If QR is a median, then:
![\dfrac{NO+MP}{2}=QR](https://tex.z-dn.net/?f=%5Cdfrac%7BNO%2BMP%7D%7B2%7D%3DQR)
We have NO = 30, MP = 9x - 42 and QR = x + 15. Substitute:
![\dfrac{30+9x-42}{2}=x+15\qquad\text{multiply both sides by 2}\\\\9x-12=2x+30\qquad\text{add 12 to both sides}\\\\9x=2x+42\qquad\text{subtract 2x from both sides}\\\\7x=42\qquad\text{divide both sides by 7}\\\\\boxed{x=6}](https://tex.z-dn.net/?f=%5Cdfrac%7B30%2B9x-42%7D%7B2%7D%3Dx%2B15%5Cqquad%5Ctext%7Bmultiply%20both%20sides%20by%202%7D%5C%5C%5C%5C9x-12%3D2x%2B30%5Cqquad%5Ctext%7Badd%2012%20to%20both%20sides%7D%5C%5C%5C%5C9x%3D2x%2B42%5Cqquad%5Ctext%7Bsubtract%202x%20from%20both%20sides%7D%5C%5C%5C%5C7x%3D42%5Cqquad%5Ctext%7Bdivide%20both%20sides%20by%207%7D%5C%5C%5C%5C%5Cboxed%7Bx%3D6%7D)
<h3>Answer: A. 6</h3>
Answer:
x=5
Step-by-step explanation:
-5x+6(5x-18)=17
-5x+30x-108=17
25x-108=17
25x=17+108
25x=125
x=5
Answer: <span>A square inscribed in a circle.
</span>
Justification:
Note that by making two perpendicular lines that intersect each other in the center of the circle, he obtains 4 equidistant points on the circumference.
So, joining each pair of neighbouring points, the image will reveal 4 congruent sides joining at right angles (90°). This is the image of a square with the four vertices on the circumference.