36 is below y=x^2
so we do
the area under the curve bounded by the lines and the x axis is 72 square units
See the attached picture for the solution:
