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2 years ago

Order these three values from least, middle, and greatest.

Mathematics

1 answer:

padilas [110]2 years ago

7 0

Answer:

answer correct order 1,3,2

Step-by-step explanation:

65% of 80 is 52

82% of 50 is 41

and 170% of 30 is 51

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The exponential model A=302.5e^0.0211 describes the population, A, of a country in millions, t years after 2003. Use the model

Naddik [55]

The population of the country in 2003 is 302.5 million

<h3>How to determine the population of the country in 2003.</h3>

From the question, we have the following parameters that can be used in our computation:

Exponential model, A = 302.5e^0.0211t

Where i is the number of years after 2003

In the year 2003, the value of t is 0

i.e. 0 years after 2003

So, we have

A = 302.5e^(0.0211 * 0)

Evaluate the products

A = 302.5 * 1

So, we have the following result

A = 302.5

Hence, the population is 302.5 million

Read more about exponential models at

brainly.com/question/27161222

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1 year ago

What’s $1.50 with a 20% markup?

Alona [7]

Answer:

$1.80

Step-by-step explanation:

3 0

2 years ago

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The graph shows the function f(x) = |x-h[ + k. What is

Olegator [25]

Answer:

K=-2.5

Step-by-step explanation:

because the line is in the middle of 2 and 3 and is below which means that it is negative

8 0

3 years ago

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Solve the equation 6(z + 8.5) = 123

kakasveta [241]

I believe it’s z = 12

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2 years ago

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A gas is said to be compressed adiabatically if there is no gain or loss of heat. When such a gas is diatomic (has two atoms per

Tems11 [23]

Answer:

The pressure is changing at $\frac{dP}{dt}=3.68$

Step-by-step explanation:

Suppose we have two quantities, which are connected to each other and both changing with time. A related rate problem is a problem in which we know the rate of change of one of the quantities and want to find the rate of change of the other quantity.

We know that the volume is decreasing at the rate of $\frac{dV}{dt}=-4 \:{\frac{cm^3}{min}}$ and we want to find at what rate is the pressure changing.

The equation that model this situation is

$PV^{1.4}=k$

Differentiate both sides with respect to time t.

$\frac{d}{dt}(PV^{1.4})= \frac{d}{dt}k\\$

The Product rule tells us how to differentiate expressions that are the product of two other, more basic, expressions:

$\frac{d}{{dx}}\left( {f\left( x \right)g\left( x \right)} \right) = f\left( x \right)\frac{d}{{dx}}g\left( x \right) + \frac{d}{{dx}}f\left( x \right)g\left( x \right)$

Apply this rule to our expression we get

$V^{1.4}\cdot \frac{dP}{dt}+1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}=0$

Solve for $\frac{dP}{dt}$

$V^{1.4}\cdot \frac{dP}{dt}=-1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}\\\\\frac{dP}{dt}=\frac{-1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}}{V^{1.4}} \\\\\frac{dP}{dt}=\frac{-1.4\cdot P \cdot \frac{dV}{dt}}{V}}$

when P = 23 kg/cm2, V = 35 cm3, and $\frac{dV}{dt}=-4 \:{\frac{cm^3}{min}}$ this becomes

$\frac{dP}{dt}=\frac{-1.4\cdot P \cdot \frac{dV}{dt}}{V}}\\\\\frac{dP}{dt}=\frac{-1.4\cdot 23 \cdot -4}{35}}\\\\\frac{dP}{dt}=3.68$

The pressure is changing at $\frac{dP}{dt}=3.68$ .

7 0

3 years ago