Answer:
12?
Step-by-step explanation:
wouldnt the width just be either 12 or 16? because theres nothing else to multiply by
Answer:
6.63324958071
Step-by-step explanation:
Check the picture below.
so the picture has a rectangle that is 8 units high and 12 units wide, and it has a couple of "empty" trapezoids, with a height of 5 and "bases" of 9 and 3.
now, if we just take the whole area of the rectangle and then subtract the area of those two trapezoids, what's leftover is the blue area.
![\textit{area of a trapezoid}\\\\ A=\cfrac{h(a+b)}{2}~~ \begin{cases} h=height\\ a,b=\stackrel{parallel~sides}{bases}\\[-0.5em] \hrulefill\\ h=5\\ a=9\\ b=3 \end{cases}\implies \begin{array}{llll} A=\cfrac{5(9+3)}{2}\implies A=30 \end{array} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{\large Areas}}{\stackrel{rectangle}{(12\cdot 8)}~~ -~~\stackrel{\textit{two trapezoids}}{2(30)}}\implies 96-60\implies 36](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%7D%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20h%3D5%5C%5C%20a%3D9%5C%5C%20b%3D3%20%5Cend%7Bcases%7D%5Cimplies%20%5Cbegin%7Barray%7D%7Bllll%7D%20A%3D%5Ccfrac%7B5%289%2B3%29%7D%7B2%7D%5Cimplies%20A%3D30%20%5Cend%7Barray%7D%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%7Brectangle%7D%7B%2812%5Ccdot%208%29%7D~~%20-~~%5Cstackrel%7B%5Ctextit%7Btwo%20trapezoids%7D%7D%7B2%2830%29%7D%7D%5Cimplies%2096-60%5Cimplies%2036)
Answer:
I am not sure if this was the correct method to solve the question, however it gives the right results if you check it.
Step-by-step explanation:
3x+250+100=1250
3x= 900
x=300
yoga: 300, cross fit: 550 (300+250), gym: 400(300+100)
Answer:
m is the slope and b is the y-intercept. Therefore, for the equation y=x+5 , the slope is 1 , and the y-intercept is 5
Step-by-step explanation:
