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levacccp [35]
3 years ago
5

A school playground is in the shape of a rectangle 500 feet long and 100 feet wide. If fencing costs $12 per? yard, what will it

cost to place fencing around the? playground?
Mathematics
1 answer:
Tresset [83]3 years ago
8 0

Answer:

Perimeter = 2(l + b)= 2( 500 + 100)= 2( 600 ) = 1200ftconvert the feet onto yard1 yard = 3feet? = 1200ft=%281200%2A1%29%2F3=400 yardsCost of fencing per yard = $12Therefore cost of fencing 400 yards = 12 * 400= $ 3600Regards    -_-

Step-by-step explanation:

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Answer: 1 < x < 2

Step-by-step explanation:

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Find the shortest distance from point A to the given line.
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2 years ago
Y is inversely proportional to x. When y=25, x=4 Work out y when x=20
ipn [44]

Answer:

y=5 when x=20

Step-by-step explanation:

Using the format for inverse proportions: y=k/x

To find k, substitute 25 in for y and 4 for x to find:

25=k/4 ⇒ k=100

Using this we can substitute k and x in for 100 and 20 to find:

y=100/20 ⇒ y=5

8 0
3 years ago
Which expression is equivalent to StartRoot 120 x EndRoot?.
Irina-Kira [14]

Here for 120, the expression for the start root and end root is given  

\sqrt{120} =2\sqrt{30}

<h3>What will be the expression for start root and enroot of 120?</h3>

Here we need to find \sqrt{120}

By simplifying the given expression, we get:

120=2\times2\times2\times3\times5

120=(2^{2} )\times3\times5

by taking square roots on both sides we get

\sqrt{120} =2\sqrt{30}

Thus for 120, the expression for the start root and end root is given  \sqrt{120} =2\sqrt{30}

To know more about the Square roots follow

brainly.com/question/124481

5 0
1 year ago
A stadium has 45,000 seats. Seats sell for $28 in Section A, $24 in Section B, and $20 in Section C. If section C held 300 fewer
White raven [17]

Answer:

Let's define the variables:

A = number of seats in section A.

B = number of seats in section B.

C = number of seats in section C.

We have the equations:

A + B + C = 45,000.

C - 300 = B/2

A*$28 + B*$24 + C*$20 = $1,139,200

This is a system of equations, the first step to solve this is to isolate one variable in one of the equations, and then replace it in the others.

I will isolate C in the second equation:

C = B/2 + 300.

Now let's replace this in the other two equations:

A + B + B/2 + 300 = 45,000

A*$28 + B*$24 + (B/2 + 300)*$20 = $1,139,200

Let's simplify these equations:

A + B*(3/2) = 44,700

A*$28 + B*$34 + $6,000 = $1,139,200

Now let's isolate A in the first equation:

A = 44,700 - B*(3/2)

Let's replace this in the other equation:

(44,700 - B*(3/2))*$28 + B*$34 + $6,000 = $1,139,200

Now let's solve this for B.

-B*$8 + $1,252,200 = $1,139,200

-B*$8 =  $1,139,200 - $1,252,200  = -$113,000

B = $113,000/8 = 14,125

Now we can replace that in the equations:

A =  44,700 - B*(3/2) =  44,700 - 14,125*(3/2) = 23,512.5, that we should round up to 23,513.

And C = B/2 + 300 = 7362.5

As we rounded the previous one up, we should round this one down to 7362.

3 0
2 years ago
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