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3 years ago

Please help and hurry

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Crank3 years ago

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Answer:

3,2

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Use induction to prove: For every integer n > 1, the number n5 - n is a multiple of 5.

nignag [31]

Answer:

we need to prove : for every integer n>1, the number $n^{5}-n$ is a multiple of 5.

1) check divisibility for n=1, $f(1)=(1)^{5}-1=0$ (divisible)

2) Assume that $f(k)$ is divisible by 5, $f(k)=(k)^{5}-k$

3) Induction,

$f(k+1)=(k+1)^{5}-(k+1)$

$=(k^{5}+5k^{4}+10k^{3}+10k^{2}+5k+1)-k-1$

$=k^{5}+5k^{4}+10k^{3}+10k^{2}+4k$

Now, $f(k+1)-f(k)$

$f(k+1)-f(k)=k^{5}+5k^{4}+10k^{3}+10k^{2}+4k-(k^{5}-k)$

$f(k+1)-f(k)=k^{5}+5k^{4}+10k^{3}+10k^{2}+4k-k^{5}+k$

$f(k+1)-f(k)=5k^{4}+10k^{3}+10k^{2}+5k$

Take out the common factor,

$f(k+1)-f(k)=5(k^{4}+2k^{3}+2k^{2}+k)$ (divisible by 5)

add both the sides by f(k)

$f(k+1)=f(k)+5(k^{4}+2k^{3}+2k^{2}+k)$

We have proved that difference between $f(k+1)$ and $f(k)$ is divisible by 5.

so, our assumption in step 2 is correct.

Since $f(k)$ is divisible by 5, then $f(k+1)$ must be divisible by 5 since we are taking the sum of 2 terms that are divisible by 5.

Therefore, for every integer n>1, the number $n^{5}-n$ is a multiple of 5.

3 0

3 years ago

A triangle has sides with lengths of 5 millimeters, 12 millimeters, and 14 millimeters. Is it a right triangle?

dmitriy555 [2]

PFA FOR THE ANSWER. YOU CAN USE THE PYTHAGOREAN EQUATION (I.E a2 + b2 =

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3 years ago

Someone please help me with this math question!

stepladder [879]

$1.25x + 2.5 \ \textless \ 2.75x - 6.5 \\ \\ 2.5 \ \textless \ 2.75x - 6.5 - 1.25x \ / \ subtract \ 1.25x \ both \ sides \\ \\ 2.5 \ \textless \ 1.5x - 6.5 \ / \ simplify \\ \\ 2.5 + 6.5 \ \textless \ 1.5x \ / \ add \ 6.5 \ both \ sides \\ \\ 9 \ \textless \ 1.5x \ / \ simplify \\ \\ 6 \ \textless \ x \ / \ divide \ both \ sides \ 1.5 \\ \\ x \ \textgreater \ 6 \ / \ change \ sides \\ \\ Answer: x \ \textgreater \ 6 (A)$