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gogolik [260]
3 years ago
12

Please help and show work

Mathematics
1 answer:
GrogVix [38]3 years ago
7 0

Answer:

Step-by-step explanation:

This is a right triangle trig problem. The base of the right triangle is the distance that Donna if from the flagpole; the flagpole is the side opposite the reference angle which was given as 26, and we are looking for the height of the flagpole, h. The trig ratio that uses the side opposite over the side adjacent is the tangent ratio, specifically:

tan26=\frac{h}{123} and

123tan(26) = h so

h = 60.0 rounded to the nearest tenth. But that is only the height from her line of vision and up, not the whole height. In order to find the whole height, we have to add in her height up to her line of vision which is 5.3 feet. Therefore, the height of the flagpole is

60.0 + 5.3 = 65.3 feet.

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The two-dimensional net of a rectangular prism is shown at the left. What is the prism’s surface area? square units
Tasya [4]

Answer:

28 units²  

Step-by-step explanation:

Assume that your net has the dimensions shown below.

Its area is

2 yellow rectangles = 2(4 × 2) = 2 × 8 = 16 units²

2 grey rectangles    =  2(4 × 1) = 2 × 4  =  8 units²

2 green rectangles  = 2(2 × 1)  = 2 × 2  = <u> 4 units²</u>

                                                 TOTAL = 28 units²

The prism's surface area is 28 units².

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3 years ago
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The area of a rectangle is 40 square feet. What could be the perimeter of the rectangle? how is this problem solved.
schepotkina [342]

A possible perimeter for the rectangle would be 28

Area=L x W

40/10=4

Width=4, Length=10

Perimeter=L+L+W+W

4+4+10+10=28

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3 years ago
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If your teacher tells you to questions 38 through 51 in your math book for homework ,how many questions is that ?
eduard
That would be 13 questions!
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2 years ago
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The braking distance, in feet of a car a Travling at v miles per hour is given.
irakobra [83]

The braking distance is the distance the car travels before coming to a stop after the brakes are applied

a. The braking distances are as follows;

  • The braking distance at 25 mph, is approximately <u>63.7 ft.</u>
  • The braking distance at 55 mph,  is approximately <u>298.35 ft.</u>
  • The braking distance at 85 mph,  is approximately <u>708.92 ft.</u>

b. If the car takes 450 feet to brake, it was traveling with a speed of 98.211 ft./s

Reason:

The given function for the braking distance is D = 2.6 + v²/22

a. The braking distance if the car is going 25 mph is therefore;

25 mph = 36.66339 ft./s

D = 2.6 + \dfrac{36.66339^2}{22} = 63.7 \ ft.

At 25 mph, the braking distance is approximately <u>63.7 ft.</u>

At 55 mph, the braking distance is given as follows;

55 mph = 80.65945  ft.s

D = 2.6 + \dfrac{80.65945^2}{22} \approx 298.35 \ ft.

At 55 mph, the braking distance is approximately <u>298.35 ft.</u>

At 85 mph, the braking distance is given as follows;

85 mph = 124.6555 ft.s

D = 2.6 + \dfrac{124.6555^2}{22} \approx 708.92 \ ft.

At 85 mph, the braking distance is approximately <u>708.92 ft.</u>

b. The speed of the car when the braking distance is 450 feet is given as follows;

450 = 2.6 + \dfrac{v^2}{22}

v² = (450 - 2.6) × 22 = 9842.8

v = √(9842.2) ≈ 98.211 ft./s

The car was moving at v ≈ <u>98.211 ft./s</u>

Learn more here:

brainly.com/question/18591940

8 0
1 year ago
Line parallel to y=1/2x-5
Sveta_85 [38]

Answer:

1/2

Step-by-step explanation:

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