Answer:
c
Step-by-step explanation:
According to the picture, angles 1 and 2 have the same measure because they are corresponding angles. Make equal both expressions for 1 and 2 and find x.
![\begin{gathered} 50-2x=110-42x \\ 42x-2x=110-50 \\ 40x=60 \\ x=\frac{60}{40} \\ x=\frac{3}{2} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%2050-2x%3D110-42x%20%5C%5C%2042x-2x%3D110-50%20%5C%5C%2040x%3D60%20%5C%5C%20x%3D%5Cfrac%7B60%7D%7B40%7D%20%5C%5C%20x%3D%5Cfrac%7B3%7D%7B2%7D%20%5Cend%7Bgathered%7D)
x is 3/2. Replace this value in both expression to find the measure of each angle.
![\begin{gathered} \measuredangle1=50-2x \\ \measuredangle1=50-2(\frac{3}{2}) \\ \measuredangle1=50-3 \\ \measuredangle1=47 \\ \measuredangle2=110-42x \\ \measuredangle2=110-42(\frac{3}{2}) \\ \measuredangle2=110-63 \\ \measuredangle2=47 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%5Cmeasuredangle1%3D50-2x%20%5C%5C%20%5Cmeasuredangle1%3D50-2%28%5Cfrac%7B3%7D%7B2%7D%29%20%5C%5C%20%5Cmeasuredangle1%3D50-3%20%5C%5C%20%5Cmeasuredangle1%3D47%20%5C%5C%20%5Cmeasuredangle2%3D110-42x%20%5C%5C%20%5Cmeasuredangle2%3D110-42%28%5Cfrac%7B3%7D%7B2%7D%29%20%5C%5C%20%5Cmeasuredangle2%3D110-63%20%5C%5C%20%5Cmeasuredangle2%3D47%20%5Cend%7Bgathered%7D)
The measure of these angles is 47°
The answer to this question is 11.
Let x be that number:
x +10 -2 +4 + 3 - 1 - 7 +1 = 0
x +8 = 0
and x = - 8
Given the coordinates of the image of line segment RT to be R'(-2,-4) and T'(4.4), if the image produced was dilated by a scale factor of 12 centered at the origin, to get the coordinate of the end point, we will simply multiply the x and y coordinates of by the factor of 12 as shown:
For R' with coordinate R'(-2,-4), the coordinates of endpoint of the pre-image will be:
R = 12R'
R = 12(-2, -4)
R = (-24, -48)
For T' with coordinate T'(4,4), the coordinates of endpoint of the pre-imagee will be:
T = 12T'
T = 12(4, 4)
T = (48, 48)
Hence the coordinate of the endpoint of the preimage will be at R(-24, -48) and T(48, 48)