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3 years ago

What is -4(2x+3)=5-(8x+1)

Mathematics

1 answer:

coldgirl [10]3 years ago

4 0

$Solution, solve\:for\:x,\:-4\left(2x+3\right)=5-\left(8x+1\right)\quad :\quad \mathrm{No\:Solution}$

$Steps:$

$\mathrm{Expand\:}-4\left(2x+3\right),\\-4\left(2x+3\right),\\\mathrm{Apply\:the\:distributive\:law}:\quad \:a\left(b+c\right)=ab+ac,\\a=-4,\:b=2x,\:c=3,\\-4\cdot \:2x+\left(-4\right)\cdot \:3,\\\mathrm{Apply\:minus-plus\:rules},\\+\left(-a\right)=-a,\\-4\cdot \:2x-4\cdot \:3,\\\mathrm{Simplify}\:-4\cdot \:2x-4\cdot \:3,\\-8x-12$

$\mathrm{Expand\:}5-\left(8x+1\right),\\5-\left(8x+1\right),\\-\left(8x+1\right),\\5-8x-1,\\\mathrm{Simplify}\:5-8x-1,\\-8x+4$

$-8x-12=-8x+4$

$\mathrm{Add\:}12\mathrm{\:to\:both\:sides},\\-8x-12+12=-8x+4+12$

$\mathrm{Simplify}, \\-8x=-8x+16$

$\mathrm{Add\:}8x\mathrm{\:to\:both\:sides},\\-8x+8x=-8x+16+8x$

$\mathrm{Simplify},\\0=16$

$\mathrm{The\:sides\:are\:not\:equal}$

$\mathrm{The\:Correct\:Answer\:is\:\mathrm{No\:Solution}}$

$\mathrm{Hope\:This\:Helps!!!}$

$\mathrm{-Austint1414}$

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Step-by-step explanation:

The given trigonometric equation is $\cos(-\theta)=\frac{\sqrt{3} }{3}$ .

We can either use the Pythagorean identity or the right angle triangle to solve for $\sin\theta$ .

According to the Pythagorean identity,

$\cos^2\theta+\sin^2\theta=1$

Recall that, the cosine function is an even function, therefore

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$\Rightarrow \cos(\theta)=\frac{\sqrt{3} }{3}$ .

We substitute this value in to the above Pythagorean identity to get;

$(\frac{\sqrt{3}}{3})^2+\sin^2\theta=1$

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$\Rightarrow \sin^2\theta=1-\frac{3}{9}$

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