Answer:
British broadcasting corporation
Step-by-step explanation:
The answer is A.First we substitute 8 into the given equation.
![s = 10 \sqrt{4(8)} +15](https://tex.z-dn.net/?f=s%20%3D%2010%20%5Csqrt%7B4%288%29%7D%20%2B15%20)
Multiply 8 and 5, then find the square root of 40. Finally multiply it by 10.
![s = 10 \sqrt{40} +15](https://tex.z-dn.net/?f=s%20%3D%2010%20%5Csqrt%7B40%7D%20%2B15%20)
![s = 63.245 +15](https://tex.z-dn.net/?f=s%20%3D%2063.245%20%2B15%20)
Then add 15.
And the answer will be 78.245.
Now the question says that s is in
thousands, so the real answer would be about 75,000.
Let number of cranberries did Carissa eat on Thanksgiving Thursday = x
Given that she ate 7 cranberries on each following day so that seems she is following arithmetic sequence.
Whose first term is a = x
common difference d = 7
Given that on following Wednesday night, she had eaten a
total of 161 cranberries for the whole week.
Wednesday means on 7th day from thanksgiving day.
Then sum of 7 terms of the sequence is 161
S7=161
Sum of arithmetic sequcen is given by formula:
![S_n=\frac{n}{2}(2a+(n-1)d)](https://tex.z-dn.net/?f=%20S_n%3D%5Cfrac%7Bn%7D%7B2%7D%282a%2B%28n-1%29d%29%20)
![S_7=\frac{7}{2}(2*x+(7-1)*7)](https://tex.z-dn.net/?f=%20S_7%3D%5Cfrac%7B7%7D%7B2%7D%282%2Ax%2B%287-1%29%2A7%29%20)
![161=\frac{7}{2}(2x+6*7)](https://tex.z-dn.net/?f=%20161%3D%5Cfrac%7B7%7D%7B2%7D%282x%2B6%2A7%29%20)
![161=\frac{7}{2}(2x+42)](https://tex.z-dn.net/?f=%20161%3D%5Cfrac%7B7%7D%7B2%7D%282x%2B42%29%20)
![161*2=7(2x+42)](https://tex.z-dn.net/?f=%20161%2A2%3D7%282x%2B42%29%20)
![322=14x+294](https://tex.z-dn.net/?f=%20322%3D14x%2B294%20)
322-294=14x
28=14x
2=x
Hence Carissa ate 2 cranberries on Thanksgiving Thursday.
Now we have to find the number of day on which Carissa will eat 499 cranberries.
So we will use nth term formula
![t_n=a+(n-1)*d](https://tex.z-dn.net/?f=%20t_n%3Da%2B%28n-1%29%2Ad%20)
![499=2+(n-1)*7](https://tex.z-dn.net/?f=%20499%3D2%2B%28n-1%29%2A7%20)
![499=2+7n-7](https://tex.z-dn.net/?f=%20499%3D2%2B7n-7%20)
![499=7n-5](https://tex.z-dn.net/?f=%20499%3D7n-5%20)
![499+5=7n](https://tex.z-dn.net/?f=%20499%2B5%3D7n%20)
![504=7n](https://tex.z-dn.net/?f=%20504%3D7n%20)
72=n
Hence final answer is 72nd day.
![\bf \left.\qquad \qquad \right.\textit{negative exponents}\\\\ a^{-{ n}} \implies \cfrac{1}{a^{ n}} \qquad \qquad \cfrac{1}{a^{ n}}\implies a^{-{ n}} \qquad \qquad a^{{{ n}}}\implies \cfrac{1}{a^{-{{ n}}}}](https://tex.z-dn.net/?f=%5Cbf%20%5Cleft.%5Cqquad%20%5Cqquad%20%5Cright.%5Ctextit%7Bnegative%20exponents%7D%5C%5C%5C%5C%0Aa%5E%7B-%7B%20n%7D%7D%20%5Cimplies%20%5Ccfrac%7B1%7D%7Ba%5E%7B%20n%7D%7D%0A%5Cqquad%20%5Cqquad%0A%5Ccfrac%7B1%7D%7Ba%5E%7B%20n%7D%7D%5Cimplies%20a%5E%7B-%7B%20n%7D%7D%0A%5Cqquad%20%5Cqquad%20%0Aa%5E%7B%7B%7B%20%20n%7D%7D%7D%5Cimplies%20%5Ccfrac%7B1%7D%7Ba%5E%7B-%7B%7B%20%20n%7D%7D%7D%7D)
so, in short, if you move a factor from the bottom to the top, or the other way around, from the top to the bottom, then you change the sign of the exponent.
The side lengths would be 19,21,23
19+21+23 = 63
So the longest side length would be 23.