PointQ′(prime)is the image of Q(-4,7)Q(−4,7)Q, left parenthesis, minus, 4, comma, 7, right parenthesis under a translation by 444 units to the right and 222 units down. What are the coordinates of Q'(prime)?

Hii the coordinates of Q' are (0,5). I did this in Khan Academy and got it correct!

Step-by-step explanation:

1. In order to move the point 4 units to the right, we need to add 4 the the x-coordinate:

-4 + 4 = 0

2. In order to move the points 2 units down, we need to subtract 2 from its y-coordinate:

7 - 2 = 5

3. The coordinates of Q' are (0, 5).

SO THE ANSWER IS ( 0, 5)

1,8,15,22,29,36,43,50

step-by-step explanation:

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)