Triangle JKL was rotated 90° clockwise about the origin. Triangle J′K′L′ is graphed below. What were the coordinates of the pre-image point J?

The coordinates of the triangle JKL are (1,2),(4,2) and (3,5)

We need to determine the coordinates of the triangle J'K'L' rotated 360° clockwise with the origin.

Coordinates of the triangle J'K'L':

The rule to rotate the triangle 360° clockwise with the origin is given by

(x,y) => (x,y)

The coordinates of the point J' is given by

(1,2) => (1,2)

The coordinates of the point K' is given by

(4,2) => (4,2)

The coordinates of the point L' is given by

(3,5) => (3,5)

Therefore, the coordinates of the triangle J'K'L' are (1,2), (4,2) and (3,5)

step-by-step explanation:

there is a formula i can give you:

it is r^2*n*sin(360/n) divided by 2 where r represents the "radius" and n is the number of sides.

plug them in:

(10sqrt(3))^2*6*sin(360/6) divided by 2

100*3*6*sin(60) divided by 2

900*sin(60)

900*sqrt(3)/2

450*sqrt(3)

approx is 779.4 cm^2

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)