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2 years ago

Multiply (2x+5)(4x^2-2x+3)

Mathematics

1 answer:

shtirl [24]2 years ago

3 0

Answer:

8x^3+16x^2-4x+15

Step-by-step explanation:

Distribute and combine like terms.

Hope this helped :)

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Determine the range of this function

hodyreva [135]

The answer is b) {y|y≤5}

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3 years ago

What is the value of this expression? (37)^0 0 3^7 1 30^7

Likurg_2 [28]

Answer:

Anything to the 0th power is 1.

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2 years ago

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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

Find the rate of change: (18, 4.5) (20, 5) (22, 5.5) (24, 6)

Korvikt [17]

Answer:

The rate of change for the

X is +2

The rate of change for the

Y is + 0.5

Step-by-step explanation:

No need

4 0

2 years ago

Read 2 more answers

A. 3.4

seropon [69]

A is your answer 3.4

Use Pythagorean Theroum

10.2 squared = 9.6 squared + x squared
104= 92.6+x squared
x squared = 11.4
x= 3.37
round x to 3.4

Hope that helps!

4 0

3 years ago