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Irina-Kira [14]
3 years ago
7

How I Devide in long Division all steps?

Mathematics
2 answers:
lara [203]3 years ago
6 0
Long division is like a bus  stop method if you know

so example the set out would be like this  

koban [17]3 years ago
4 0
1)  Set up the equation. On a piece of paper, write the dividend (number being divided) on the right, under the division symbol, and the divisor (number doing the division) to the left on the outside...

2) Divide the first digit...

3) Divide the first two digits...

4) Write or enter the first digit of the quotient...

Hope this helps,
xXharleyquinn04Xx

I would be happy to help with more if you need!
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A class tossed coins and recorded 165 heads and 172 tails. What is the experimental probability of tails?
Nata [24]
165+172=337

172/337 is the experimental probability of tails.

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5 0
2 years ago
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Kira made 4 identical necklaces, each having beads and a pendant. The total cost of the beads and pendants for all 4 necklaces w
snow_lady [41]

Answer: 2.10

Step-by-step explanation: 2.10 * 4 + 6

7 0
2 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

3 0
3 years ago
Mhhhhhhhhhh welppp i need help
Dahasolnce [82]

Answer:

It does not pass the vertical Line test, because the vertical line test is when you put a line through a something like this circle and if it passes throug it more the once it is not a line.

Step-by-step explanation:

Hope this help.

6 0
3 years ago
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If the range of the function f(x) = 4x – 3 is {11.4, 15, 17, 29}, what is its domain?
malfutka [58]
The range of the function is the y-component or the ordinate of the sets of points. The domain of the function meanwhile is the x-component ot the abscissa of the sets of points. In this case, we have the function <span>f(x) = 4x – 3 and given each y, we get the x. Subsituting each y to the function we get the domain equal to {3.535, 4.5, 5, 8}.</span>
5 0
3 years ago
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