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3 years ago

Find the reciprocal of 3/20

1 answer:

5 0

20/3
taking the reciprocal is the same as flipping the fraction upside, and the 2 fractions should then multiply together to make 1

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Consider the right triangle. Determine whether each equation is correct. Select yes or no for each question.

joja [24]

Step-by-step explanation:

Given

Opp=8

Adj=6

Hyp=10

Answer

1)Sin<A=Opp/Hyp

8/10=4/5... Yes

2)Cos<A=Opp/Adj

8/6=4/3... No

Given

Opp=6

Adj= 8

Hyp= 10

Answer

1)Sin<B= Opp/Hyp

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2)Cos<B= Opp/Adj

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3 years ago

What’s 7 to the 8 power?

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4 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

3 years ago