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3 years ago

1. Jack says "six less than a number is tour" is written as 6-n = 4. Jane says he is incorrect and

Mathematics

1 answer:

seropon [69]3 years ago

4 0

Answer:

well it really depends on how and what number you put for the n like I put 10 and so the way jane said it should be doesnt work because 10-6 just doesnt make sense and you could do it a different way but if you had do 10-6 it wouldve been in the negatives and it wouldnt have been 4 so in this case we have more choices so I would say jack is right because it is more reasonable to do 6-n=4 and instead of the n we put a 10

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2 years ago

a street light is at the top of a 18 ft tall pole. a woman 6 ft tall walks away from the pole with a speed of 4 ft/sec along a s

shutvik [7]

The most appropriate choice for similarity of triangles will be given by -

Speed of tip of the shadow of woman = 6 ft/s

What are similar triangles?

Two triangles are said to be similar, if the corrosponding angles of the triangles are same and the corrosponding sides of the triangles are in the same ratio.

Here,

The diagram has been attached here

Let the distance of woman from the pole be x ft and the distance of tip of the shadow to the pole be y ft.

Height of street light = 18 ft

Height of woman = 6ft

The two triangles are similar [As height of woman is parallel to the height of pole]

$\frac{y - x}{6}=\frac{y}{18}\\18y - 18x = 6y\\18y - 6y = 18x\\12y = 18x\\y = \frac{18}{12}x\\y = \frac{3}{2}x\\$

To find the speed, we have to differentiate both sides with respect to time 't'

$\frac{dy}{dt} =\frac{3}{2}\frac{dx}{dt}\\\frac{dy}{dt}=\frac{3}{2} \times 4\\\frac{dy}{dt} = 6$

Speed of tip of her shadow = 6 ft

To learn more about similarity of triangles, refer to the link-

brainly.com/question/14285697

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8 months 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

Distance 1:

- 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 (I)

Distance 2:

- 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 (II)

If we apply (I) + (II)

- 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 (III)

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 III) 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)

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

3 0

3 years ago