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2 years ago

Find the product.

Mathematics

1 answer:

IrinaK [193]2 years ago

4 0

(2n+1)(2n-1)(n+5)
=(4n^2+2n-2n-1)(n+5)
=(4n^2-1)(n+5)
=4n^3-n+20n^2-5
4n^3+20n^2-n-5

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PLEASE HELP WILL MARK BRAINLIEST!!(plz show work too)

trapecia [35]

$\huge \bf༆ Answer ༄$

Let the capacity of bus be x students

And van be y students, now ;

From the given statements we get two equations ~

${ \qquad{ \sf{ \dashrightarrow}}} \: \: \sf \:2x + 10y = 260 \: \: \: \: \: (1)$

${ \qquad{ \sf{ \dashrightarrow}}} \: \: \sf \:13x + 5y = 670 \: \: \: \: \: (2)$

multiply the equation (2) with 2 [ it won't change the values ]

${ \qquad{ \sf{ \dashrightarrow}}} \: \: \sf \:2x + 10y = 260 \: \: \: \: \: \: \: \: \: (1)$

${ \qquad{ \sf{ \dashrightarrow}}} \: \: \sf \:26x + 10y = 1340 \: \: \: \: \: (3)$

Now, deduct equation (1) from equation (3)

${ \qquad{ \sf{ \dashrightarrow}}} \: \: \sf \:26x + 10y - 2x - 10y = 1340 - 260$

${ \qquad{ \sf{ \dashrightarrow}}} \: \: \sf \:24x = 1080$

${ \qquad{ \sf{ \dashrightarrow}}} \: \: \sf \:x = 1080 \div 24$

${ \qquad{ \sf{ \dashrightarrow}}} \: \: \sf \:x = 45$

Therefore each bus can carry (x) = 45 students

Now, plug the value of x in equation (1) to find y ~

${ \qquad{ \sf{ \dashrightarrow}}} \: \: \sf \:2x + 10y = 260$

${ \qquad{ \sf{ \dashrightarrow}}} \: \: \sf \:(2 \times 45) + 10y = 260$

${ \qquad{ \sf{ \dashrightarrow}}} \: \: \sf \:90 + 10y = 260$

${ \qquad{ \sf{ \dashrightarrow}}} \: \: \sf \:10y = 260 - 90$

${ \qquad{ \sf{ \dashrightarrow}}} \: \: \sf \:10y = 170$

${ \qquad{ \sf{ \dashrightarrow}}} \: \: \sf \:y = 170 \div 10$

${ \qquad{ \sf{ \dashrightarrow}}} \: \: \sf \:y = 17$

Hence, each van can carry (y) = 17 students in total.

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<em>Brainliest, please!</em>

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