What u can do is do 3/63=5/x. so u have to get x all by itself.
3/63=5/x
cross multiply
63×5 and 3×x
315=3x
divide by 3 to the number 315
ANSWER:105
(a) m∠RQS = (1/2)m∠QPR = 37°
(b) m∠QPR = m(arc QR) = 74° . . . . . an arc has the same measure as its central angle
Answer:
f ≤ 1
Step-by-step explanation:
I hope this helps, I got a few people to help me and really only remembered the answer so sorry I can't teach you much but I have the answer
1.
![(3^3+5x^2+3x-7)+(8x-6x^2+6)=\\\\=27\ \underline{+\,5x^2}\ \underline{ \underline{+\,3x}}-7\ \underline{\underline{\,+8x}}\ \underline{-\,6x^2}+6=\\\\=-x^2+11x+26](https://tex.z-dn.net/?f=%283%5E3%2B5x%5E2%2B3x-7%29%2B%288x-6x%5E2%2B6%29%3D%5C%5C%5C%5C%3D27%5C%20%5Cunderline%7B%2B%5C%2C5x%5E2%7D%5C%20%5Cunderline%7B%20%5Cunderline%7B%2B%5C%2C3x%7D%7D-7%5C%20%5Cunderline%7B%5Cunderline%7B%5C%2C%2B8x%7D%7D%5C%20%5Cunderline%7B-%5C%2C6x%5E2%7D%2B6%3D%5C%5C%5C%5C%3D-x%5E2%2B11x%2B26)
or if you mean (3x^3+5x^2+3x-7)+(8x-6x^2+6):
![(3x^3+5x^2+3x-7)+(8x-6x^2+6)=\\\\=3x^3\ \underline{+\,5x^2}\ \underline{ \underline{+\,3x}}-7\ \underline{\underline{\,+8x}}\ \underline{-\,6x^2}+6=\\\\=3x^3-x^2+11x-1](https://tex.z-dn.net/?f=%283x%5E3%2B5x%5E2%2B3x-7%29%2B%288x-6x%5E2%2B6%29%3D%5C%5C%5C%5C%3D3x%5E3%5C%20%5Cunderline%7B%2B%5C%2C5x%5E2%7D%5C%20%5Cunderline%7B%20%5Cunderline%7B%2B%5C%2C3x%7D%7D-7%5C%20%5Cunderline%7B%5Cunderline%7B%5C%2C%2B8x%7D%7D%5C%20%5Cunderline%7B-%5C%2C6x%5E2%7D%2B6%3D%5C%5C%5C%5C%3D3x%5E3-x%5E2%2B11x-1)
2.
![(8x-4x^2+3)-(x^3+7x^2+3x-8)=\\\\=\underline{8x}\ \underline{\underline{-\ 4x^2}}+3-x^3\ \underline{\underline{-\,7x^2}}\ \underline{-\,3x}+8=\\\\=-x^3-11x^2+5x+10](https://tex.z-dn.net/?f=%288x-4x%5E2%2B3%29-%28x%5E3%2B7x%5E2%2B3x-8%29%3D%5C%5C%5C%5C%3D%5Cunderline%7B8x%7D%5C%20%5Cunderline%7B%5Cunderline%7B-%5C%204x%5E2%7D%7D%2B3-x%5E3%5C%20%5Cunderline%7B%5Cunderline%7B-%5C%2C7x%5E2%7D%7D%5C%20%5Cunderline%7B-%5C%2C3x%7D%2B8%3D%5C%5C%5C%5C%3D-x%5E3-11x%5E2%2B5x%2B10)