Check the picture below.
now, we're making an assumption that, the two blue shaded region are equal in shape, and thus if that's so, that area above the 14 is 6 and below it is also 6, 14 + 6 + 6 = 26.
so hmm if we simply get the area of the trapezoid and subtract the area of the yellow triangle and the area of the cyan triangle, what's leftover is what we didn't subtract, namely the shaded region.
![\textit{area of a trapezoid}\\\\ A=\cfrac{h(a+b)}{2}~~ \begin{cases} h~~=height\\ a,b=\stackrel{parallel~sides}{bases~\hfill }\\[-0.5em] \hrulefill\\ h=15\\ a=14\\ b=26 \end{cases}\implies A=\cfrac{15(14+26)}{2}\implies A=300 \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{\Large Areas}}{\stackrel{trapezoid}{300}~~ - ~~\stackrel{yellow~triangle}{\cfrac{1}{2}(26)(9)}~~ - ~~\stackrel{cyan~triangle}{\cfrac{1}{2}(15)(6)}} \\\\\\ 300~~ - ~~117~~ - ~~45\implies 138\qquad \textit{blue shaded area}](https://tex.z-dn.net/?f=%5Ctextit%7Barea%20of%20a%20trapezoid%7D%5C%5C%5C%5C%20A%3D%5Ccfrac%7Bh%28a%2Bb%29%7D%7B2%7D~~%20%5Cbegin%7Bcases%7D%20h~~%3Dheight%5C%5C%20a%2Cb%3D%5Cstackrel%7Bparallel~sides%7D%7Bbases~%5Chfill%20%7D%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20h%3D15%5C%5C%20a%3D14%5C%5C%20b%3D26%20%5Cend%7Bcases%7D%5Cimplies%20A%3D%5Ccfrac%7B15%2814%2B26%29%7D%7B2%7D%5Cimplies%20A%3D300%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7B%5CLarge%20Areas%7D%7D%7B%5Cstackrel%7Btrapezoid%7D%7B300%7D~~%20-%20~~%5Cstackrel%7Byellow~triangle%7D%7B%5Ccfrac%7B1%7D%7B2%7D%2826%29%289%29%7D~~%20-%20~~%5Cstackrel%7Bcyan~triangle%7D%7B%5Ccfrac%7B1%7D%7B2%7D%2815%29%286%29%7D%7D%20%5C%5C%5C%5C%5C%5C%20300~~%20-%20~~117~~%20-%20~~45%5Cimplies%20138%5Cqquad%20%5Ctextit%7Bblue%20shaded%20area%7D)
5+5=10
y=5
b=5
the answer is 5 plus 5
Answer:
for number 8 it is the multiplictivie inverse because you just do 8/9 x 2 so it would be correct and for number 9 it is not the multiplicative inverse because it is not reversing it also can i pls get brainliest
Step-by-step explanation: