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6. If ACB is a right angle, m<1 = 4y, and m<2 = 2y + 18, find m<2. (1 point) 12º 48º 36º 42º

Mathematics
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here's a hint.. \[\angle ACB = m\angle 1 + m\angle 2 = 90\] does that help? or do you need more help?
More please, if you don't mind.

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Other answers:

what is the value of m<1 according to the problem?
4y?
right. and m<2 is?
2y+18
right. and i said m<1 + m<2 = 90 so 4y + 2y + 18 = 90 solve for y and tell me what you get
y=12?
So 12 is the answer?
nope. that's just y
m<2 = 2y + 18 so to solve m<2 you have to substitute y into 2y + 18 make sense?
Okay so it would be 4(12) +2(12)+18...?
no...substitute y into 2y + 18 only
so 2(12)+18
right
which equals 42?
right
so 42 is my answer?
yes
Thank you!! :)
welcome
Oh I have one more question, that goes to the same photo above. 4. What is another name for <1? C DCA ACB CDA I choose CDA as my answer but I'm not sure if that's right.
or it might be DCA, I'm not sure.
|dw:1348283028935:dw| that's how m<1 looks right?
Yes
can you identify the letter of the middle of that angle?
C?
right. C is the middle. and what are the endpoints?
A and D
So C would be inthe middle, right? so it would be DCA?
right
Thanks!

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