


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
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

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If you're having problems understanding this answer, try seeing it through your browser: brainly.com/question/2150237
Tags: <em>trigonometry trig function cosecant csc double angle identity geometry</em>
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Answer:
-3/7 > b
Step-by-step explanation:
2/7> b+5/7
Subtract 5/7 from each side
2/7-5/7> b+5/7-5/7
-3/7 > b
Answer:
x = 0
y = -12
Step-by-step explanation:
Plug in what y is equal to into the second equation
-3(x - 12) = 2x + 36
Distribute the -3
-3x + 36 = 2x + 36
-36 -36
-3x = 2x
They cannot be equal so x = 0
Plug in 0 into the y equation
y = 0 - 12
y = -12