The sides of a square are 3 cm long. one vertex of the
square is at (2,0) on a square coordinate grid marked in
centimeter units. which of the following points could
also be a vertex of the square?
f. (−4, 0)
g. ( 0, 1)
h. ( 1,−1)
j. ( 4, 1)
k. ( 5, 0)

 The required point that could also be a vertex of the square is K(5, 0).

Step-by-step explanation:  Given that the sides of a square are 3 cm long and one vertex of the  square is at (2,0) on a square coordinate grid marked in  centimeter units.

We are to select the co-ordinates of the point that could also be a vertex of the square.

To be a vertex of the given square, the distance between the point and the vertex at (2, 0) must be 3 cm.

Now, we will be suing the distance formula to calculate the lengths of the segment from the point to the vertex (2, 0).

If the point is F(-4, 0), then the length of the line segment will be

If the point is G(0, 1), then the length of the line segment will be

If the point is H(1, -1), then the length of the line segment will be

If the point is J(4, 1), then the length of the line segment will be

If the point is K(5, 0), then the length of the line segment will be

Thus, the required point that could also be a vertex of the square is K(5, 0).

the answer is a

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)