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Please help me on this khan academy

Please help me on this khan academy-example-1

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Answer:

THERE YA GO BUDDY

Explanation:

To find the equation of a line that is perpendicular to y = -0.3x + 6 and passes through the point (3,-8), we can follow these steps:

1. Determine the slope of the given line. The equation y = -0.3x + 6 is in slope-intercept form (y = mx + b), where the coefficient of x (-0.3 in this case) represents the slope. So, the slope of the given line is -0.3.

2. The slope of a line perpendicular to another line is the negative reciprocal of the slope of the given line. In this case, the negative reciprocal of -0.3 is 1/0.3 or approximately 3.33 (rounded to two decimal places).

3. Now that we have the slope of the perpendicular line, we can use the point-slope form of a linear equation, which is y - y1 = m(x - x1). Plug in the values for the slope (m = 3.33) and the coordinates of the given point (x1 = 3, y1 = -8) to get the equation.

y - (-8) = 3.33(x - 3)

4. Simplify the equation.

y + 8 = 3.33x - 9.99

5. Rearrange the equation in slope-intercept form (y = mx + b) to get the final equation.

y = 3.33x - 17.99

Therefore, the equation of the line that is perpendicular to y = -0.3x + 6 and passes through the point (3,-8) is y = 3.33x - 17.99.

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