The point J(-2, 1) is rotated 270° counterclockwise around the origin. What are the coordinates of the resulting point, J′?

The coordinates of the point J' is (1,2)

Explanation:

Given that the point J(-2,1) is rotated about 270° about the origin.

Coordinates of the resulting point J':

Since, we know that, when rotating a point 270° counterclockwise about the origin, the point (x,y) becomes (y,-x) which means we switch the places of x and y coordinates and make x as negative.

Therefore, the transformation is given by

Using the above transformation, we have,

(-2,1) ⇒ (1,-(-2)) ⇒ (1,2)

Therefore, the resulting point of the coordinates of the point J' is (1,2)

(1,2)

Step-by-step explanation:

(1,2)

270° anticlockwise rotation is the same as 90° clockwise rotation

Which maps:

(x,y) ---> (y,-x)

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)