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Serga [27]
3 years ago
5

the speeders scccer team charged $12 to wash each car at a fundraiser car wash. the team collected a total of $672 by the end of

the day. how many cars did the team wash???
Mathematics
1 answer:
snow_lady [41]3 years ago
4 0
All u have do do is add 12 ×56 and u will get 627 so they will wash 56 cars
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Expand and simplify (x+1)(x+6)
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x^{2}+7x+6

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3 years ago
This question has three parts. Answer the parts in order.
nalin [4]

Answer:

The area of the smallest section is A_{1}=100yd^{2}

The area of the largest section is A_{2}=625yd^{2}

The area of the remaining section is A_{3}=250yd^{2}

Step-by-step explanation:

Please see the picture below.

1. First we are going to name the side of the larger square as x.

As the third section shares a side with the larger square and the four sides of a square are equal, we have the following:

- Area of the first section:

A_{1}=10yd*10yd

A_{1}=100yd^{2}

- Area of the second section:

A_{2}=x^{2} (Eq.1)

- Area of the third section:

A_{3}=width*length

A_{3}=10yd*x (Eq.2)

2. The problem says that the total area of the enclosed field is 975 square yards, and looking at the picture below, we have:

A_{1}+A_{2}+A_{3}=975yd^{2}

Replacing values:

100+x^{2}+10x=975

Solving for x:

x^{2}+10x-875=0

x=\frac{-10+\sqrt{100+(4*875)}}{2}

x=\frac{-10+\sqrt{3600}}{2}

x=\frac{-10+60}{2}

x=25

3. Replacing the value of x in Eq.1 and Eq.2:

- From Eq.1:

A_{2}=25^{2}

A_{2}=625yd^{2}

- From Eq.2:

A_{3}=10*25

A_{3}=250yd^{2}

3 0
3 years ago
i was doing this task in math and I came across this: x-12x=0.88x . can someone plz explain why is this equal to this? I underst
valkas [14]

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1x-12x=0.88x

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-11.88x=0x

-11.88x=0

Isolate x

3 0
2 years ago
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