The answer is B sorry i was wrong
<h2>
Explanation:</h2>
The nth term of an arithmetic series (
) and the sum of an arithmetic series (Sum), for n terms, can be found as:
![a_{n}=a_{1}+d(n-1) \\ \\ Sum=\frac{n}{2}[2a_{1}+(n-1)d] \\ \\ \\ Where: \\ \\ a_{1}:First \ term \\ \\ d:Common \ difference \\ \\ n=Number \ of \ term](https://tex.z-dn.net/?f=a_%7Bn%7D%3Da_%7B1%7D%2Bd%28n-1%29%20%5C%5C%20%5C%5C%20Sum%3D%5Cfrac%7Bn%7D%7B2%7D%5B2a_%7B1%7D%2B%28n-1%29d%5D%20%5C%5C%20%5C%5C%20%5C%5C%20Where%3A%20%5C%5C%20%5C%5C%20a_%7B1%7D%3AFirst%20%5C%20term%20%5C%5C%20%5C%5C%20d%3ACommon%20%5C%20difference%20%5C%5C%20%5C%5C%20n%3DNumber%20%5C%20of%20%5C%20term)
So, in this exercise:
![a_{1}=a=9 \\ \\ d=4 \\ \\ n=16 \\ \\ \\ Sum=\frac{16}{2}[2(9)+(16-1)4] \\ \\ Sum=8[18+(15)4] \\ \\ Sum=8[18+60] \\ \\ Sum=8[78] \\ \\ \boxed{Sum=624}](https://tex.z-dn.net/?f=a_%7B1%7D%3Da%3D9%20%5C%5C%20%5C%5C%20d%3D4%20%5C%5C%20%5C%5C%20n%3D16%20%5C%5C%20%5C%5C%20%5C%5C%20Sum%3D%5Cfrac%7B16%7D%7B2%7D%5B2%289%29%2B%2816-1%294%5D%20%5C%5C%20%5C%5C%20Sum%3D8%5B18%2B%2815%294%5D%20%20%5C%5C%20%5C%5C%20Sum%3D8%5B18%2B60%5D%20%5C%5C%20%5C%5C%20Sum%3D8%5B78%5D%20%5C%5C%20%5C%5C%20%5Cboxed%7BSum%3D624%7D)
<h2>Learn more:</h2>
Missing numbers in triomino: brainly.com/question/10510270
#LearnWithBrainly
i think this is the answer Simplifying h(t) = -16t2 + 20t + 6 Multiply h * t ht = -16t2 + 20t + 6 Reorder the terms: ht = 6 + 20t + -16t2 Solving ht = 6 + 20t + -16t2 Solving for variable 'h'. Move all terms containing h to the left, all other terms to the right. Divide each side by 't'. h = 6t-1 + 20 + -16t Simplifying h = 6t-1 + 20 + -16t Reorder the terms: h = 20 + 6t-1 + -16t
