Solve the inequality. Graph the solution

1.  y > -3/5 and y < 7/5        Direction to  graph 1 below.

2. r > 1 and r < 13/3              Direction to  graph 2 below

Step-by-step explanation:

|5y - 2| + 5 > 0

Case 1    Positive ||                                 Case 2  Negative ||

5y - 2 + 5 > 0                                       -(5y - 2) + 5 > 0

5y +3 > 0                                             -5y + 2 +5 > 0

5y > -3                                                  -5y + 7 > 0

y .> -3/5                                               -5y > -7

                                                             -y > -7/5

                                                             y < 7/5   y value negative, change sign

The graph 1: open circle at -3/5;  close circle at 7/5; shade between these two numbers.  

2.    |8-3r|< 5

       Case I                                           Case II

8 - 3r < 5                                                   -(8-3r) < 5

-3r < 5 - 8                                                   -8 + 3r < 5

-3r < -3                                                         3r < 5 + 8  

-r < -3/3                                                        3r <  13

r >1       r is negative, change                        r < 13/3

            the direction of the sign

Graph 2: Open circle at 1; open circle at 13/3;  shade between the two numbers.                    

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.