Answer:
Part A: OC. 16π square inches
Part B: OC. 36π square inches
Step-by-step explanation:
<u>Part A:</u>
Given
Diameter of circular picture = d = 8 inches
We need to find the radius first to find the area
So,
Rdius = r = d/2 = 8/2 = 4 inches

Therefore, option C is the correct answer..
<u>Part B: </u>
As two inches frame is added around the picture, the diameter will become 8+4 =12 inches
The new radius will be:
r = 12/2 = 6
So,

Therefore, option C is correct ..
The matrix represents the vertices of the rectangle after it is scaled by a factor of 3 is,
[ 0 12 12 0]
[ 0 0 12 6]
<h2>
We have to determine</h2>
Which matrix represents the vertices of the rectangle after it is scaled by a factor of 3?
<h3>According to the question</h3>
The vertices of the triangle are
The scale factor is 3.
<h3>The vertices of the triangle are scaled with the factor of 3 so the vertices are get multiplied by 3.</h3>
Therefore,
The vertices of the rectangle after scaled by the factor of 3 are,

The vertices of the rectangle after it are scaled by a factor of 3.
Therefore,
The matrix represents the vertices of the rectangle after it is scaled by a factor of 3 is,
[ 0 12 12 0]
[ 0 0 12 6]
To know more about Matrix click the link given below.
brainly.com/question/2489305
Answer:hello the answer is C. 7$
Step-by-step explanation:
Answer:
• David
,
• 4 miles
Explanation:
In the graph:
The given locations are:
• Owen's House, A(11,3)
,
• David's House, B(15,13)
,
• School, C(3,18)
We determine both Owen's and David's distance from the school using the distance formula.

Owen's distance from school (AC)
![\begin{gathered} AC=\sqrt[]{(3-11)^2+(18-3)^2} \\ =\sqrt[]{(-8)^2+(15)^2} \\ =\sqrt[]{64+225} \\ =\sqrt[]{289} \\ AC=17\text{ miles} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20AC%3D%5Csqrt%5B%5D%7B%283-11%29%5E2%2B%2818-3%29%5E2%7D%20%5C%5C%20%3D%5Csqrt%5B%5D%7B%28-8%29%5E2%2B%2815%29%5E2%7D%20%5C%5C%20%3D%5Csqrt%5B%5D%7B64%2B225%7D%20%5C%5C%20%3D%5Csqrt%5B%5D%7B289%7D%20%5C%5C%20AC%3D17%5Ctext%7B%20miles%7D%20%5Cend%7Bgathered%7D)
David's distance from school (BC)
![\begin{gathered} BC=\sqrt[]{(3-15)^2+(18-13)^2} \\ =\sqrt[]{(-12)^2+(5)^2} \\ =\sqrt[]{144+25} \\ =\sqrt[]{169} \\ BC=13\text{ miles} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20BC%3D%5Csqrt%5B%5D%7B%283-15%29%5E2%2B%2818-13%29%5E2%7D%20%5C%5C%20%3D%5Csqrt%5B%5D%7B%28-12%29%5E2%2B%285%29%5E2%7D%20%5C%5C%20%3D%5Csqrt%5B%5D%7B144%2B25%7D%20%5C%5C%20%3D%5Csqrt%5B%5D%7B169%7D%20%5C%5C%20BC%3D13%5Ctext%7B%20miles%7D%20%5Cend%7Bgathered%7D)
We see from the calculations that David lives closer to the school, and by 4 miles.
The graph below is attached for further understanding: