KatherinePierce1864
6. If ACB is a right angle, m<1 = 4y, and m<2 = 2y + 18, find m<2. (1 point)
12º
48º
36º
42º
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lgbasallote
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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?
KatherinePierce1864
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More please, if you don't mind.
lgbasallote
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what is the value of m<1 according to the problem?
lgbasallote
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right. and m<2 is?
lgbasallote
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right. and i said m<1 + m<2 = 90
so 4y + 2y + 18 = 90
solve for y and tell me what you get
KatherinePierce1864
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So 12 is the answer?
lgbasallote
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nope. that's just y
lgbasallote
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m<2 = 2y + 18
so to solve m<2 you have to substitute y into 2y + 18
make sense?
KatherinePierce1864
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Okay so it would be 4(12) +2(12)+18...?
lgbasallote
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no...substitute y into 2y + 18 only
lgbasallote
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right
lgbasallote
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right
KatherinePierce1864
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so 42 is my answer?
lgbasallote
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yes
lgbasallote
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welcome
KatherinePierce1864
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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.
KatherinePierce1864
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or it might be DCA, I'm not sure.
lgbasallote
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|dw:1348283028935:dw|
that's how m<1 looks right?
lgbasallote
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can you identify the letter of the middle of that angle?
lgbasallote
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right. C is the middle. and what are the endpoints?
KatherinePierce1864
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So C would be inthe middle, right? so it would be DCA?
lgbasallote
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right