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The equation of a linear function in point-slope form is y – y1 = m(x – x1). Harold correctly wrote the equation y = 3(x – 7) using a point and the slope. Which point did Harold use?

When Harold wrote his equation, the point he used was (7, 3).
When Harold wrote his equation, the point he used was (0, 7).
When Harold wrote his equation, the point he used was (7, 0).
When Harold wrote his equation, the point he used was (3, 7).

Respuesta :

I believe answer is 7,0/ hope this helps

Answer:

When Harold wrote his equation, the point he used was (7,0).

Step-by-step explanation:

The equation of a linear function in point-slope form :

[tex]y-y_1=m(x-x_1)[/tex]  --A

Where m is the slope

[tex](x_1,y_1)[/tex] are passing points

Harold's Equation : [tex]y = 3(x – 7)[/tex]

On comparing Harold's Equation with A

[tex]x_1=7[/tex]

[tex]y_1=0[/tex]

Thus the passing point is (7,0).

Hence Option C is correct.

When Harold wrote his equation, the point he used was (7,0).

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