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Zarrin [17]
2 years ago
6

Linear relationships are important to understand because they are common in the world around you. For example, all rates and rat

ios are linear relationships. Miles per gallon is a common rate used to describe the number of miles a car can travel on one gallon of gasoline. Dollars per gallon, or the price of gas, is a linear relationship as well. What other relationships can you think of that are linear? How do they affect your everyday life?
​
Mathematics
1 answer:
emmasim [6.3K]2 years ago
6 0

Two other examples of linear relationships are changes of units and finding the total cost for buying a given item x times.

<h3>Other examples of linear relationships?</h3>

Two examples of linear relationships that are useful are:

Changes of units:

These ones are used to change between units that measure the same thing. For example, between kilometers and meters.

We know that:

1km = 1000m

So if we have a distance in kilometers x, the distance in meters y is given by:

y = 1000*x

This is a linear relationship.

Another example can be for costs, if we know that a single item costs a given quantity, let's say "a", then if we buy x of these items the total cost will be:

y = a*x

This is a linear relationship.

So linear relationships appear a lot in our life, and is really important to learn how to work with them.

If you want to learn more about linear relationships, you can read:

brainly.com/question/4025726

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Y=-3x+4 solved for y
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Y= -3x+4   should be 0

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The exponent indicates how many times the ( blank) is used as a factor.
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How many extraneous solutions does the equation below have? StartFraction 9 Over n squared 1 EndFraction = StartFraction n 3 Ove
BigorU [14]

The equation has one extraneous solution which is n ≈ 2.38450287.

Given that,

The equation;

\dfrac{9}{n^2+1} =\dfrac{n+3}{4}

We have to find,

How many extraneous solutions does the equation?

According to the question,

An extraneous solution is a solution value of the variable in the equations, that is found by solving the given equation algebraically but it is not a solution of the given equation.

To solve the equation cross multiplication process is applied following all the steps given below.

\rm \dfrac{9}{n^2+1} =\dfrac{n+3}{4}\\\\9 (4) = (n+3) (n^2+1)\\\\36 = n(n^2+1) + 3 (n^2+1)\\\\36 = n^3+ n + 3n^2+3\\\\n^3+ n + 3n^2+3 - 36=0\\\\n^3+ 3n^2+n -33=0\\

The roots (zeros) are the  x  values where the graph intersects the x-axis. To find the roots (zeros), replace  y

with  0  and solve for  x. The graph of the equation is attached.

n  ≈  2.38450287

Hence, The equation has one extraneous solution which is n  ≈  2.38450287

For more information refer to the link.

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Is (9, 90) a solution to the equation y = 10x?<br>yes​
Galina-37 [17]

Use the substitution method

(9,90) y= 10x

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90= 90

Answer is yes

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