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jenyasd209 [6]
3 years ago
5

The distance between itahari and Damak is

Mathematics
1 answer:
Ludmilka [50]3 years ago
8 0

Answer:

8 km by motorbike.

Step-by-step explanation:

The distance between itahari and Damak is  42 km.

Mr. Dhamala travelled 18.325 km by a  bus / 15.675 km by a taxi and the remaining distance by a motorbike.

Let he covered x distance by motorbike. So,

Remaining distance = Total distance -(distance covered by bus + distance by taxi)

= 42 -(18.325 +15.675 )

= 42-34

= 8 km

So, he travelled 8 km by motorbike.

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2 kg=2000 gram

The ratio of 2000 and 750 is

2000:750=400:150=80:30=8:3

Answer: The ratio is 8:3

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Write a function using cotangent that would create the graph of y = tan(x).
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Silva set up a dog-walking business. He earned $60 in June. He is projected to earn 110% more in July. How much is Silva project
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Elasticity of Demand The demand function for a certain brand of backpacks is
Trava [24]

Answer:

See explanation

Step-by-step explanation:

Solution:-

- The demand function of a certain brand is given as price P a function of x quantity of goods ( in hundred ) demanded per month. The relation is:

                           P ( x ) = 50 Ln ( 50 / x ).

- The point price elasticity ( E ) of demand is given by:

                           E = \frac{P}{x}*\frac{dP}{dx}  

- Where, dP / dx : is the rate of change of price ( P ) with each hundred unit of good ( x ) is demanded.

- To determine the " dP / dx " by taking the first derivative of the given relation:

                          P ( x ) = 50 Ln ( 50 / x ).

                          d P ( x ) / dx = [ 50*x / 50 ] * [ -1*50 / x^2 ]

                                              = - 50 / x

- Hence the point price elasticity of demand is given by:

                          E = - ( P / x ) * ( 50 / x )

                          E = -50*P / x^2    

- For an inelastic demand, ! E ! is < 1:

                          ! -50*P / x^2 ! < 1

                          50*P / x^2 < 1

                          P < x^2 / 50

- For an unitary demand, ! E ! is =  1:

                          ! -50*P / x^2 !  = 1

                          50*P / x^2 = 1

                          P = x^2 / 50

- For an inelastic demand, ! E ! is > 1:

                          ! -50*P / x^2 ! > 1

                          50*P / x^2 > 1

                          P > x^2 / 50

2)

If the unit price is increased slightly from $50, will the revenue increase or decrease?

- We see from the calculated demand sensitivity d P / dt:

                          d P ( x ) / dx = - 50 / x

- We see that as P increases the from P = $50, the quantity of goods demanded would be:

                          50 = 50 ln(50/x)

                           1 = Ln ( 50 / x )

                           50/x = e

                           x = 50 / e

Then,

                          d P ( x ) / dx = - 50 / ( 50 / e )

                          d P ( x ) / dx = - e

- We see that if price slightly increases from $ 50 then the quantity demanded would decrease by e (hundreds ) goods.

- The decrease in the quantity demanded is higher than the increase in price. The revenue is given by the product of price P ( x ) and x:

                Revenue R ( x ) = P ( x ) * x

                                = 50*x*ln(50/x)

Then the product of price and quantity goods also decreases; hence, revenue decreases.

8 0
3 years ago
How to solve for 40! / 10!30!
nlexa [21]
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