The answer would be 241.2
we know the segment QP is an angle bisector, namely it divides ∡SQR into two equal angles, thus ∡1 = ∡2, and ∡SQR = ∡1 + ∡2.
![\bf \begin{cases} \measuredangle SQR = \measuredangle 1 + \measuredangle 2\\\\ \measuredangle 2 = \measuredangle 1 = 5x-7 \end{cases}\qquad \qquad \stackrel{\measuredangle SQR}{7x+13} = (\stackrel{\measuredangle 1}{5x-7})+(\stackrel{\measuredangle 2}{5x-7}) \\\\\\ 7x+13 = 10x-14\implies 13=3x-14\implies 27=3x \\\\\\ \cfrac{27}{3}=x\implies 9=x \\\\[-0.35em] ~\dotfill\\\\ \measuredangle SQR = 7(9)+13\implies \measuredangle SQR = 76](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Bcases%7D%20%5Cmeasuredangle%20SQR%20%3D%20%5Cmeasuredangle%201%20%2B%20%5Cmeasuredangle%202%5C%5C%5C%5C%20%5Cmeasuredangle%202%20%3D%20%5Cmeasuredangle%201%20%3D%205x-7%20%5Cend%7Bcases%7D%5Cqquad%20%5Cqquad%20%5Cstackrel%7B%5Cmeasuredangle%20SQR%7D%7B7x%2B13%7D%20%3D%20%28%5Cstackrel%7B%5Cmeasuredangle%201%7D%7B5x-7%7D%29%2B%28%5Cstackrel%7B%5Cmeasuredangle%202%7D%7B5x-7%7D%29%20%5C%5C%5C%5C%5C%5C%207x%2B13%20%3D%2010x-14%5Cimplies%2013%3D3x-14%5Cimplies%2027%3D3x%20%5C%5C%5C%5C%5C%5C%20%5Ccfrac%7B27%7D%7B3%7D%3Dx%5Cimplies%209%3Dx%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cmeasuredangle%20SQR%20%3D%207%289%29%2B13%5Cimplies%20%5Cmeasuredangle%20SQR%20%3D%2076)
Answer:
√2.37+ 3 √9 = √5. 3 is the last one
x+9 =+25
Step-by-step explanation:
(x+9)2= 25 This means x= 3.5 and the square root of x = √3.5 = 1.87 square root of 9 x 2 = √18 =4.24
We can also see 18+7 =25
√5.65 √5
this is 25 = 5 so it's the last one.
as 2.37+ 3 √9 = √5. 3
Answer:
150 degrees
Step-by-step explanation:
- Let the supplement = x
- So the angle you want is 5x
- Then [x + 5x] = 180 because they are supplementary.
- Solving. 6x = 180. Then x = 30.
- So the angle is 5x = 5×30 = 150 degrees