PLS HELP ME! A ladder 25ft long leans against a vertical wall with it's foot on level ground 7ft from the base of the wall. If the foot is pulled away from the wall at the rate 2ft/s , Express the distance (yft)of the top of the ladder above the ground as a function the time t second in moving

hope it will help you.....

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.

(23*2)-12=x     x=34

step-by-step explanation: