Answer :

Supernova

Answer:

[tex]\fbox{y = -3x + 5}[/tex]

Step-by-step explanation:

You are given the slope and the y-intercept of the line; so you can substitute these values into slope-intercept form: [tex]y=mx+b[/tex];

  • where [tex]m= \ $slope = -3[/tex]
  • and [tex]y-$int = 5[/tex]

Plugging these values into slope-intercept form gives:

  • [tex]y=(-3)x+(5)[/tex]
  • [tex]y=-3x+5[/tex]

Another way to find the slope-intercept form of a line given the slope and a point:

We are given the slope and a point that the line passes through, so we can use the point-slope equation to find the slope-intercept form of the line.

The point that the line passes through is the y-intercept: [tex](0,5)[/tex].

Point-slope form:

  • [tex]y-y_1=m(x-x_1)[/tex]

where [tex](x_1, \ y_1)[/tex] are the coordinates of the point that the line passes through and [tex]m= $ slope of the line.[/tex]

Substitute [tex]m=-3[/tex] and [tex](0,5)[/tex] into the point-slope form equation.

  • [tex]y-(5)=-3(x-(0))[/tex]

Simplify the equation on both sides.

  • [tex]y-5=-3(x)[/tex]

Add 5 to both sides of the equation.

  • [tex]y=-3x+5[/tex]

This is in slope-intercept form: [tex]y=mx+b[/tex], so we are done.

The answer is [tex]D) \ $y = -3x+5[/tex].

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