Type the correct answer in each box. use numerals instead of words.
you reflect triangle pqr, with coordinates p(-4, -4), q(-1, -3), and r(-3, -1), across the x-axis, across the y-axis, and across the x-axis again to form triangle p′q′r′.

after these reflections, the coordinates of p′ will be (,

After these reflections, the coordinates of P′ will be (4 , -4)

Step-by-step explanation:

* Lets revise some transformation

- If point (x , y) reflected across the x-axis

∴ Its image is (x , -y)

- If point (x , y) reflected across the y-axis

∴ Its image is (-x , y)

* Lets solve the problem

- The triangle PQR with coordinates P(-4, -4), Q(-1, -3), and R(-3, -1)

- The triangle is reflected across the x-axis

∵ Δ PQR is reflected across the x-axis

∴ All y-coordinates of the vertices P, Q , R reversed their signs

∴ The new points will be (-4 , 4) , (-1 , 3) , (-3 , 1)

- The new vertices will reflected across the y-axis

∴ All x-coordinates of the new vertices reversed their signs

∴ The new points will be (4 , 4) , (1 , 3) , (3 , 1)

- The new vertices will reflected across the x-axis to form Δ P'Q'R'

∴ All y-coordinates of the new vertices reversed their signs

∴ P' = (4 , -4) , Q' = (1 , -3) , R' = (3 , -1)

* After these reflections, the coordinates of P′ will be (4 , -4)

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.

3 is greater then both of those numbers