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2 years ago

Give the domain and range of the relation.

Mathematics

1 answer:

igor_vitrenko [27]2 years ago

6 0

Answer: A Domain: {-5, 0, 4, 6} Range: {-9,0, 9, 13}

Step-by-step explanation:

Domain is the x-values, least to greatest.

Range is the y-values, least to greatest.

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alina1380 [7]

Answer:

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Step-by-step explanation:

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2 years ago

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2 years ago

Read 2 more answers

A child walks 3/10 of a mile in 1\5 hour. At this rate, how far will the child walk in 1 hour

defon

Answer:

1½ mile

Step-by-step explanation:

Given

3/10 mile = ⅕ hour

Required

Determine the distance in 1 hour

Since rate is constant, we have:

3/10 mile = ⅕ hour

Multiply both sides by 5

5 * 3/10 mile = ⅕ hour * 5

15/10 mile = 5/5 hour

3/2 mile = 1 hour

1½ mile = 1 hour

Hence,the child will walk 1½ mile in 1 hour

8 0

3 years ago

In a certain town there were 235 robberies last year. This year the number of robberies has gone down 33%. How many robberies we

HACTEHA [7]

Answer:

157 is the answer

Step-by-step explanation:

If you go on google and type 235- 33% you get 157.45 and to the nearest whole number is 157 hope that helped.

6 0

3 years ago

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