Check the picture below.
so we really have 2 equal triangles with 3 rectangles, so let's get the area of each and sum them up.
![\stackrel{\textit{two equal triangles}}{2\left[ \cfrac{1}{2}(8)(6) \right]}~~+~~\stackrel{back}{(6\cdot 4)}~~ +~~\stackrel{bottom}{(8\cdot 4)}~~+~~\stackrel{top}{(10\cdot 4)}\implies 144](https://tex.z-dn.net/?f=%5Cstackrel%7B%5Ctextit%7Btwo%20equal%20triangles%7D%7D%7B2%5Cleft%5B%20%5Ccfrac%7B1%7D%7B2%7D%288%29%286%29%20%5Cright%5D%7D~~%2B~~%5Cstackrel%7Bback%7D%7B%286%5Ccdot%204%29%7D~~%20%2B~~%5Cstackrel%7Bbottom%7D%7B%288%5Ccdot%204%29%7D~~%2B~~%5Cstackrel%7Btop%7D%7B%2810%5Ccdot%204%29%7D%5Cimplies%20144)
I think it will be 91 cm because the area of the square is 49 and the area of the triangles is 14
49+14+14+14=91cm
I would love to help, which problem?
Answer:
We use the formula:
S 10 = 8 · [(1/2)^9 - 1] / (1/2 - 1 );
Then, S 10 = 8 · ( 1/512 - 1 ) / ( 1/2 - 1 );
S 10 = 8 · ( - 511 / 512 ) / (-1 / 2 );
S 10 = ( - 511 / 64 ) · ( - 2 );
S 10 = + 511 / 32;
S 10 = 15.96;
Step-by-step explanation:
First you'd substitute x= y-3 in the first equation.
10(y-3) -10y= 1
10y - 30 -10y= 1
The y's cancel each other out, so you're left with -30= 1. That means there's no solution :)