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3 years ago

1/8 the sum of 23 and 17 show your work

Mathematics

1 answer:

seraphim [82]3 years ago

3 0

First we take the sum of 23 and 17...

23 + 17 = 40

Now we take 1/8 of 40. In math "of" means ×

1/8 × 40= 40/8 = 5

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

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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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<h3>How to use composition between two function</h3>

Let be f and g two functions, there is a composition of f with respect to g when the domain of f is equal to the range of g. In this question, the domain variable of the function V(r) is replaced by substitution.

If we know that V(r) = (4/3) · π · r³ and r(t) = (1/4) · t², then the composite function is:

V(t) = (4/3) · π · [(1/4) · t²]³

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V(t) = (1/48) · π · t⁶

There are two true statements:

When the function is composed with r, the composite function is V(t) = (1/48) · π · t⁶.
V(r(6)) shows that the volume is 972π cubic inches after 6 seconds.

To learn on composition between functions: brainly.com/question/12007574

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