Write a function that describes a relationship between two quantities - HiSET

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Question

is a linear function defined on the set of real numbers. Four values of are given in the following table:

$\begin{matrix} \underline{x}& \underline{f(x)} \ 13 & 4 \ 10 & 6 \ 7 & 8 \ 4 & 10 \end{matrix}$

Give the definition for .

Answer

By the two-point formula, the equation of a line through points $(x_{1}, y_{1}), (x_{2}, y_{2})$ is

$y-y_{1} = m(x- x_{1})$ ,

where $m = \frac{x_{2}-x_{1}}{y_{2}-y_{1}}$ , the slope of the line. The choice of the two points from the given four is arbitrary, so we will select the middle two ordered pairs. Substituting $x_{1}= 10,y_{1} =6, x_{2}= 7, y_{2}= 8$ , then

$m = \frac{x_{2}-x_{1}}{y_{2}-y_{1}}= \frac{7-10}{8-6} = -\frac{3}{2}$

The first formula becomes

$y-10 = -\frac{3}{2}(x-6)$

Since , solve for . First, distribute $-\frac{3}{2}$ throughout the difference at right:

$y-10 = -\frac{3}{2} x- \left (-\frac{3}{2} \right )6$

$y-10 = -\frac{3}{2} x- \left (-9 \right )$

$y-10 = -\frac{3}{2} x+9$

Isolate by adding 10 to both sides:

$y-10+ 10 = -\frac{3}{2} x+9+ 10$

$y = -\frac{3}{2} x+19$

Replacing, the definition of is

$f(x) = -\frac{3}{2} x+19$ .

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Question

is a linear function defined on the set of real numbers. Four values of are given in the following table:

$\begin{matrix} \underline{x}& \underline{f(x)} \ 13 & 4 \ 10 & 6 \ 7 & 8 \ 4 & 10 \end{matrix}$

Give the definition for .

Answer

By the two-point formula, the equation of a line through points $(x_{1}, y_{1}), (x_{2}, y_{2})$ is

$y-y_{1} = m(x- x_{1})$ ,

where $m = \frac{x_{2}-x_{1}}{y_{2}-y_{1}}$ , the slope of the line. The choice of the two points from the given four is arbitrary, so we will select the middle two ordered pairs. Substituting $x_{1}= 10,y_{1} =6, x_{2}= 7, y_{2}= 8$ , then

$m = \frac{x_{2}-x_{1}}{y_{2}-y_{1}}= \frac{7-10}{8-6} = -\frac{3}{2}$

The first formula becomes

$y-10 = -\frac{3}{2}(x-6)$

Since , solve for . First, distribute $-\frac{3}{2}$ throughout the difference at right:

$y-10 = -\frac{3}{2} x- \left (-\frac{3}{2} \right )6$

$y-10 = -\frac{3}{2} x- \left (-9 \right )$

$y-10 = -\frac{3}{2} x+9$

Isolate by adding 10 to both sides:

$y-10+ 10 = -\frac{3}{2} x+9+ 10$

$y = -\frac{3}{2} x+19$

Replacing, the definition of is

$f(x) = -\frac{3}{2} x+19$ .

Compare your answer with the correct one above

Question

is a linear function defined on the set of real numbers. Four values of are given in the following table:

$\begin{matrix} \underline{x}& \underline{f(x)} \ 13 & 4 \ 10 & 6 \ 7 & 8 \ 4 & 10 \end{matrix}$

Give the definition for .

Answer

By the two-point formula, the equation of a line through points $(x_{1}, y_{1}), (x_{2}, y_{2})$ is

$y-y_{1} = m(x- x_{1})$ ,

where $m = \frac{x_{2}-x_{1}}{y_{2}-y_{1}}$ , the slope of the line. The choice of the two points from the given four is arbitrary, so we will select the middle two ordered pairs. Substituting $x_{1}= 10,y_{1} =6, x_{2}= 7, y_{2}= 8$ , then

$m = \frac{x_{2}-x_{1}}{y_{2}-y_{1}}= \frac{7-10}{8-6} = -\frac{3}{2}$

The first formula becomes

$y-10 = -\frac{3}{2}(x-6)$

Since , solve for . First, distribute $-\frac{3}{2}$ throughout the difference at right:

$y-10 = -\frac{3}{2} x- \left (-\frac{3}{2} \right )6$

$y-10 = -\frac{3}{2} x- \left (-9 \right )$

$y-10 = -\frac{3}{2} x+9$

Isolate by adding 10 to both sides:

$y-10+ 10 = -\frac{3}{2} x+9+ 10$

$y = -\frac{3}{2} x+19$

Replacing, the definition of is

$f(x) = -\frac{3}{2} x+19$ .

Compare your answer with the correct one above

Question

is a linear function defined on the set of real numbers. Four values of are given in the following table:

$\begin{matrix} \underline{x}& \underline{f(x)} \ 13 & 4 \ 10 & 6 \ 7 & 8 \ 4 & 10 \end{matrix}$

Give the definition for .

Answer

By the two-point formula, the equation of a line through points $(x_{1}, y_{1}), (x_{2}, y_{2})$ is

$y-y_{1} = m(x- x_{1})$ ,

where $m = \frac{x_{2}-x_{1}}{y_{2}-y_{1}}$ , the slope of the line. The choice of the two points from the given four is arbitrary, so we will select the middle two ordered pairs. Substituting $x_{1}= 10,y_{1} =6, x_{2}= 7, y_{2}= 8$ , then

$m = \frac{x_{2}-x_{1}}{y_{2}-y_{1}}= \frac{7-10}{8-6} = -\frac{3}{2}$

The first formula becomes

$y-10 = -\frac{3}{2}(x-6)$

Since , solve for . First, distribute $-\frac{3}{2}$ throughout the difference at right:

$y-10 = -\frac{3}{2} x- \left (-\frac{3}{2} \right )6$

$y-10 = -\frac{3}{2} x- \left (-9 \right )$

$y-10 = -\frac{3}{2} x+9$

Isolate by adding 10 to both sides:

$y-10+ 10 = -\frac{3}{2} x+9+ 10$

$y = -\frac{3}{2} x+19$

Replacing, the definition of is

$f(x) = -\frac{3}{2} x+19$ .

Compare your answer with the correct one above

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