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ss7ja [257]
4 years ago
9

3x^3y + 12xy − 9x^2y − 36y Rewrite the expression completely factored. Show the steps of your work.

Mathematics
1 answer:
Sergeeva-Olga [200]4 years ago
4 0
Factor 3 out of 12:
3x^3y + 3 • 4xy - 9x^2y - 36y

Factor out 3 out of -9:
3x^3y + 3(4xy) + 3 • -3x^2y - 36y

Factor 3 out of -36:
3(x^3y) + 3(4x7) + 3(-3x^2y) + 3(-12)y

Factor 3 out of 3 (x^3y) + 3(-12y):
3(x^2y + 4xy) + 3(-3x^2y) + 3(-12y)

Factor 3 out of 3 (x^3y + 4xy) + 3(-3x^2y):
3(x^2y + 4xy - 3x^2y) + 3(-12y)

Factor 3 out of
3 (x^3y + 4xy - 3x^2y) + 3 (-12y):
3 (x^3y + 4xy - 3x^2y - 12y)
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Set up an equation and solve the following problem.
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Answer:

The speed of Dave is 42 miles per hour

The speed of Kent is 46 miles per hour .

Step-by-step explanation:

Given as :

The distance cover by Dave = d = 210 miles

The time taken by Dave = t hour

The speed of Dave = s miph

<u>Again</u>

The distance cover by Kent = D = 230 miles

The time taken by Kent = T hour

The speed of Kent = S = (s + 4 ) miph

<u>For Dave</u>

Time = \dfrac{\textrm Distance}{\textrm Speed}

So, t = \dfrac{\textrm d miles}{\textrm s miph}

Or, t = \dfrac{\textrm 210 miles}{\textrm s miph}

<u>For Kent</u>

Time = \dfrac{\textrm Distance}{\textrm Speed}

So, T = \dfrac{\textrm D miles}{\textrm S miph}

Or, T = \dfrac{\textrm 230 miles}{\textrm (s + 4) miph}

∵ Time taken by both is same

So, t = T

Or,  \dfrac{\textrm 210 miles}{\textrm s miph} = \dfrac{\textrm 230 miles}{\textrm (s + 4) miph}

Or, 210 × (s + 4) = 230 × s

Or, 210 × s + 210 × 4 = 230 × s

Or, 210 × 4 = 230 × s -210 × s

Or, 210 × 4 = 20 × s

∴  s = \dfrac{840}{20}

i.e s = 42 miph

So, The speed of Dave = s = 42 miles per hour

Again

The speed of Kent = S = (s + 4 ) miph

i.e S = 42 + 4

or, S = 46 miph

So, The speed of Kent = S = 46 miles per hour

Hence,The speed of Dave is 42 miles per hour

And The speed of Kent is 46 miles per hour . Answer

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