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3 years ago

Solve for a.

Mathematics

2 answers:

vesna_86 [32]3 years ago

8 0

Answer:

C) all real numbers

Step-by-step explanation:

4(a + 3) = 12 + 4a

open the parenthesis

4*a + 3*4 = 12 + 4a

4a + 12 = 12 + 4a

subtract 4a from both sides of the equation

12 = 12

since 12 = 12 regardless of what x is the solution is all real numbers

arlik [135]3 years ago

3 0

Answer:

All real numbers

Step-by-step explanation:

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DaniilM [7]

Answer:

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Step-by-step explanation:

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mixer [17]

Answer:

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Step-by-step explanation:

Given:

$\sin (x-90^{\circ})+\cos(x+270^{\circ})=-1$

First, note that

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So, the equation is

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Multiply this equation by $\frac{\sqrt{2}}{2}:$

$-\dfrac{\sqrt{2}}{2}\cos x+\dfrac{\sqrt{2}}{2}\sin x= -\dfrac{\sqrt{2}}{2}\\ \\\dfrac{\sqrt{2}}{2}\cos x-\dfrac{\sqrt{2}}{2}\sin x=\dfrac{\sqrt{2}}{2}\\ \\\cos 45^{\circ}\cos x-\sin 45^{\circ}\sin x=\dfrac{\sqrt{2}}{2}\\ \\\cos (x+45^{\circ})=\dfrac{\sqrt{2}}{2}$

The general solution is

$x+45^{\circ}=\pm \arccos \left(\dfrac{\sqrt{2}}{2}\right)+2\pi k,\ \ k\in Z\\ \\x+\dfrac{\pi }{4}=\pm \dfrac{\pi }{4}+2\pi k,\ \ k\in Z\\ \\x=-\dfrac{\pi }{4}\pm \dfrac{\pi }{4}+2\pi k,\ \ k\in Z\\ \\\left[\begin{array}{l}x=2\pi k,\ \ k\in Z\\ \\x=-\dfrac{\pi }{2}+2\pi k,\ k\in Z\end{array}\right.$

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