the circle has a diameter of 12, thus its radius is half that, or 6.
![\bf \textit{area of a circle}\\\\ A=\pi r^2~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=6 \end{cases}\implies A=\pi 6^2\implies A=36\pi \\\\\\ A \approx 113.0973355\implies A=\stackrel{\textit{rounded up}}{113.1}](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Barea%20of%20a%20circle%7D%5C%5C%5C%5C%20A%3D%5Cpi%20r%5E2~~%20%5Cbegin%7Bcases%7D%20r%3Dradius%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20r%3D6%20%5Cend%7Bcases%7D%5Cimplies%20A%3D%5Cpi%206%5E2%5Cimplies%20A%3D36%5Cpi%20%5C%5C%5C%5C%5C%5C%20A%20%5Capprox%20113.0973355%5Cimplies%20A%3D%5Cstackrel%7B%5Ctextit%7Brounded%20up%7D%7D%7B113.1%7D)
Answer: A dilation
Step-by-step explanation:
A dilation is when you change the scale factor of an object to take up more or less space to it into the desired area.
Answer:
Graph the function and its parent function. Then describe the transformation. 7.9(x) = x + 4. 8. f(x) = x - 6. 4. 8 x. -4. 48 x ... Write a function g whose graph represents the indicated transformation of the graph of f. ... 96)= 14x+31+2 -2 = 14x+3| ... 33. f(x) = x; translation 3 units down followed by a vertical shrink by a factor of.
Step-by-step explanation:
Hope this helps!
We are given polynomial:
.
We need to explain why the binomial (x + 2) IS a factor of this polynomial expression and why the binomial (x + 1) IS NOT a factor of this polynomial expression.
Let us set first factor equal to 0 and solve for x.
x+2=0
x=-2.
Plugging x=-2 in given polynomial, we get



<em>Because x=-2 gives 0 on plugging in given polynomial, so it's factor of given polynomial expression.</em>
Now, let us check second factor x+1=0
x=-1.
Plugging x=-1 in given polynomial, we get

=-5+8+7-6.
= -4.
<em>Because x=-1 doesn't gives 0 on plugging in given polynomial, so it's not a factor of given polynomial expression.</em>
Stokes' theorem says the integral of the curl of
over a surface
with boundary
is equal to the integral of
along the boundary. In other words, the flux of the curl of the vector field is equal to the circulation of the field, such that

We have


Parameterize the ellipse
by

with
and
.
Take the normal vector to
to be

Then the flux of the curl is
