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3 years ago

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Five pounds of cherries cost $14.95. Write and solve an equation to determine how much each pound of cherries costs.

2 answers:

harina [27]3 years ago

5 0

2.99 each hope that HELPSSSSSSS

larisa [96]3 years ago

3 0

Answer:each pound is $2.99

Step-by-step explanation:

divide $14.95 by 5

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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.

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3 years ago

Of the following, which is the solution to 2x^2-5x=-8

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the answer is B these are the steps of getting your answer

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3 years ago

Find the inverse of the function. Show work. <br> g(x)= - 3/x-2 +2

4vir4ik [10]

Answer:

$\displaystyle g^{-1}(x)=\frac{-7+2x}{x-2}$

Step-by-step explanation:

We are given the function:

$\displaystyle g(x)=-\frac{3}{x-2}+2$

Let's find the inverse of g.

Call y=g(x):

$\displaystyle y=-\frac{3}{x-2}+2$

We need to solve for x. Multiply both sides by x-2 to eliminate denominators:

$y(x-2)=-3+2(x-2)$

Operate:

$yx-2y=-3+2x-4$

Collect the x's to the left side and the rest to the right side of the equation:

$yx-2x=-3-4+2y$

Factor the left side and operate on the right side:

$x(y-2)=-7+2y$

Solve for x:

$\displaystyle x=\frac{-7+2y}{y-2}$

Interchange variables:

$\displaystyle y=\frac{-7+2x}{x-2}$

Call y as the inverse function:

$\boxed{\displaystyle g^{-1}(x)=\frac{-7+2x}{x-2}}$

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3 years ago