, 25.01.2023Kimberlytdb

# Pick one of the remaining choices and explain on the lines below, why thatchoice contains two expressions that are not equivalent for any value of r.

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.

Cdoes not have two variables

If two expressions have the same coefficients of the variable, then the variable can be any number

For the first choice

3(3r + 3) = 9r + 9 === 1st expression

6r + 6 2nd expression

If we equate them

9r + 9 = 6r + 6

If we solve the equation we will find just one value of r

9r + 9 - 6r = 6r - 6r + 6

3r + 9 = 6

3r + 9 - 9 = 6 - 9

3r = -3

r = -1

But in the other choices, the two expressions have the same coefficients of r

Then when we equate them they will cancel each other

Then r can be any value

Because it does not affect the equation

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