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In order for the data in the table to represent a linear function with a rate of change of +5, what must be the value of m?

m = 3
m = 8
m = 18
m = 33

In order for the data in the table to represent a linear function with a rate of change of 5 what must be the value of m m 3 m 8 m 18 m 33 class=

Respuesta :

Matheng
The table to represent a linear function with a rate of change of +5

so, m = 13 + 5 = 18

The correct answer is option 3 ⇒ m = 18

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another solution :
The table represents a linear function
the general equation of the linear function is ⇒ y = ax+b
where a is the slope and b is constant
By using the first and the third points from table to find the equation of the linear function
at x = 3  ⇒ y = 13   and   at x = 5 ⇒ y = 23
∴ 13 = 3a+b → (1)

23 = 5a+b → (2)
solve (1) and (2) to find a and b
So,  a = 5  and b = -2
∴ y = 5x - 2

After that by substitution with x = 4 int the equation of y
∴ y = 5*4-2 = 18

∴ m = 18

A linear function has a constant slope. The value of m for specified function is m = 18.

What is a linear relationship and how to find it from two value pairs?

A linear relationship between two variables is always writable in the form [tex]y = mx + c[/tex]

This is the equation of straight line if we plot it on XY coordinate plane.

Suppose the given points are [tex](x_1, y_1)[/tex] and [tex](x_2, y_2)[/tex] , then the equation of the straight line joining both two points is given by

[tex](y - y_1) = \dfrac{y_2 - y_1}{x_2 - x_1} (x -x_1)[/tex]

Since, we're given with two value pairs

[tex](x_1, y_1) = (3,13)\\\: \rm and \: \\(x_2, y_2) = (5.23)[/tex]

Then, we have the equation of the linear relationship between x and y as

[tex](y - y_1) = \dfrac{y_2 - y_1}{x_2 - x_1} (x -x_1)\\\\\\y -13 = \dfrac{23-13}{5-3}(x-3) \\y - 13 = 5x - 15\\y = 5x - 2[/tex]

Since this relationship should be true for [tex](4,m)[/tex] too, thus,

[tex]y = 5x - 2\\m = 5(4) - 2 = 18[/tex]

The value of m for the given condition is

Option C: m = 18

Learn more about straight line here;
https://brainly.com/question/2503591

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