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3 years ago

A _______ is a collection of values.

Mathematics

1 answer:

uranmaximum [27]3 years ago

3 0

A data summation is a collection of values.

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-2/5n = -30<br> What does n=

STatiana [176]

N= 75 (let me know if you need more help)

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4 years ago

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On the way to her babysitting job, Mary rode her bike for the first mile. She stopped at a traffic light and waited a few minute

Aloiza [94]

The answer is b, hope this helps

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3 years ago

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500 divide 10/2 and divide 5 and times x divide 5

coldgirl [10]

(500 / (10/2) / 5 ) * x / 5 = 20 * x / 5 = (20 / 5) * x = 4 * x

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3 years ago

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.

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3 years ago

Find the area of the rhombus please or teach me how to find the area

nevsk [136]

A=pq divided by 2.
I hoped this helped a little

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3 years ago