Question Write an equation of the line passing through the point A(0, 3) that is perpendicular to the line y = −1/2x−6

The equation would be y=2x+3 I am pretty sure

Step-by-step explanation:

This is because you take the slope which is 2 from the line y=-1/2x-6. To find the slope perpendicular to the line, you need to find the exact opposite of the slope: -1/2, and in this case, it would be 2.

(Ex: if it was -3/4, you would need to use a slope of 4/3 since it is the opposite)

Then you take that slope (2) and the point (0, 3) and put it into "y-y1=m(x-x1)"

y-y1=m(x-x1)

y-3=2(x-0)

-distribute the 2-

y-3=2x-0

-add 3 on both sides-

so equation would be y=2x+3

Not sure if this is all correct, but I hope it helps! :)

hello from mrbilldoesmath!

answer:   (7x + 6y) ( 49x^2 - 42xy + 36y^2)

discussion:

in general, the factorization of a^3 + b^3 is

a^3 + b^ 3 = (a + b) ( a^2 - ab + b^2)

in our case note that 343 = 7^3 and 216 = 6^3 so the equation becomes:

343x^3 + 216y^3 = (7x)^3 + (6y)^3

if you think of a (above) as 7x and b as 6y the factorization becomes

343x^3 + 216y^3 =

((7x + 6y) (   (7x)^2 - 7x*6y + (6y)^2)   =

(7x + 6y) ( 49x^2 - 42xy + 36y^2)

you,

mrb

(a + b) (a^2 - a b + b^2)