Resuelve las siguientes inecuaciones y exprese la respuesta en su notacion de intervalos y grafica
A:7×-8 <6
B:5×-7 <_3×+5
C:7×-3 <10×+4
D:3×-1>_5×+3
E:2×+7 <_-1
F:7×-3>4×+9
G:10×-5 <9
H:15×+×-6×<6+9+5

720

step-by-step explanation:

first i did work 2nd i did work and 3rd i did work

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.

table 3 represents a direct variation function. : )