Write an equation in slope-intercept form for the line parallel to y=5*x-2 that passes through the point (6,-1).

1 answer:

5 0

Y=5x-31

You can use point slope form y-y1=m(x-x1), plug in the point (6,-1) to it, then solve and make it slope-intercept form

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$\dfrac{5}{x^3-12x^2+47x-60}$

Step-by-step explanation:

Though it is not what you have written, we think you want to simplify ...

$\dfrac{2(a-3)}{(a-4)(a-5)}+\dfrac{(a-1)}{(3-a)(a-4)}+\dfrac{(a-2)}{(5-a)(a-3)}\\\\=\dfrac{2(a-3)^2}{(a-3)(a-4)(a-5)}+\dfrac{-(a-1)(a-5)}{(a-3)(a-4)(a-5)}+\dfrac{-(a-2)(a-4)}{(a-3)(a-4)(a-5)}\\\\=\dfrac{2(a^2-6a+9)-(a^2-6a+5)-(a^2-6a+8)}{x^3-12x^2+47x-60}\\\\=\boxed{\dfrac{5}{x^3-12x^2+47x-60}}$

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