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laila [671]
2 years ago
7

I need help on 9,10,11 please

Mathematics
1 answer:
Vitek1552 [10]2 years ago
8 0

9) The True Distance is 209 ft

10) The True Distance is; 66.5 ft

11) The True Height is; 57 ft

<h3>How to Interpret the Scale of a Map?</h3>

We are given that the scale of the map is 1 in : 38 ft

This means that 1 inch represents 38 ft

9) Now, we want to find the true distance of 5.5 inches from the map. Thus;

True distance = (5.5 * 38)/1

True Distance = 209 ft

10) Now, we want to find the true distance of 1.75 inches from the map. Thus;

True distance = (1.75 * 38)/1

True Distance = 66.5 ft

11) We are given the scale is 1 in : 85 ft.

Thus, the true height of 1.5 inches is;

True height = (1.5 * 38)/1

True Height = 57 ft

Read more about Scale of a Map at; brainly.com/question/5244920

#SPJ1

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8

Step-by-step explanation:

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My brother needs help with his math homework and I don't understand this?need help.
Olin [163]
To answer the problems correctly, you have to follow and pay close attention to whether or not it says "Area" or "Perimeter". For example, for the first problem, area of 63 square units, you would have to create a shape thats sides multiply out to equal 63 units squared. He could draw a square with 9 units by 7 units. This would work because 9x7=63. For perimeter, all four sides must add up to equal 38 units. Feel free to get creative with the shape on this one! If you have any other questions, just ask. Hope I helped some. :)
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3 years ago
Add.<br> =<br> -5+ (-5) = 0<br> .<br> - 3 + 6
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-30

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A rectangular field is 100 meters wide and 140meters long. Give the length and width of another rectangular field that has the s
sertanlavr [38]

Answer:

The length of rectangle 2  = 80 m

The width of rectangle 2  = 160 m

Step-by-step explanation:

The width of rectangle 1  = 100 m

The length of rectangle 1  = 140 m

Now the PERIMETER OF A RECTANGLE = 2 (L+B)

So, here the Perimeter of R 1 =  2 ( 100 + 140)  = 2 x 240  = 480 m

Also, AREA OF THE RECTANGLE = LENGTH x WIDTH

So, here the Area of R 1 =  100 x 140  = 14,000  sq m  

Now, in the Rectangle  2:

Let us assume the width of rectangle 2  = x

Assume the length of rectangle 2  = y  

Now,Perimeter of rectangle 2= Perimeter of rectangle 1

⇒ 2 (x + y)  = 480 m ......... (1)

Also, Area of rectangle 2   <  Area of rectangle 1

⇒ x y  < 14,000  .....  (2)

Now, solving both  the equations, we get

(x +y ) = 240  ⇒  y = 240 - x

x y < 14,000

Substituting y = 240 - x in (2), we get:

x (240-x) < 14,000

or, 0< x^2 -240x +14,000

Solving the above equation, we get

x < 140

Now, if we take x  = 80, then y = 160

So, for x = 80, y = 160,

Perimeter = 2 ( 80 + 160) = 240 m

Area = x y  = 80 x 160  = 12,800 < 14,000

Hence, one possible pair of solution for (x,y) =(80,160)

6 0
3 years ago
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