Evaluate 8j -k+148j−k+14 when j=0.25j=0.25 and k=1k=1.

15

Step-by-step explanation:

8j -k+14

We want to find the value when j = .25 and k=1

Substitute the values in

8(.25) -1 +14

2 -1+14

1+14

15

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.

answer: d. $20

step-by-step explanation:

just took the test