Find the equation of the line that contains the point ( -3, -1 ) and is parallel to the line 6x+5y=11 write the equation in slope intercept form

To find the equation of the line, you must first of all find the slope in the given equation,how do you do that? You just need to make y the subject of the formula and then compare the resulting equation to y = mx + c, knowing fully well that m represent the slope.

i.e from the equation 5y = 11 - 6x

                                      y = 11/5 - 6x/5

i,e the slope is -6/5

Note: for two lines to be parallel, they must have the same slope

i.e m1 = m2

using the formula y - y1 = m ( x - x 1) to find the equation of the line

it implies : y - (-1) = -6/5 ( x - {-3})

                 y + 1 = -6/5 ( x +3)

multiply through by 5

5 ( y+1) = -6 ( x +3)

5y +5 = -6x - 18

      5y = -6x -18-5

       5y = -6x -23

      To write it in slope intercept form, divide through by 5

        y = -6x/5  - 23/5

Step-by-step explanation:

f(x) = x^3 -2x^2 -3x

x( x^2 -2x-3)

x( x^2 +x -3x -3)

x( x(x+1) -3(x+1))

x( x-3)( x+1)

so zeroes are 0, 3,-1

its (2) option

only 10/60 isn't equivalent

hope this : )