Answer:
48 + 8x = 9
Step-by-step explanation:
4 +
x = 
multiply through by 12 ( the LCM of 3 and 4 ) to clear the fractions
48 + 8x = 9
Answer: (C) 13,120
<u>Step-by-step explanation:</u>
Given the sequence {4, 12, 36, 108, ... , 8748} we know that the first term (a) is 4 and the ratio (r) is 
Input the values above into the Sum formula:

Answer:
A. 2, 2, 5
Step-by-step explanation:
<u>DE:</u>
BA = 10
10 * 1/5 = 2
<u>EF:</u>
BC = 10
10 * 1/5 = 2
<u>DF:</u>
AC = 25
25 * 1/5 = 5
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)