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Hitman42 [59]
3 years ago
8

Solve for x: |2x + 12| = 18

Mathematics
2 answers:
Mashcka [7]3 years ago
7 0
2x + 12 = 18
18 - 12 = 6
2x = 6
x = 3
kumpel [21]3 years ago
7 0
So you want to drop the absolute value signs and you need to make two equations:

2x + 12 = 18 OR 2x + 12 = -18
subtract both sides by 12 for both equations
2x = 6 OR 2x = -30
divide both sides by 2 for both equations
x = 3 OR x = -15

i hope this helps!!

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<em>Hello there! And thank you for asking your question here on brainly . com! The number one asking and answering site.</em>

Answer: C. 34

Explanation: The series shown is reducing by 5 digits. 54 - 5 = 49, 49 - 5 = 44, and so on. When we're at 39, just subtract by 5. 39 - 5 = 34. 34 is the answer.

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\csc A - \sec A = \dfrac 83 + \dfrac{8}{\sqrt{55}}\\\\\csc A - \sec A = \dfrac 83 - \dfrac{8}{\sqrt{55}}

Step by step explanation:

\text{Given that,}\\\\~~~~~~\sin A = \dfrac 38 \\\\\implies \sin^2 A = \dfrac 9{64}\\\\\implies  1 - \cos^2 A = \dfrac{9}{64}\\\\\implies \cos ^2 A = 1 - \dfrac 9{64}\\\\\implies \cos^2 A = \dfrac{55}{64}\\\\\implies \cos A =\pm\sqrt{\dfrac{55}{64}}\\ \\\implies \cos A = \pm\dfrac{\sqrt{55}}8\\\\

\implies \dfrac 1{\cos A} = \pm\dfrac{8}{\sqrt{55}}

\text{Now,}\\\\\csc A - \sec A\\\\=\dfrac{1}{\sin A}- \dfrac{1}{\cos A}\\\\=\dfrac 83 -\left(\pm \dfrac 8{\sqrt {55}} \right)\\ \\\text{Hence,}\\\\\csc A - \sec A = \dfrac 83 + \dfrac{8}{\sqrt{55}}\\\\\csc A - \sec A = \dfrac 83 - \dfrac{8}{\sqrt{55}}

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