Answer:
4.65
Step-by-step explanation:
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)
1) You'll have to get raid of the fraction by multiplying 10ac by 11, x/11 by 11 which the 11's cancels out and -3 multiplied by 11 .
2)Now you have 10ac -x = -33. Add x on both sides of the equal sign you get 10ac = -33 + x
3) Divide both sides by 10c to get "a" by itself, so "a" = -33+x divided by 10c
I think the answer would be C to this problem
Answer:
7/12
Step-by-step explanation:
Simplification is something we can do if both the numerator and denominator are divisible by the same number.
In 14/24, both numbers are divisible by 2.
So, we divide both the numerator and denominator by 2. We get 7/12.
Therefore 14/24 = 7/12.
