Answer:
4in = r
Step-by-step explanation:
Different problem if the x was a variable in the question
formula for area of circle = pi x r^2.
So if the area of a circle is 16 pi in^2, it would be an obvious estimate that the radius is the square root of 16 which is 4 inches.
If the x was a variable, you would find the square root of 16x which is 
which would make it 4
as the radius.
Answer is 33 remaining chocolate cupcakes and 22 remaining vanilla cupcakes.
Answer:
Complementary Angles
Explanation:
In Euclidean geometry, an angle is the figure formed by two rays, called the sides of the angle, sharing a common endpoint, called the vertex of the angle.[1] Angles formed by two rays lie in the plane that contains the rays. Angles are also formed by the intersection of two planes. These are called dihedral angles. Two intersecting curves define also an angle, which is the angle of the tangents at the intersection point. For example, the spherical angle formed by two great circles on a sphere equals the dihedral angle between the planes containing the great circles.
An angle formed by two rays emanating from a vertex.
Check the picture below.
you can pretty much just count off the grid the units for JK and MI.
now, let's check how long are KI and JM
![\bf \textit{distance between 2 points}\\ \quad \\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) K&({{ -4}}\quad ,&{{ 4}})\quad % (c,d) I&({{ -2}}\quad ,&{{ 3}}) \end{array}\qquad % distance value d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2} \\\\\\ KI=\sqrt{[-2-(-4)]^2+[3-4]^2}\implies KI=\sqrt{(-2+4)^2+(3-4)^2} \\\\\\ KI=\sqrt{2^2+(-1)^2}\implies KI=\sqrt{4+1}\implies \boxed{KI=\sqrt{5}}\\\\ -------------------------------](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bdistance%20between%202%20points%7D%5C%5C%20%5Cquad%20%5C%5C%0A%5Cbegin%7Barray%7D%7Blllll%7D%0A%26x_1%26y_1%26x_2%26y_2%5C%5C%0A%25%20%20%28a%2Cb%29%0AK%26%28%7B%7B%20-4%7D%7D%5Cquad%20%2C%26%7B%7B%204%7D%7D%29%5Cquad%20%0A%25%20%20%28c%2Cd%29%0AI%26%28%7B%7B%20-2%7D%7D%5Cquad%20%2C%26%7B%7B%203%7D%7D%29%0A%5Cend%7Barray%7D%5Cqquad%20%0A%25%20%20distance%20value%0Ad%20%3D%20%5Csqrt%7B%28%7B%7B%20x_2%7D%7D-%7B%7B%20x_1%7D%7D%29%5E2%20%2B%20%28%7B%7B%20y_2%7D%7D-%7B%7B%20y_1%7D%7D%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0AKI%3D%5Csqrt%7B%5B-2-%28-4%29%5D%5E2%2B%5B3-4%5D%5E2%7D%5Cimplies%20KI%3D%5Csqrt%7B%28-2%2B4%29%5E2%2B%283-4%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0AKI%3D%5Csqrt%7B2%5E2%2B%28-1%29%5E2%7D%5Cimplies%20KI%3D%5Csqrt%7B4%2B1%7D%5Cimplies%20%5Cboxed%7BKI%3D%5Csqrt%7B5%7D%7D%5C%5C%5C%5C%0A-------------------------------)
![\bf \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) J&({{ -7}}\quad ,&{{ 4}})\quad % (c,d) M&({{ -8}}\quad ,&{{ 3}}) \end{array}\qquad % distance value \\\\\\ JM=\sqrt{[-8-(-7)]^2+[3-4]^2}\implies JM=\sqrt{(-8+7)^2+(3-4)^2} \\\\\\ JM=\sqrt{(-1)^2+(-1)^2}\implies \boxed{JM=\sqrt{2}}](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Barray%7D%7Blllll%7D%0A%26x_1%26y_1%26x_2%26y_2%5C%5C%0A%25%20%20%28a%2Cb%29%0AJ%26%28%7B%7B%20-7%7D%7D%5Cquad%20%2C%26%7B%7B%204%7D%7D%29%5Cquad%20%0A%25%20%20%28c%2Cd%29%0AM%26%28%7B%7B%20-8%7D%7D%5Cquad%20%2C%26%7B%7B%203%7D%7D%29%0A%5Cend%7Barray%7D%5Cqquad%20%0A%25%20%20distance%20value%0A%5C%5C%5C%5C%5C%5C%0AJM%3D%5Csqrt%7B%5B-8-%28-7%29%5D%5E2%2B%5B3-4%5D%5E2%7D%5Cimplies%20JM%3D%5Csqrt%7B%28-8%2B7%29%5E2%2B%283-4%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0AJM%3D%5Csqrt%7B%28-1%29%5E2%2B%28-1%29%5E2%7D%5Cimplies%20%5Cboxed%7BJM%3D%5Csqrt%7B2%7D%7D)
so, add all sides, and that's the perimeter of the trapezoid.
The vertical uprights on both sides of the number 8 mean to take the absolute value.
To get the absolute value of a number is really simple. If it negative, then make it positive by taking away the minus sign. If it positive, then do nothing.
Now we, -3 and 8 and of course the solution is '<' because -3 is less than 8.
Hope this helps!
