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)
<h3>
Answer: 8</h3>
Reason:
Replace p with 2 and evaluate.

Think of a cube that is 2 units along each side. It's volume is 2*2*2 = 8 cubic units. Repeated multiplication can be shortened to using exponents.
Another example: 
Answer:
1. 4x - 29=11
2. (3x - 4) (2) = 46
3. 3/5x + 8 = 50
4. The difference of 51 and the sum of x and 5.
5. The product of 3 and y minus eight.
Step-by-step explanation:
If the endpoints of a diameter are (6,3) and (2,1) the midpoint is the center of the circle so:
(x,y)=((6+2)/2, (3+1)/2)=(4,2)
Now we need to find the radius....the diameter is:
d^2=(6-2)^2+(3-1)^2
d^2=16+4
d^2=20 since d=2r, r=d/2, and r^2=d^2/4 so
r^2=5
The standard form of the circle is (x-h)^2+(y-k)^2=r^2 and we know:
(h,k)=(4,2) from earlier so:
(x-4)^2+(y-2)^2=5
Answer:
19:
4 < third side < 14
20:
3 < third side < 13
Step-by-step explanation:
hint: go to http://www.17 28.org/trianinq.htm
(fix the space between 17 and 28)
you're welcome xx