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m_a_m_a [10]
8 months ago
7

I need some help to tell whether a table is a linear or non linear function

Mathematics
1 answer:
Doss [256]8 months ago
6 0

A linear equation has the form:

y=mx+b

Let's find a linear equation using the table, and let's check if it satisfy all the points:

\begin{gathered} x=0,y=-1 \\ -1=b \\ ------ \\ x=1,y=1 \\ 1=m+b \\ 1=m-1 \\ m=2 \end{gathered}

so:

y=2x-1

Let's check the other points:

\begin{gathered} x=-3 \\ y=2(-3)-1=-6-1=-7_{\text{ }}True \\ ------ \\ x=-2 \\ y=2(-2)-1=-4-1=-5_{\text{ }}True \\ ------ \\ x=-1 \\ y=2(-1)-1=-2-1=-3_{\text{ }}True \end{gathered}

Therefore, we can conclude it is a linear function.

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2 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.

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3 years ago
Write the converse of this statement.if two angles are both obtuse, the two angles are equal.
Aliun [14]

To form the <u>converse</u> of the <u>conditional statement</u>, interchange the <u>hypothesis</u> and the <u>conclusion</u>.

Given conditional statement: "if two angles are both obtuse, the two angles are equal".

Here,

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Then the converse statement will be: "If two angles are equal, then two angles are both obtuse".

8 0
3 years ago
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Answer:

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Step-by-step explanation:

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