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sesenic [268]
3 years ago
11

A ramp is 35 ft long rises to a platform. The bottom is 15 ft from the foot of the ramp. Find x, the angle of elevation of the r

amp. Round your answer to the nearest tenth of a degree.
Mathematics
1 answer:
sineoko [7]3 years ago
4 0

Answer:

The angle of elevation of the ramp is 64.60°

Step-by-step explanation:

Given;

length of the ramp, L = 35 ft

distance of the platform to the foot of the ramp, d = 15 ft

The length of the ramp forms the hypotenuse side of this right angled triangle;

The angle of elevation of the ramp is in angle between the hypotenuse and adjacent side of the triangle.

Cos x =  adjacent / hypotenuse

Cos x = 15 / 35

Cos x = 0.4286

x = Cos⁻¹ (0.4286)

x = 64.62

x = 64.60°

Therefore, the angle of elevation of the ramp is 64.60°

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Answer:

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Please help! Will give brainly, 50 points!! I'm stuck with this question and I don't get it!
pishuonlain [190]

Answer: The answer is x^2 + 4x - 3

Step-by-step explanation:

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100 points!! Brainliest if correct!!
Elena-2011 [213]

surface area (S) of a right rectangular solid is:

S = 2*L*W + 2*L*H + 2*W*H (equation 1)

where:

L = length

W = width

H = height

-----

you have:

L = 7

W = a

H = 4

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formula becomes:

S = 2*7*a + 2*7*4 + 2*a*4

simplify:

S = 14*a + 56 + 8*a

combine like terms:

S = 22*a + 56

-----

answer is:

S = 22*a + 56 (equation 2)

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to prove, substitute any value for a in equation 2:

let a = 15

S = 22*a + 56 (equation 2)

S = 22*15 + 56

S = 330 + 56

S = 386

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since a = 15, then W = 15 because W = a

go back to equation 1 and substitute 15 for W:

S = 2*L*W + 2*L*H + 2*W*H (equation 1)

where:

L = length

W = width

H = height

-----

you have:

L = 7

W = 15

H = 4

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equation 1 becomes:

S = 2*7*15 + 2*7*4 + 2*15*4

perform indicated operations:

S = 210 + 56 + 120

S = 386

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surface area is the same using both equations so:

equations are good.

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S = 22*a + 56

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3 years ago
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valentinak56 [21]

Answer:

None of the options is the answer to the question

Step-by-step explanation:

we know that

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f(x)=a(x-h)^{2}+k

where

(h,k) is the vertex of the parabola

case A) we have

f(x)=x^{2}+4x-11

In this case the x-coordinate of the vertex will be negative

therefore

case A is not the solution

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f(x)=-2x^{2}+16x-35

This case is a vertical parabola open downward (the vertex is a maximum)

The vertex is the point (4,-3) <u>but is not a minimum</u>

see the attached figure

therefore

case B is not the solution

case C) we have

f(x)=x^{2}-4x+5

Convert into vertex form

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f(x)-5=x^{2}-4x

Complete the square. Remember to balance the equation by adding the same constants to each side

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The vertex is the point (2,1)

therefore

case C is not the solution

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f(x)=2x^{2}-16x+35  

Convert into vertex form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

f(x)-35=2x^{2}-16x

Factor the leading coefficient

f(x)-35=2(x^{2}-8x)

Complete the square. Remember to balance the equation by adding the same constants to each side

f(x)-35+32=2(x^{2}-8x+16)

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Rewrite as perfect squares

f(x)-3=2(x-4)^{2}

f(x)=2(x-4)^{2}+3 --------> vertex form

The vertex is the point (4,3)

therefore

case D is not the solution

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