Please help with these 2 !! Thank you

Because we have been given the slope and y-intercept, we can create an equation in slope-intercept form.

y = mx + b

m = the slope

b = y-intercept.

In the first problem,

m = -4

b = 10

So, our equation looks like: y = -4x + 10

In the second problem,

m = 1/3

b = -6

So, our equation looks like: y = 1/3x - 6

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.

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step-by-step explanation: