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SVEN [57.7K]
3 years ago
8

A cube cut half along with diagnol of 20?

Mathematics
1 answer:
zavuch27 [327]3 years ago
8 0

Answer:

the answer is 42

Step-by-step explanation:

you round it

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Mezclas de alimentos. Un chef repostero creó una solución de azúcar de 50 onzas constituida por 34% de azúcar con una solución d
Kay [80]

Answer:

Las cantidades empleadas para su preparación son: 15 onzas de solución al 20 % y 35 onzas de solución al 40 %.

Step-by-step explanation:

Podemos estimar la proporción de ingredientes mediante el siguiente promedio ponderado:

0.34= \frac{0.2\cdot m_{A} + 0.4\cdot m_{B}}{m_{A}+m_{B}} (1)

Donde:

m_{A} - Masa de la solución al 20 %, en onzas.

m_{B} - Masa de la solución al 40 %, en onzas.

Podemos simplificar la formula como sigue:

0.2\cdot x + 0.4\cdot (1-x) = 0.34 (2)

Donde x es la proporción de la solución al 20 % dentro de la solución final, sin unidades.

Ahora resolvemos para x en (2):

0.4-0.2\cdot x = 0.34

0.2\cdot x = 0.06

x = \frac{0.06}{0.2}

x = 0.3

Este resultado quiere decir que la solución al 34 % es el resultado de 30 % de la solución al 20 % y 70 % de la solución al 40 %. Si conocemos que la solución final tiene una masa de 50 onzas, entonces las cantidades empleadas para su preparación son: 15 onzas de solución al 20 % y 35 onzas de solución al 40 %.

4 0
3 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

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4 years ago
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1/12.5 = 0.08, or 8%

We’re given that he’ll increase his distance by the same percentage from week 2 to 3, so to find his week 3 distance, we can find 8% of the week 2 distance and add that on. 8% of 13.5 miles is 0.08 x 13.5 = 1.08 miles, so by week 3, he’ll be running 13.5 + 1.08 = 14.58 miles.
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Pat is making 5 stools with circular seats that are each 12 inches in diameter. She wants to paint the tops of all the stools br
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Start by finding the area of one of the stools with this formula:
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