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Serjik [45]
3 years ago
10

I have no clue Someone Please Help!!!!!

Mathematics
1 answer:
erma4kov [3.2K]3 years ago
7 0

Answer:

D

Step-by-step explanation:

we have to see the value of y on x=-2

which is 2

so g(-2)=2

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What is the reliability of a two-component product if the components are in parallel, with individual reliabilities of 0.95, and
user100 [1]

The reliability of a two-component product if the components are in parallel is 0.99.

In this question,

The probability of failure-free operation of a system with several parallel elements is always higher than that of the best element in the system. Reliability can be increased if the same function is done by two or more elements arranged in parallel.

A system contains two components that are arranged in parallel, they are 0.95 and 0.80.

Therefore the system reliability can be calculated as follows

⇒ 1 - ( 1 - 0.95 ) × ( 1 - 0.80 )

⇒ 1 - (0.05 × 0.20)

⇒ 1 - 0.01

⇒ 0.99

Hence we can conclude that the reliability of a two-component product if the components are in parallel is 0.99.

Learn more about reliability of components here

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1 year ago
There are 5 people in a room, i came and killed 4. how many remain?
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Two because after you came you and that other person are still in the room
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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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Ansley wants to drink at least 64 ounces of water per day, but no more than 72 ounces. How many ounces of water per day might sh
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64,65,66,67,68,69,70,71,72
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34kurt

1. EFG

2. F

3. FA and FB Take a moment to look at this one. The two rays are pointing in opposite directions. They are 180o apart. They share a common point at F.

FE and FG does the same thing.

4. No G is on a line that has only one point on the plane and that point is F.

5, AC has an infinite number of points on the plane.

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