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mel-nik [20]
11 months ago
6

Determine the x- and y-intercepts of the graph ofx−4y=8Then plot the intercepts to graph the equation.

Mathematics
1 answer:
Over [174]11 months ago
8 0

The x-intercept is the point where the graph cuts the x-axis, and the y-intercept is the point where the graph cuts the y-axis.

The x-axis is the line y = 0 and the y-axis is the line x = 0. To find the intercept between each axis and our graph, we just need to evaluate our function at x = 0 and y = 0.

Calculating the x-intercept, we have

\begin{gathered} x-4\cdot0=8 \\ x=8 \end{gathered}

The x-intercept is (8, 0).

Calculating the y-intercept, we have

\begin{gathered} 0-4y=8 \\ y=-\frac{8}{4} \\ y=-2 \end{gathered}

The y-intercept is (0, -2).

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garik1379 [7]
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On a linear X temperature scale, water freezes at −115.0°X and boils at 325.0°X. On a linear Y temperature scale, water freezes
belka [17]

Answer:

The current temperature on the X scale is 1150 °X.

Step-by-step explanation:

Let is determine first the ratio of change in X linear temperature scale to change in Y linear temperature scale:

r = \frac{\Delta T_{X}}{\Delta T_{Y}}

r = \frac{325\,^{\circ}X-(-115\,^{\circ}X)}{-25\,^{\circ}Y - (-65.00\,^{\circ}Y)}

r = 11\,\frac{^{\circ}X}{^{\circ}Y}

The difference between current temperature in Y linear scale with respect to freezing point is:

\Delta T_{Y} = 50\,^{\circ}Y - (-65\,^{\circ}Y)

\Delta T_{Y} = 115\,^{\circ}Y

The change in X linear scale is:

\Delta T_{X} = r\cdot \Delta T_{Y}

\Delta T_{X} = \left(11\,\frac{^{\circ}X}{^{\circ}Y} \right)\cdot (115\,^{\circ}Y)

\Delta T_{X} = 1265\,^{\circ}X

Lastly, the current temperature on the X scale is:

T_{X} = -115\,^{\circ}X + 1265\,^{\circ}X

T_{X} = 1150\,^{\circ}X

The current temperature on the X scale is 1150 °X.

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The expression when m=6 and n=4
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