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3 years ago

Multiply the polynomials. (8x2 + 6x + 8)(6x – 5)

Mathematics

2 answers:

makvit [3.9K]3 years ago

6 0

Answer:

48x^3 -4x^2 + 18x -40

Step-by-step explanation:

multiply 8x^2 by 6x and -5 ➡ 48x^3 - 40x^2

multiply 6x by 6x and -5 ➡ 36x^2 -30x

multiply 8 by 6x and -5 ➡ 48x -40

now add the like terms

48x^3 - 40x^2 + 36x^2 -30x + 48x -40 ➡48x^3 -4x^2 + 18x -40

kodGreya [7K]3 years ago

3 0

Answer:

48x^3 -4x^2 + 18x -40

Step-by-step explanation: apexvs

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marshall27 [118]

Answer:

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Step-by-step explanation:

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2 years ago

The triangular prism has a square base face of 8 cm. The perpendicular height from

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2 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.

3 0

2 years ago

The quotient of 24 and the sum of<br> -4 and<br> -8

pychu [463]

Answer:

0.75

Step-by-step explanation:

-4 x -8 = 32

24 divided by 32 = 0.75

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3 years ago

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Answer:

Point a is correct.

Step-by-step explanation:

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3 years ago