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kirill115 [55]
3 years ago
11

Explain how low-, and medium-, and high density housing characteristics represent different patterns of residential land use.

Geography
1 answer:
Fittoniya [83]3 years ago
7 0

Explanation:

The quantity of people differs in each type of housing: High-density housing has the highest number of people per geographic unit. Then medium-density housing has less people than the first one, and finally low-density housing has the smallest number of people.

- High-density housing. An example for this are apartment buildings with many storeys in a city.

- Medium-density housing: for example, row houses and townhomes in a midtown urban zone.

Note that residential, retail and commercial buildings together are classified as high to medium density housing. The reason is that the housing are usually apartments and condos.

- Low-density housing: for example, a large area with a few farms and houses.

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The gravitational pull of the sun and moon on the earth also causes waves. Although the sun's gravity also contributes to the waves, but its effects are smaller and serve mostly to vary the heights of high and low tide.

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3 years ago
What is the value of X6? <br><br> Show the solution.
wariber [46]

Answer : The value of x_6 is \sqrt{7}.

Explanation :

As we are given 6 right angled triangle in the given figure.

First we have to calculate the value of x_1.

Using Pythagoras theorem in triangle 1 :

(Hypotenuse)^2=(Perpendicular)^2+(Base)^2

(x_1)^2=(1)^2+(1)^2

x_1=\sqrt{(1)^2+(1)^2}

x_1=\sqrt{2}

Now we have to calculate the value of x_2.

Using Pythagoras theorem in triangle 2 :

(Hypotenuse)^2=(Perpendicular)^2+(Base)^2

(x_2)^2=(1)^2+(X_1)^2

(x_2)^2=(1)^2+(\sqrt{2})^2

x_2=\sqrt{(1)^2+(\sqrt{2})^2}

x_2=\sqrt{3}

Now we have to calculate the value of x_3.

Using Pythagoras theorem in triangle 3 :

(Hypotenuse)^2=(Perpendicular)^2+(Base)^2

(x_3)^2=(1)^2+(X_2)^2

(x_3)^2=(1)^2+(\sqrt{3})^2

x_3=\sqrt{(1)^2+(\sqrt{3})^2}

x_3=\sqrt{4}

Now we have to calculate the value of x_4.

Using Pythagoras theorem in triangle 4 :

(Hypotenuse)^2=(Perpendicular)^2+(Base)^2

(x_4)^2=(1)^2+(X_3)^2

(x_4)^2=(1)^2+(\sqrt{4})^2

x_4=\sqrt{(1)^2+(\sqrt{4})^2}

x_4=\sqrt{5}

Now we have to calculate the value of x_5.

Using Pythagoras theorem in triangle 5 :

(Hypotenuse)^2=(Perpendicular)^2+(Base)^2

(x_5)^2=(1)^2+(X_4)^2

(x_5)^2=(1)^2+(\sqrt{5})^2

x_5=\sqrt{(1)^2+(\sqrt{5})^2}

x_5=\sqrt{6}

Now we have to calculate the value of x_6.

Using Pythagoras theorem in triangle 6 :

(Hypotenuse)^2=(Perpendicular)^2+(Base)^2

(x_6)^2=(1)^2+(X_5)^2

(x_6)^2=(1)^2+(\sqrt{6})^2

x_6=\sqrt{(1)^2+(\sqrt{6})^2}

x_6=\sqrt{7}

Therefore, the value of x_6 is \sqrt{7}.

5 0
2 years ago
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