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kumpel [21]
2 years ago
14

You volunteer to help drive children at a charity event to the zoo, but you can fit only 6 of the 15 children present in your va

n How many different groups of 6 children can you drive?​
Mathematics
1 answer:
shtirl [24]2 years ago
6 0

<u><em>However if you are asking how many different combinations, there can be 5005 different combinations.</em></u>

Answer:

<u><em>2 different groups of 6 children.</em></u>

Step-by-step explanation:

<u><em>There are 15 children </em></u><u><em>you need to bring. </em></u><u><em>You can fit 6 children in each van.</em></u>

<u><em>One van : 6 children</em></u>

<u><em>Two vans : 12 children</em></u>

<u><em>Three vans : 18 children</em></u>

<u><em>If you take </em></u><u><em>two vans, that will take 12 children</em></u><u><em>. But you need to fit 15.</em></u>

<u><em>If you take </em></u><u><em>three vans, that will take 18 children.</em></u><u><em> This is more than enough.</em></u>

<u><em>15 children - 6 = 9 children</em></u>

<u><em>Thats one group with 9 children left to take</em></u>

<u><em>9 children - 6 = 3 children</em></u>

<u><em>Thats two groups with 3 children left to take</em></u>

<u><em>3 children - 6 = -3.</em></u>

<u><em>This doesn't work. In this case, </em></u><u><em>there will be 3 more empty seats in the last group.</em></u>

<u><em>Since the question asks how many groups of </em></u><u><em>6 </em></u><u><em>children you can drive. It would be </em></u><u><em>two. </em></u><u><em>Becuase </em></u><u><em>there are 2 groups of 6 and one of 3</em></u><u><em>.</em></u>

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Peter's farm has 160 meters of fencing, and he wants to fence a rectangular field thatborders a straight river. He needs no fenc
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ANSWER

3200 m²

EXPLANATION

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Replace L with the expression we found from the perimeter,

A=W(160-2W)=160W-2W^2

The area is given by a quadratic function whose leading coefficient is negative, which means that the graph is a downward parabola and, therefore, the vertex is the maximum value of the area.

The x-coordinate of the vertex is given by,

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W_{vertex}=\frac{-160}{2(-2)}=\frac{160}{4}=40m

And the length when W = 40 is,

L=160-2W=160-2\cdot40=160-80=80m

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A=WL=40m\cdot80m=3200m^2

Hence, the largest area of Peter's farm that can be fenced is 3200 square meters.

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