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4 years ago

What is equivalent to the expression 4 / -3 to the square root of 64

Mathematics

1 answer:

satela [25.4K]4 years ago

6 0

$\sqrt{a}=b\iff b^2=a\ for\ a\geq0\ and\ b\geq0\\\\\dfrac{4}{-3\sqrt{64}}=\dfrac{\not4^1}{-3(\not8_2)}=-\dfrac{1}{6}$

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Read 2 more answers

Given two linear equations: y=3x+3 and y=-3x-5. Select all or one that apply.

sergejj [24]

Answer:

Step-by-step explanation:

We have the two linear equations:

$\text{ Line 1: }y=3x+3$

$\text{ Line 2: }y=-3x-5$

And we want to determine the relationship between them.

Let's go through each of the answer choices.

Parallel?

Remember that parallel lines have the same slopes.

The slope of Line 1 is 3, while the slope is Line 2 is -3.

Since they have different slopes, they are not parallel.

Perpendicular?

Perpendicular lines have slopes who are negative reciprocals of each other.

For example, the negative reciprocal of 2 will be -1/2. And the negative reciprocal of -1/2 is 2.

The slope of Line 1 is 3. The negative reciprocal of 3 is -1/3. This is <em>not</em> the slope of Line 2.

We can do it for Line 2 just to confirm. The slope of Line 2 is -3. The negative reciprocal of -3 is 1/3.

So, because their slopes are not negative reciprocals of each other, the two lines are not congruent.

Both?

No two lines can be both parallel and perpendicular simultaneously, so this is out of the question.

[N]either?

Yes indeed. They lines are neither parallel nor perpendicular, so this is our only choice.

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4 years ago

Find the exact value of the trigonometric expression.

Dmitrij [34]

$\bf \textit{Half-Angle Identities}
\\\\
sin\left(\cfrac{\theta}{2}\right)=\pm \sqrt{\cfrac{1-cos(\theta)}{2}}
\qquad 
cos\left(\cfrac{\theta}{2}\right)=\pm \sqrt{\cfrac{1+cos(\theta)}{2}}\\\\
-------------------------------\\\\
\cfrac{1}{12}\cdot 2\implies \cfrac{1}{6}\qquad therefore\qquad \cfrac{\pi }{12}\cdot 2\implies \cfrac{\pi }{6}~thus~ \cfrac{\quad \frac{\pi }{6}\quad }{2}\implies \cfrac{\pi }{12}$

$\bf sec\left( \frac{\pi }{12} \right)\implies sec\left( \cfrac{\frac{\pi }{6}}{2} \right)\implies \cfrac{1}{cos\left( \frac{\frac{\pi }{6}}{2} \right)}\impliedby \textit{now, let's do the bottom}
\\\\\\
cos\left( \cfrac{\frac{\pi }{6}}{2} \right)=\pm\sqrt{\cfrac{1+cos\left( \frac{\pi }{6} \right)}{2}}\implies \pm\sqrt{\cfrac{1+\frac{\sqrt{3}}{2}}{2}}\implies \pm\sqrt{\cfrac{\frac{2+\sqrt{3}}{2}}{2}}$

$\bf \pm \sqrt{\cfrac{2+\sqrt{3}}{4}}\implies \pm\cfrac{\sqrt{2+\sqrt{3}}}{\sqrt{4}}\implies \pm \cfrac{\sqrt{2+\sqrt{3}}}{2}
\\\\\\
therefore\qquad \cfrac{1}{cos\left( \frac{\frac{\pi }{6}}{2} \right)}\implies \cfrac{2}{\sqrt{2+\sqrt{3}}}$

which simplifies thus far to

$\bf \cfrac{2}{\sqrt{2+\sqrt{3}}}\cdot \cfrac{\sqrt{2+\sqrt{3}}}{\sqrt{2+\sqrt{3}}}\implies \cfrac{2\sqrt{2+\sqrt{3}}}{2+\sqrt{3}}\implies \cfrac{2\sqrt{2+\sqrt{3}}}{2+\sqrt{3}}\cdot \stackrel{conjugate}{\cfrac{2-\sqrt{3}}{2-\sqrt{3}}}
\\\\\\
\cfrac{2\sqrt{2+\sqrt{3}}(2-\sqrt{3})}{2^2-(\sqrt{3})^2}\implies \cfrac{2\sqrt{2+\sqrt{3}}(2-\sqrt{3})}{1}$

$\bf 2\sqrt{2+\sqrt{3}}(2-\sqrt{3})\impliedby \stackrel{\textit{keep in mind that}}{(2-\sqrt{3})=(\sqrt{2-\sqrt{3}})(\sqrt{2-\sqrt{3}})}
\\\\\\
2\sqrt{2+\sqrt{3}}\cdot \sqrt{2-\sqrt{3}}\cdot \sqrt{2-\sqrt{3}}
\\\\\\
2\sqrt{(2+\sqrt{3})(2-\sqrt{3})\cdot (2-\sqrt{3})}
\\\\\\
2\sqrt{[2^2-(\sqrt{3})^2]\cdot (2-\sqrt{3})}\implies 2\sqrt{1(2-\sqrt{3})}
\\\\\\
2\sqrt{2-\sqrt{3}}$

7 0

4 years ago

Use the equation n= b-4 to find the value of n when b=9

Valentin [98]

Put 9 in for b

N=9-4

N=5

6 0

3 years ago