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Katen [24]
3 years ago
11

6. Elizabeth drew a rectangle on a coordinate grid. How could she transform the rectangle to produce an image that is not congru

ent to the original rectangleA. Reflect it over the y-axis
B. Translate it 3 units right and 6 units up
c. Dilate it by a scale factor of 5
D. All of the above​
Mathematics
1 answer:
marusya05 [52]3 years ago
3 0

Answer:

C I think

Step-by-step explanation:

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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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Determine whether the function f(x) = 3x4 is even or odd.
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Answer:

i am 4 th

Step-by-step explanation:

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