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How do you write a line parallel to 2y=5x+10 and goes through point (6,12)?

User Lemex
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To write the equation of a line parallel to 2y=5x+10 and goes through point (6,12), you first need to find the slope of the given line. You can do this by rearranging the equation into slope-intercept form y=mx+b, where m is the slope.

Rearranging 2y=5x+10 gives y=(5/2)x+5. So the slope of the given line is 5/2. Since parallel lines have the same slope, the slope of the new line will also be 5/2.

Next, you can use point-slope form to find the equation of the new line. The point-slope form is y-y1=m(x-x1), where (x1,y1) is a point on the line and m is the slope. Plugging in the point (6,12) and the slope 5/2 gives y-12=(5/2)(x-6).

Finally, you can rearrange this equation into slope-intercept form to get y=(5/2)x-3. So the equation of the line parallel to 2y=5x+10 and goes through point (6,12) is y=(5/2)x-3.

User Kothar
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