Find the equation of the linear function represented by the table below in slope-intercept form.

Mathematics

1 answer:

Ilya [14]3 years ago

7 0

Answer:

the desired equation is y = x + 4.

Step-by-step explanation:

Slope is defined as m = rise over run. Run is the change (usually an increase) in x, and rise an increase or decrease in y.

We see, in the table, that if x increases from 1 to 6 (a 'run' of 5), y increases from 5 to 10 (a 'rise' of 5). Thus, it's immediately apparent that the slope is m = rise / run = 5 / 5, or just 1.

Using the slope-intercept form of the equation of a straight line, y = mx + b, and the point (1, 5), we calculate b:

5 = 1(1) + b, or b = 4.

Therefore the desired equation is y = x + 4.

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Which is the approximate solution to the system y = 0.5x + 3.5 and y = − 2/3 x + 1/3 shown on the graph? (–2.7, 2.1) (–2.1, 2.7)

AlekseyPX

Answer:

The approximate solution to the system is $(-2.7, 2.1)$ .

Step-by-step explanation:

To solve the system of equations $\begin{bmatrix}y=0.5x+3.5\\ y=-\frac{2}{3}x+\frac{1}{3}\end{bmatrix}$ you must:

$\mathrm{Rationalize\:equations}\\\\\begin{bmatrix}y=\left(\frac{1}{2}\right)x+\left(\frac{7}{2}\right)\\ y=-\frac{2}{3}x+\frac{1}{3}\end{bmatrix}$

$\mathrm{Subsititute\:}y=-\frac{2}{3}x+\frac{1}{3}\\\\\begin{bmatrix}-\frac{2}{3}x+\frac{1}{3}=\frac{1}{2}x+\frac{7}{2}\end{bmatrix}$

$\mathrm{Isolate}\:x\:\mathrm{for}\:-\frac{2}{3}x+\frac{1}{3}=\frac{1}{2}x+\frac{7}{2}\\\\-\frac{2}{3}x=\frac{19}{6}+\frac{1}{2}x\\\\-\frac{7}{6}x=\frac{19}{6}\\\\6\left(-\frac{7}{6}x\right)=\frac{19\cdot \:6}{6}\\\\-7x=19\\\\x=-\frac{19}{7}\approx-2.7$

$\mathrm{For\:}y=-\frac{2}{3}x+\frac{1}{3}\\\\\mathrm{Subsititute\:}x=-\frac{19}{7}\\\\y=-\frac{2}{3}\left(-\frac{19}{7}\right)+\frac{1}{3}\\\\y=\frac{15}{7}\approx 2.1$