Y + R + V = 42
but 4 YEARS AGO : (R-4) +(V-4) = Y
Replace Y in the 1st equation by: (R-4) +(V-4)
(R-4) +(V-4) + R + V = 42
2R - 2V - 8 = 42; 2R -2V = 50, or R+V=25
IF R+V =25 AND Y + R +V = 42, then Y + 25 = 42 and Y =17
The range is the set of all y-coordinates.
R = {2, 6, 8}
Answer:
12
Step-by-step explanation:
Answer: A. "Segment AD bisects angle CAB." is the right answer.
Step-by-step explanation:
Given : In ΔABC ,AC≅AB.
⇒∠ACB=∠CBA....(1) (∵ angles opposite to equal sides of a triangle are equal )
Now in ΔACD and ΔABD
AD=AD (common)....(2)
Here we need one more statement to prove the triangles congruent that is only statement (A) fits in it.
If AD bisects ∠CAB then ∠CAD=∠BAD..(3)
Now again Now in ΔACD and ΔABD
∠ACB=∠CBA [from (1)]
AD=AD [common]
∠CAD=∠BAD [from (3)]
So by ASA congruency criteria ΔADC≅ΔABD.
X is left and right
hmm
see the aproximate slope
see endpoints
ok, so from about (0,3) to (35,30)
slpoe is 27/35
so
y=(27/35)x+3 about
so when x=
ok, so according to the garph, when x=35, y=30, not 20
ok, so when x=40
y=(27/35)x+3
y=(27/35)*40+3
y=33.8571 about for x=40
