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Ne4ueva [31]
2 years ago
10

I don't understand this please help

Mathematics
2 answers:
inna [77]2 years ago
8 0
$9.10 do you need to know explanation?
LUCKY_DIMON [66]2 years ago
7 0

Answer:

$9.10

Step-by-step explanation:

you just need to multiply 6.50 by 2 and then take 30% off of it which is 9.10

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PLEASE HELP ME WITH THIS....<br> BRAINLIEST AVAILABLE
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4 0
3 years ago
Use mathematical induction to prove
Alex17521 [72]

Prove\ that\ the\ assumption \is \true for\ n=1\\1^3=\frac{1^2(1+1)^2}{4}\\ 1=\frac{4}{4}=1\\

Formula works when n=1

Assume the formula also works, when n=k.

Prove that the formula works, when n=k+1

1^3+2^3+3^3...+k^3+(k+1)^3=\frac{(k+1)^2(k+2)^2}{4} \\\frac{k^2(k+1)^2}{4}+(k+1)^3=\frac{(k+1)^2(k+2)^2}{4} \\\frac{k^2(k^2+2k+1)}{4}+(k+1)^3=\frac{(k^2+2k+1)(k^2+4k+4)}{4} \\\frac{k^4+2k^3+k^2}{4}+k^3+3k^2+3k+1=\frac{k^4+4k^3+4k^2+2k^3+8k^2+8k+k^2+4k+4}{4}\\\\\frac{k^4+2k^3+k^2}{4}+k^3+3k^2+3k+1=\frac{k^4+6k^3+13k^2+12k+4}{4}\\\frac{k^4+2k^3+k^2}{4}+\frac{4k^3+12k^2+12k+4}{4}=\frac{k^4+6k^3+13k^2+12k+4}{4}\\\frac{k^4+6k^3+13k^2+12k+4}{4}=\frac{k^4+6k^3+13k^2+12k+4}{4}\\

Since the formula has been proven with n=1 and n=k+1, it is true. \square

7 0
2 years ago
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