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Molodets [167]
3 years ago
6

New York City has a population of 8.55 million and 300 square mile. Manhattan has 1.64 million and 23 square miles. How much gre

ater is the population of Manhattan than New York City?
Mathematics
1 answer:
harkovskaia [24]3 years ago
4 0

Given :

New York City has a population of 8.55 million and 300 square mile area .

Manhattan has 1.64 million and 23 square miles area .

To Find :

How much greater is the population density of Manhattan than New York City.

Solution :

Population density of New York City :

D_{NY}=\dfrac{8.55}{300}=0.029 \ million/mile^2\\\\D_{M}=\dfrac{1.64}{23}=0.071 \ million/mile^2

Now ,

D_{M}-D_{NY}\\\\(0.071-0.029)=0.042\ million/miles^2

Therefore , population density of Manhattan is 0.042\ million/miles^2 greater than New York City .

Hence , this is the required solution .

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Kirk's family drove 36 miles in 50 minutes, which was 48% of the distance to his grandmother's house. What was the total distanc
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Answer:

Option D is correct.

75 miles is the total distance in miles to Kirk's grandmother's house

Step-by-step explanation:

Let x represents the total distance in miles to Kirk's grandmother's house.

As per the statement:

Kirk's family drove 36 miles in 50 minutes, which was 48% of the distance to his grandmother's house

⇒ Distance Kirk's family drove = 36 miles.

then;

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36 = \frac{48}{100}x

Multiply both sides by 100 we get;

3600 = 48x

Divide both sides by 48 we have;

75 miles = x

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We have seen that isosceles triangles have two sides of equal length. The angles opposite these sides have the same measure. Use
Naddik [55]

Question has missing figure, the figure is in the attachment.

Answer:

The measure of ∠1 is 65°.

The measure of ∠2 is 65°.

The measure of ∠3 is 50°.

The measure of ∠4 is 115°.

The measure of ∠5 is 65°.

Step-by-step explanation:

Given,

We have an isosceles triangle which we can named it as ΔABC.

In which Length of AB is equal to length of BC.

And also m∠B is equal to m∠C.

ext.m∠C= 115°(Here ext. stands for exterior)

We have to find the measure of angles angles 1 through 5.

Solution,

For ∠1.

∠1 and ext.∠C makes straight angle, and we know that the measure of straight angle is 180°.

So, we can frame this in equation form as;

\angle1+ext.\angle C=180\°

On putting the values, we get;

\angle 1+115\°=180\°\\\\\angle1=180\[tex]\therefore m\angle2=65\°-115\°=65\°[/tex]

Thus the measure of ∠1 is 65°.

For ∠2.

Since the given triangle is an isosceles triangle.

So, m\angle1=m\angle2

Thus the measure of ∠2 is 65°.

For ∠3.

Here ∠1, ∠2 and ∠3 are the three angles of the triangle.

So we use the angle sum property of triangle, which states that;

"The sum of all the angles of a triangle is equal to 180°".

\therefore \angle1+\angle2+\angle3=180\°

Now we put the values and get;

65\°+65\°+\angle3=180\°\\\\130\°+\angle3=180\°\\\\\angle3=180\°-130\°=50\°

Thus the measure of ∠3 is 50°.

For ∠4.

∠4 and ∠2 makes straight angle, and we know that the measure of straight angle is 180°.

So, we can frame this in equation form as;

\angle2 +\angle 4 =180\°

Substituting the values of of angle 2 to find angle 4 we get;

65\°+ \angle 4 = 180\°\\\\ \angle 4 = 180\°-65\°\\\\\angle 4= 115\°

Thus the measure of ∠4 is 115°.

For ∠5.

∠4 and ∠5 makes straight angle, and we know that the measure of straight angle is 180°.

So, we can frame this in equation form as;

\angle4 +\angle 5 =180\°

Substituting the values of of angle 4 to find angle 5 we get;

115\°+ \angle 5 = 180\°\\\\ \angle 5 = 180\°-115\°\\\\\angle 5= 65\°

Thus the measure of ∠5 is 65°.

Hence:

The measure of ∠1 is 65°.

The measure of ∠2 is 65°.

The measure of ∠3 is 50°.

The measure of ∠4 is 115°.

The measure of ∠5 is 65°.

6 0
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