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GrogVix [38]
3 years ago
5

The density of hydrogen is approximately 9 × 10-5 , while the density of helium is approximately 2 × 10-4 . Which of the followi

ng is true?
A.
The density of hydrogen is approximately 20 times the density of helium.

B.
The density of helium is approximately 20 times the density of hydrogen.

C.
The density of hydrogen is approximately 2 times the density of helium.

D.
The density of helium is approximately 2 times the density of hydrogen.
Mathematics
2 answers:
svp [43]3 years ago
7 0
<span>The density of helium is approximately 2 times the density of hydrogen.</span>
Yanka [14]3 years ago
3 0

Answer:

Option D.

Step-by-step explanation:

Density of Hydrogen = 9\times 10^{-5}

Density of Helium = 2\times 10^{-4}

Now ratio of the densities = \frac{\text{Density of Helium}}{\text{Density of Hydrogen}}

= \frac{2\times 10^{-4}}{9\times 10^{-5}}

= 0.22×10^{5-4}

= 2.2

Density of Helium = 2× Density of Hydrogen

Therefore, density of Helium is approximately 2 times the density of Hydrogen.

Option D is the answer.

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Horsepower and watts are units of measure of
Leni [432]

The 2 horsepower is equal to 1492 watts.

Given that horsepower and watts are units of measure of power and they are directly proportional such that 5 horsepower is equal to 3730 watts.

Horsepower is a unit of measurement of power, or speed at which work is done, usually related to the power of an engine or motor. power is calculated by multiplying force (in pounds) by  speed (in feet per second).

Let H= horsepower, W=watt

As we are given horsepower is directly proportional.

So, the relation is W=kH where k is constant.

Here, H=5 and W=3730.

By substituting the value of W and H we will find the value of k,  we get

3730=k×5

3730/5=k

746=k

Now, substituting the value of k in relation and we get

W=746H

We have to find the power of 2 horsepower.

So, we will substitute H=2 in above relation, we get

W=746×2

W=1492 watts.

Hence, the 2 horsepower is equal to 1492 watts when 5 horsepower is equal to 3730 watts.

Learn more about horsepower from here brainly.com/question/1335872

#SPJ4

7 0
2 years ago
Find the marginal and average revenue functions associated with the demand function P= -0.3Q + 221
LekaFEV [45]

Answer:

Marginal revenue = R'(Q) = -0.6 Q + 221

Average revenue = -0.3 Q + 221

Step-by-step explanation:

As per the question,

Functions associated with the demand function P= -0.3 Q + 221, where Q is the demand.

Now,

As we know that the,

Marginal revenue is the derivative of the revenue function, R(x), which is equals the number of items sold,

Therefore,

R(Q) = Q × ( -0.3Q + 221) = -0.3 Q² + 221 Q

∴ Marginal revenue = R'(Q) = -0.6 Q + 221

Now,

Average revenue (AR) is defined as the ratio of the total revenue by the number of units sold that is revenue per unit of output sold.

Average\ revenue\ = \frac{Total\ revenue}{number\ of\ units\ sold}

Where Total Revenue (TR) equals quantity of output multiplied by price per unit.

TR = Price (P) × Total output (Q) = (-0.3Q + 221) × Q = -0.3 Q² + 221 Q

Average\ revenue\ = \frac{TR}{Q}

Average\ revenue\ = \frac{-0.3Q^{2}+221Q}{Q}

∴ Average revenue = -0.3Q + 221

5 0
3 years ago
3(2x + 1) = 6(5x - 4)
Law Incorporation [45]

Answer:

X= 9 over 8 or 9/8

Step-by-step explanation:

hopefully this helps

5 0
3 years ago
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HELPP!! 10 points 1 question . ( explain your answer )
Greeley [361]
50.25% off because 100÷199=50.25
4 0
3 years ago
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I need help setting it up and solving it but mostly setting it all up please and thank you ?!!!i need help asap!!!!!
professor190 [17]
This is the work that I did. Hope it helps

8 0
3 years ago
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