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3 years ago

14

Durning one winter snow storm in Denver,Colorado, Jesse noted that 16 inches of snow fell in 5 1/3 hour.What was the rate of sno

wfall in one hour

1 answer:

givi [52]3 years ago

3 0

16 / 5.333333... = 3 inches per hour

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A softball has a diameter of 3.82 inches. How many square inches of material is necessary to cover a softball? Use 3.14 for π

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45.82 square inches

Step-by-step explanation:

Use the formula for the surface area of a sphere, A = 4 $\pi$ r²

If the diameter is 3.82 inches, the radius is 1.91 inches.

Plug in 3.14 as $\pi$ and 1.91 as r, and solve:

A = 4 $\pi$ r²

A = 4(3.14)(1.91²)

A = 45.82

So, approximately 45.82 square inches of material are needed.

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3 years ago

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3 years ago

The length of a rectangle is 5 metres less than twice the breadth. If the perimeter is 50 meters,find the length and breadth

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<h3><u>S</u><u> </u><u>O</u><u> </u><u>L</u><u> </u><u>U</u><u> </u><u>T</u><u> </u><u>I</u><u> </u><u>O</u><u> </u><u>N</u><u> </u><u>:</u></h3>

As per the given question, it is stated that the length of a rectangle is 5 m less than twice the breadth.

Assumption : Let us assume the length as "l" and width as "b". So,

$\twoheadrightarrow \quad\sf{ Length =2(Width)-5}$

$\twoheadrightarrow \quad\sf{ \ell=(2b-5) \; m}$

Also, we are given that the perimeter of the rectangle is 50 m. Basically, we need to apply here the formula of perimeter of rectangle which will act as a linear equation here.

$\\ \twoheadrightarrow \quad\sf{ Perimeter_{(Rectangle)} = 2(\ell +b) } \\$

<em>l</em> denotes length
<em>b</em> denotes breadth

$\\ \twoheadrightarrow \quad\sf{50= 2(2b-5+b)} \\$

$\\ \twoheadrightarrow \quad\sf{50= 2(3b-5)} \\$

$\\ \twoheadrightarrow \quad\sf{50= 6b - 10} \\$

$\\ \twoheadrightarrow \quad\sf{50+10= 6b} \\$

$\\ \twoheadrightarrow \quad\sf{60= 6b} \\$

$\\ \twoheadrightarrow \quad\sf{\cancel{\dfrac{60}{6}}=b} \\$

$\\ \twoheadrightarrow \quad\underline{\bf{10\; m = Width }} \\$

Now, finding the length. According to the question,

$\twoheadrightarrow \quad\sf{ \ell=(2b-5) \; m}$

$\twoheadrightarrow \quad\sf{ \ell=2(10)-5\; m}$

$\twoheadrightarrow \quad\sf{ \ell=20-5\; m}$

$\\ \twoheadrightarrow \quad\underline{\bf{15\; m = Length }} \\$

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