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Korolek [52]
3 years ago
15

How does the total expenditures test help determine demand elasticity?

Physics
1 answer:
s2008m [1.1K]3 years ago
8 0

Answer: The three causes of demand elasticity are measured by looking at the relative change in demand as it responds to a change in price. How does the total expenditures test help determine demand elasticity? By finding out what the total expenditures are for a product, a business is able to determine the demand elasticity.

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Describe three societal needs that drive scientific endeavor and give an example of a scientific endeavor associated with each.
Trava [24]

The answer is:


1. Cure disease

2. Improve livelihoods

3. Understand nature and themselves.


1. Disease has plagued humans since the first man. Therefore, man has always used science to seek a cure for any emerging disease such as cancer today

2. Humans use science to improve livelihoods such as means of faster traveling and increasing yields in the fields

3. Science has been used to understand humans and the environment in which they live. This is evident with the numerous scientific probes that explore the earth and space collecting and analyzing data.


6 0
4 years ago
46 points :)
IgorC [24]

Answer:

Its is dividing by 2

Explanation:

It starts with 100 them it goes to 50, 25, 12.5 so its a cycle of dividing by 2

3 0
3 years ago
Read 2 more answers
A paper airplane with mass 0.1 kg is flying 1.5 m above the ground with a speed of 2 m/s. What is the total mechanical energy of
Wewaii [24]
Total mechanical energy = kinetic energy + potential energy
E = KE + PE
E = ½mv² + mgh
E = ½(0.1 kg)(2 m/s)² + (0.1 kg)(9.8 m/s²)(1.5 m)
E = 0.2 J + 1.47 J
E = 1.67 J
7 0
3 years ago
Read 2 more answers
Boyle's Law mainly involves _______.
goblinko [34]
Your answer is B, gases
6 0
4 years ago
If a system has 225 kcal of work done to it, and releases 5.00 × 102 kj of heat into its surroundings, what is the change in int
vovikov84 [41]

We can solve the problem by using the first law of thermodynamics:

\Delta U = Q-W

where

\Delta U is the change in internal energy of the system

Q is the heat absorbed by the system

W is the work done by the system on the surrounding


In this problem, the work done by the system is

W=-225 kcal=-941.4 kJ

with a negative sign because the work is done by the surrounding on the system, while the heat absorbed is

Q=-5 \cdot 10^2 kJ=-500 kJ

with a negative sign as well because it is released by the system.


Therefore, by using the initial equation, we find

\Delta U=Q-W=-500 kJ+941.4 kJ=441.4 kJ

8 0
3 years ago
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