Lesson 4 Solving One Variable Equation 2. What value makes the equation 3b + 15 = – 26 – 20 true?

b = -³¹⁄₃

Step-by-step explanation:

We can solve this Algebra equation by separating the variable and constants.

3b + 15 = -26 - 20

3b + 15 = -46

3b = -31

b = -³¹⁄₃

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step-by-step explanation:

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.