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sergeinik [125]
4 years ago
14

A ramp 18 ft long rises to a platform. The bottom of the platform is 13 ft from the foot of the ramp. Find x, the angle of eleva

tion of the ramp. Round the answer to the nearest tenth of a degree.
Mathematics
1 answer:
Varvara68 [4.7K]4 years ago
8 0

Answer:

The answer to your question is: Ф = 43.95°

Step-by-step explanation:

Data

length = 18 ft

distance bottom to is 13 ft

x = angle = ?

Formula

cos Ф = opposite side / hypotenuse  

Process

cos Ф = 13 ft / 18 ft                   substitution

cos Ф = 0.72                            simplify

Ф = cos⁻¹ (0.72)

Ф = 43.95°                                result

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What is the slope of the thing
elena-14-01-66 [18.8K]

Answer:

-2/3

Step-by-step explanation:

m=\frac{y_2-y_1}{x_2-x_1} =\frac{0-2}{3-0} =-\frac{2}{3}

7 0
2 years ago
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Plz help no guessing
ankoles [38]
Your answer would be C. 6.07 × 10^24.

This is because to add numbers in standard form, we need to get their (×10^x) to be the same, as in, we need to get 7 × 10^22 to be [value] × 10^24.

To convert ×10^22 to ×10^24, we need to divide 7 by 100, as 0.07 × 10^24 = 7 × 10^22.

Now that we have them both with the same power, we can add 6 to 0.7 to get 6.07 × 10^24.

I hope this helps! Let me know if you have any questions because my explanation was a bit strange :)
6 0
3 years ago
X-5+2x-7 find the vale pleaser!!
frosja888 [35]

Answer:

12

Step-by-step explanation:

They are the same angle so they must be equal.

So, x+5 = 2x-7

     5+7 = 2x-x

        12 = x

3 0
3 years ago
The position of an object moving vertically along a line is given by the function s(t)= −16t^2 +128t. Find the average velocity
olchik [2.2K]

Answer:

a) 48

b) 64

c) 80

d) -16h + 96

Step-by-step explanation:

To find the average velocity of an object over an interval you need to find the distance it moved during the interval and divide by the time.

So

a) To find the distance it moved, we need to know the position the object was at the start of the interval and the position it was at the end of the interval. I will suppose the position is in meters and time is in seconds.

The position at the start of the interval is s(1) = -16*1² + 128(1) = 112m

The position at the end of the interval is s(4) = -16*4² + 128(4) = 256m

So, from the start to the end of the interval, the object moved 256-112 = 144m in 4-1 = 3 seconds.

Av = 144/3 = 48 m/s

b) From a), we already know s(1) = 112.

We need to find s(3)

s(3) = -16*3² + 128*3 = -144 + 384 = 240m.

From the start to the end of the interval, the object moved 240-112 = 128m in 3-1 = 2 seconds.

Av = 128/2 = 64 m/s

c) From a), s(1) = 112

We need to find s(2)

s(2) = -16*2² + 128*2 = -64 + 256  = 192m

From the start to the end of the interval, the object moved 192-112 = 80m in 2-1 = 1 second.

Av = 80/1 = 80m/s

d) From a), s(1) = 112

We need to find s(1+h).

s(1+h) = -16*(1+h)² + 128(1+h) = -16(1 + 2h + h²) + 128 + 128h = -16 - 32h - 16h² + 128 + 128h = -16h² + 96h + 112.

From the start to the end of the interval, the object moved -16h² + 96h + 112 - 112 = (-16h² + 96h)m in 1+h-1 = h seconds.

Av = (-16h² + 96h)/h = h(-16h+96)/h = -16h + 96

7 0
4 years ago
Is line de parallel to line bc, explain/ show how you know:
coldgirl [10]
Yes, it is shown horizontal for both lines and they will never touch, therefore parallel.
5 0
3 years ago
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