What is the area of a triangle with vertices at (−2, -3) , (6, -3) , and (2, 5) ?

Hello from MrBillDoesMath!

Answer :  32

Discussion:

Label the points as follows:

A  = (-2, -3),             B = ( 6, -3),    C = ( 2.5)

Area of a triangle = (1/2) bh

where "b" is the "base" and "h" is the height of the triangle.

In our case, the base is the line AB, so the length of the base is

    b = 6 - ( -2) =  6 + 2 = 8

(6 and -2 are the x coordinates of the points B and A, respectively)

The height is the distance from the y-coordinate of point C to the base AB. That is,

5 - (-3) = 5 + 3 = 8

(5 is the y coordinate of point C;  -3 is the y coordinate of the line AB)

Finally,

Area =   (1/2) * 8 * 8 = (1/2) 64 = 32

Thank you,

MrB

i think it's ф or x=6,-6

step-by-step explanation:

-x^2=-36

x^2=-36

ф

or

if you mean by : -(x^2)=-36

it'll be: x^2=36

x=6,x=-6

hope it'll ! have a good day!

you for asking!

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)