If you are given the graph of h(x)=log9x, how could you graph j(x)=log9(x−2)? Translate each point of the graph of h(x) 2 units right.
Translate each point of the graph of h(x) 2 units up.
Translate each point of the graph of h(x) 2 units down.
Translate each point of the graph of h(x) 2 units left.

d. bc = 12 and ef = 16

step-by-step explanation:

we are given that δabc : δdef such that .

also, it is given that

so, from the options, we need that .

a.

b.

c.  

d.

as, we can see that only option d gives the same ratio as i.e. .

hence, the possible lengths are bc = 12 and ef = 16.

b) k’(6, 4); l’(3, 3)

step-by-step explanation:

a 270° clockwise rotation maps every point (x, y) to (y, -x). (it negates the x-coordinate and then switches it and the y-coordinate.)

this means the endpoints of the side congruent to kl, k' and l', would be mapped as:

k'(-4, 6)→(6, 4)

l'(-3, 3)→(3, 3)

the one below it is a function

b

step-by-step explanation: