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2 years ago

If the sum of five consecutive even integers is 290, what is the smallest of the five integers?

Mathematics

1 answer:

PilotLPTM [1.2K]2 years ago

4 0

So let n be the smallest number. To have consecutive even integers each number goes up by 2 so n, n+2, n+4, n+6, n+8.

Therefore, n+n+2+n+4+n+6+n+8=290

5n+20=290

5n=270

n=54

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\begin{array}{llll}
\textit{and now we}\\
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$\bf \left( \cfrac{n^{\frac{2}{3}\cdot \frac{2}{3}}}{5^{\frac{2}{3}}m^{\frac{4}{3}\cdot \frac{2}{3}}} \right)\left( \cfrac{8^{\frac{4}{3}}n^{5\cdot \frac{4}{3}}}{m^{4\cdot \frac{4}{3}}} \right)\implies \cfrac{n^{\frac{4}{9}}}{5^{\frac{2}{3}}m^{\frac{8}{9}}}\cdot \cfrac{8^{\frac{4}{3}}n^{\frac{20}{3}}}{m^{\frac{16}{3}}}\implies \cfrac{8^{\frac{4}{3}}n^{\frac{4}{9}+\frac{20}{3}}}{5^{\frac{2}{3}}m^{\frac{8}{9}+\frac{16}{3}}}$

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3 years ago

YALL PLEASE HELP, need to turn this in ASAP

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Answer:

I believe the answer is 1,800 :)

Step-by-step explanation:

1,500x0.20=300+1,500=1,800

Hope this helped!

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