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3 years ago

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Kaitlin has 46 coins in her coin collection . She has albums that can hold 5 coins per page. The albums have 6 pages each. How m

any of these albums will kaitlin need if she fills each page with her coins

1 answer:

Mademuasel [1]3 years ago

6 0

Answer:

Hence, kaitlin will need 2 albums if she fills each page with her coins.

Step-by-step explanation:

The given information is:

Kaitlin has 46 coins in her coin collection.

She has albums that can hold 5 coins per page.

The albums have 6 pages each.

Now we are asked to find how many albums will be needed to fill each page with her coins.

As we know that 1 album has 6 pages.

1 page can hold=5 coins.

So, 1 album can old=6×5=30 coins.

Now 46-30=16 coins are left

So we will need one ore album so that these 16 coins could be placed in it.

Hence, kaitlin will need 2 albums if she fills each page with her coins.

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Step-by-step explanation:

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Step-by-step explanation:

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The question seems not to be complete, below is a possible complete question:

Kristin earns $6.50 for every hour she babysits for the Overstreet family. Part A: Last Friday, Kristin babysat for 4.5 hours. How much money did she make during that job? $__________ Part B: Mrs. Overstreet has told Kristin that she will pay her $45.50 for tomorrow's babysitting job. How many hours will Kristin babysit then? __________ hours

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Step-by-step explanation:

Part A:

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Answer:

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3. $\frac{2}{x^2-9}-\frac{3x}{x^2-5x+6}$ = $\frac{-3x^2-7x-4}{(x+3)(x-3)(x-2)}$

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Step-by-step explanation:

1. $\frac{x}{x+3}+\frac{x+2}{x+5}$

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$=\frac{x(x+5)+(x+2)(x+3)}{(x+3)(x+5)}\\=\frac{x^2+5x+(x)(x+3)+2(x+3)}{(x+3)(x+5)} \\=\frac{x^2+5x+x^2+3x+2x+6}{(x+3)(x+5)} \\=\frac{x^2+x^2+5x+3x+2x+6}{(x+3)(x+5)} \\=\frac{2x^2+10x+6}{(x+3)(x+5)}\\$

Prove closure: The value of x≠-3 and x≠-5 because if there values are -3 and -5 then the denominator will be zero.

2. $\frac{x+4}{x^2+5x+6}*\frac{x+3}{x^2-16}$

Factors of x^2-16 = (x)^2 -(4)^2 = (x-4)(x+4)

Factors of x^2+5x+6 = x^2+3x+2x+6 = x(x+3)+2(x+3) =(x+2)(x+3)

Putting factors

$=\frac{x+4}{(x+3)(x+2)}*\frac{x+3}{(x-4)(x+4)}\\\\=\frac{1}{(x+2)(x-4)}$

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3. $\frac{2}{x^2-9}-\frac{3x}{x^2-5x+6}$

Factors of x^2-9 = (x)^2-(3)^2 = (x-3)(x+3)

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Putting factors

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Factors of x^2-5x+6 = x^2-3x-2x+6 = x(x-3)-2(x-3) = (x-2)(x-3)

Factors of x^2-16 = (x)^2 -(4)^2 = (x-4)(x+4)

$\frac{x+4}{(x-2)(x+3)}\div\frac{(x-4)(x+4)}{x+3}$

Converting ÷ sign into multiplication we will take reciprocal of the second term

$=\frac{x+4}{(x-2)(x+3)}*\frac{x+3}{(x-4)(x+4)}\\=\frac{1}{(x-2)(x-4)}$

Prove Closure: The value of x≠2 and x≠4 because if there values are 2 and 4 then the denominator will be zero.

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3 years ago