16 - 5(3t - 4) = 8(-2t + 11) <em>use distributive property</em>: <em>a(b + c) = ab + ac</em>
16 - (5)(3t) - (5)(-4) = (8)(-2t) + (8)(11)
16 - 15t + 20 = -16t + 88
36 - 15t = -16t + 88 <em>subtract 36 from both sides</em>
-15t = -16t + 52 <em>add 16t to both sides</em>
t = 52
You could go to mathtutor.com they could help you or ask ur teacher about any availiable times they may be able to help you.
The sum of given series is 44
<em><u>Solution:</u></em>
Given that we have to find the sum of the series
<em><u>Given series is:</u></em>
![\sum_{k=1}^{4}\left(2 k^{2}-4\right)](https://tex.z-dn.net/?f=%5Csum_%7Bk%3D1%7D%5E%7B4%7D%5Cleft%282%20k%5E%7B2%7D-4%5Cright%29)
Substitute k = 1 to k = 4 to find the sum of the series
When k = 1:
![\rightarrow 2(1)^2 - 4 = 2 - 4 = -2](https://tex.z-dn.net/?f=%5Crightarrow%202%281%29%5E2%20-%204%20%3D%202%20-%204%20%3D%20-2)
When k = 2:
![\rightarrow 2(2)^2 - 4 = 2(4) - 4 = 8 - 4 = 4](https://tex.z-dn.net/?f=%5Crightarrow%202%282%29%5E2%20-%204%20%3D%202%284%29%20-%204%20%3D%208%20-%204%20%3D%204)
When k = 3:
![\rightarrow 2(3)^2 - 4 = 2(9) - 4 = 18 - 4 = 14](https://tex.z-dn.net/?f=%5Crightarrow%202%283%29%5E2%20-%204%20%3D%202%289%29%20-%204%20%3D%2018%20-%204%20%3D%2014)
When k = 4:
![\rightarrow 2(4)^2 - 4 = 2(16) - 4 = 32 - 4 = 28](https://tex.z-dn.net/?f=%5Crightarrow%202%284%29%5E2%20-%204%20%3D%202%2816%29%20-%204%20%3D%2032%20-%204%20%3D%2028)
<em><u>Therefore, the sum of series is given as:</u></em>
![\sum_{k=1}^{4}\left(2 k^{2}-4\right)=-2+4+14+28=44](https://tex.z-dn.net/?f=%5Csum_%7Bk%3D1%7D%5E%7B4%7D%5Cleft%282%20k%5E%7B2%7D-4%5Cright%29%3D-2%2B4%2B14%2B28%3D44)
Thus the sum of series is 44
It’s the same number of sides no matter what I’m pretty sure sorry if not helpful
I think is a hope this helps