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emmasim [6.3K]
3 years ago
13

Find the greatest comman factor of 15 and 20

Mathematics
2 answers:
Vinil7 [7]3 years ago
6 0

Answer:

GCF(15,20) = 5

Step-by-step explanation:

Prime Factorization of 15

Prime factors of 15 are 3, 5. Prime factorization of 15 in exponential form is:

15 = 31 × 51

Prime Factorization of 20

Prime factors of 20 are 2, 5. Prime factorization of 20 in exponential form is:

20 = 22 × 51

Factors of 15

List of positive integer factors of 15 that divides 15 without a remainder.

1, 3, 5

Factors of 20

List of positive integer factors of 20 that divides 15 without a remainder.

1, 2, 4, 5, 10

Greatest Common Factor

We found the factors and prime factorization of 15 and 20. The biggest common factor number is the GCF number.

So the greatest common factor 15 and 20 is 5.

Darina [25.2K]3 years ago
4 0
The greatest common factor of 15 and 20 is 5
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Answer:

μ =  Sin θ * d₁ / (d₂ - Cos θ*d₁)

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

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⇒   - 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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