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defon
3 years ago
10

In this triangle, what is the value of X?

Mathematics
1 answer:
vovangra [49]3 years ago
3 0

Answer:

The measure of angle x is 26.1\°

Step-by-step explanation:

we know that

In the right triangle of the figure, the tangent of angle x is equal to divide the opposite side to angle x by the adjacent side to angle x

so

tan(x)=\frac{26}{53}

x=arctan(\frac{26}{53})=26.1\°

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The length of a rectangle is 19 feet longer than the width. If the perimeter is 46 feet, write an equation for this scenario
user100 [1]

Answer: The equation shall be

8 = 4W

Step-by-step explanation: The perimeter has been given as 46 feet, whereas the length and the width are not yet given. But the clue we have states that “the length is 19 feet longer than the width,” which means, if the width is W then the length shall be 19 plus W. We can express the dimensions as;

Length = 19 + W

Width = W

We can now express the perimeter as follows;

Perimeter = 2(L + W)

46 = 2(19 + W + W)

46 = 2(19 + 2W)

46 = 38 + 4W

Subtract 38 from both sides of the equation

8 = 4W

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How much interest will $2,000 earn at an annual rate of 8% in one year if the interest is compound every half a year?
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Answer:

the yearly interest would be 160.00

Step-by-step explanation:

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What is 175.42 rounded to the nearest tenth?
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175.4 because 4 is the tenth place and 2 would round down
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3 years ago
A sled is being held at rest on a slope that makes an angle theta with the horizontal. After the sled is released, it slides a d
Alenkasestr [34]

Answer:

μ =  Sin θ * d₁ / (d₂ - Cos θ*d₁)

d₂ = (d₁*Sin θ) / μ

Step-by-step explanation:

a) We apply The work-energy theorem

W = ΔE

W = - Ff*d

Ff = μ*N = μ*m*g

<em>Distance 1:</em>

- Ff*d₁ = Ef - Ei

⇒  - (μ*m*g*Cos θ)*d₁ = (Kf+Uf) - (Ki+Ui) = (Kf+0) - (0+Ui) = Kf - Ui

Kf = 0.5*m*vf² = 0.5*m*v²

Ui = m*g*h = m*g*d₁*Sin θ

then

- (μ*m*g*Cos θ)*d₁ = 0.5*m*v² - m*g*d₁*Sin θ  

⇒   - μ*g*Cos θ*d₁ = 0.5*v² - g*d₁*Sin θ   <em>(I)</em>

 

<em>Distance 2:</em>

<em />

- Ff*d₂ = Ef - Ei

⇒  - (μ*m*g)*d₂ = (0+0) - (Ki+0) = - Ki

Ki = 0.5*m*vi² = 0.5*m*v²

then

- (μ*m*g)*d₂ = - 0.5*m*v²

⇒   μ*g*d₂ = 0.5*v²     <em>(II)</em>

<em />

<em>If we apply (I) + (II)</em>

- μ*g*Cos θ*d₁ = 0.5*v² - g*d₁*Sin θ

μ*g*d₂ = 0.5*v²

 ⇒ μ*g (d₂ - Cos θ*d₁) = v² - g*d₁*Sin θ   <em>  (III)</em>

Applying the equation (for the distance 1) we get v:

vf² = vi² + 2*a*d = 0² + 2*(g*Sin θ)*d₁   ⇒   vf² = 2*g*Sin θ*d₁ = v²

then (from the equation <em>III</em>) we get

μ*g (d₂ - Cos θ*d₁) = 2*g*Sin θ*d₁ - g*d₁*Sin θ

⇒  μ (d₂ - Cos θ*d₁) = Sin θ * d₁

⇒   μ =  Sin θ * d₁ / (d₂ - Cos θ*d₁)

b)

If μ is a known value

d₂ = ?

We apply The work-energy theorem again

W = ΔK   ⇒   - Ff*d₂ = Kf - Ki

Ff = μ*m*g

Kf = 0

Ki = 0.5*m*v² = 0.5*m*2*g*Sin θ*d₁ = m*g*Sin θ*d₁

Finally

- μ*m*g*d₂ = 0 - m*g*Sin θ*d₁   ⇒   d₂ = Sin θ*d₁ / μ

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