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Ghella [55]
3 years ago
5

The sand pit for the long jump has a width of 2.75 meters and a length of 9.54 meters. Just in case it rains, the principal want

s to cover the sand pit with a piece of plastic the night before the event. How many square meters of plastic will the principal need to cover the sand pit?
Mathematics
1 answer:
Salsk061 [2.6K]3 years ago
7 0

Answer:

The principal will need 26.24 m² of plastic to cover the sand pit.

Step-by-step explanation:

In order to calculate the amount of plastic, we need to find out the area of the sandpit. The shape of the sandpit is a rectangle, so we can use the formula:

A=W×H

where,

W is the width of the sandpit

L is the length of the sandpit

Therefore,

A=2.75 m × 9.54 m = 26.24 m²

The amount of plastic needed is 26.24 m²

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step by step explanation

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2 years ago
For the function f(x) = x + 4, what is the ordered pair for the point on the graph when x = 3p?
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Thank you.
6 0
2 years ago
Rewrite the expression in the form y^ny n y, start superscript, n, end superscript. \dfrac{1}{y^{^{\scriptsize\dfrac54}}}= y 4 5
just olya [345]

Answer:

The expression can be rewritten in y^n form as following:

⇒  y^{-\frac{5}{4}}

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Step-by-step explanation:

Given expression:

\dfrac{1}{y^{^{\scriptsize\dfrac54}}}

To rewrite the expression in the form of y^n.

Solution:

By property of exponents :

\frac{1}{x^a}=x^{-a}

<em>So, we can apply this property to the given expression.</em>

We have:

\dfrac{1}{y^{^{\scriptsize\dfrac54}}}

⇒  y^{-\frac{5}{4}}

The above expression is in the form of y^n where n=-\frac{5}{4}

8 0
3 years ago
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