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Elenna [48]
3 years ago
5

Prove the following statement.

Mathematics
1 answer:
gayaneshka [121]3 years ago
6 0

Answer:

You can prove this statement as follows:

Step-by-step explanation:

An odd integer is a number of the form 2k+1 where k\in \mathbb{Z}. Consider the following cases.

Case 1. If k is even we have: (2k+1)^{2}=(2(2s)+1)^{2}=(4s+1)^{2}=16s^2+8s+1=8(2s^2+s)+1.

If we denote by m=2s^2+2 we have that (2k+1)^{2}=8m+1.

Case 2. if k is odd we have: (2k+1)^{2}=(2(2s+1)+1)^{2}=(4s+3)^{2}=16s^2+24s+9=16s^{2}+24s+8+1=8(2s^{2}+3s+1)+1.

If we denote by m=2s^{2}+24s+1 we have that (2k+1)^{2}=8m+1

This result says that the remainder when we divide the square of any odd integer by 8 is 1.

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\bf \textit{Cofunction Identities}
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sin\left(\frac{\pi}{2}-\theta\right)=cos(\theta)
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3 years ago
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Step-by-step explanation:

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padilas [110]

Answer:

d) 2x²(6x³ + 3x + 4)

Step-by-step explanation:

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