5 and 6 find the least common multiple of each set of number

A region with a 5-mile radius has a population density of about 3100 people per square mile. Find the number of people who live

Gwar [14]

Answer:

243,000 people

Step-by-step explanation:

First, what we need to do is to calculate the area of the region using the 5-mile radius

Mathematically, we can calculate this as the area of a circle

Area = pi * r^2

Our r here is 5 miles

Area = pi * 5^2 = pi * 25 = 78.54 square miles

Now to get the population, we know that per square mile we have 3,100.

Hence per 78.54 square mile, we have 78.54 * 3100 = 243,273 people

To the nearest thousand, 243,000

4 0

3 years ago

What is the simplified form of the quantity of x plus 3, all over 4 + the quantity of x plus 2, all over 4?

Naily [24]

x+3
4+(x+2)
4
or
x+3 * 1
x+6 4

x+3
4x+24

4 0

3 years ago

Linear programming: You're making fruit baskets. Each day you have 240 oranges, 270 bananas, and 320 apples. Arrangement A earns

8_murik_8 [283]

Answer:

知りません

Step-by-step explanation:

5 0

3 years ago

Write an expression that equals 56 include a n exponent

Ket [755]

56^1 is a easy way to do a exponent I think

3 0

2 years ago

You want to place a towel bar that is 24 1/4 centimeters long in the center of a door that is 70 1/3 centimeters wide. How far s

Aleksandr [31]

Answer:

The towel bar should be placed at a distance of $23\frac{1}{24}\ cm$ from each edge of the door.

Step-by-step explanation:

Given:

Length of the towel bar = $24\frac14\ cm$

Now given length is in mixed fraction we will convert in fraction.

To Convert mixed fraction into fraction Multiply the whole number part by the fraction's denominator, then Add that to the numerator, then write the result on top of the denominator.

$24\frac14\ cm$ can be Rewritten as $\frac{97}{4}\ cm$

Length of the towel bar = $\frac{97}{4}\ cm$

Length of the door = $70 \frac13\ cm$

$70 \frac13\ cm$ can be Rewritten as $\frac{211}{3}\ cm$

Length of the door = $\frac{211}{3}\ cm$

We need to find the distance bar should be place at from each edge of the door.

Solution:

Let the distance of bar from each edge of the door be 'x'.

So as we placed the towel bar in the center of the door it divides into two i.e. '2x'

Now we can say that;

$\frac{97}{4}+2x=\frac{211}{3}\\\\2x=\frac{211}{3}-\frac{97}{4}$

Now we will take LCM to make the denominators common we get;

$2x=\frac{211\times4}{3\times4}-\frac{97\times3}{4\times3}\\\\\\2x= \frac{844}{12}+\frac{281}{12}$

Now denominators are common so we will solve the numerators.

$2x =\frac{844-291}{12}\\\\2x=\frac{553}{12}\\\\x=\frac{553}{12\times2} =\frac{553}{24}$

Or $x=23\frac{1}{24}\ cm$

Hence The towel bar should be placed at a distance of $23\frac{1}{24}\ cm$ from each edge of the door.

8 0

3 years ago

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