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2 years ago

5x+3

Mathematics

1 answer:

Leya [2.2K]2 years ago

4 0

To write this sum as a fraction, you can use the following formula: 5x/4 + 3/4.

<h3>How to transform a fraction into a sum?</h3>

To transform a fraction into a sum we must perform the following procedure:

$\frac{5x + 3}{4}$ = $\frac{5x}{4} + \frac{3}{4}$

According to the above, all we have to do is divide the fraction using each digit of the numerator and put the same denominator.

Learn more about fractions in: brainly.com/question/10354322

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A paint box contains 12 bottles of different colors. If we choose equal quantities of 3 different colors at random, how many col

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Step-by-step explanation:

You are choosing 3 from a total of 12. Order does not matter. So you are working with combinations. The answer symbolically is

12C3

12C3 = 12!/(9!3!)

12C3 = 12 * 11 * 10 * 9!/(9! 3!)

12C3 = 12 * 11 * 10/6

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Which figures demonstrate a single reflection?<br><br> Select each correct answer.

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Please see the attached image below, to find more information about the graph

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2 years ago

Help! Prove the equality<br><br>arccos √(2/3) - arccos (1+√6)/(2*√3) = π/6

stealth61 [152]

Answer:

<em>Proof in the explanation</em>

Step-by-step explanation:

<u>Trigonometric Equalities</u>

Those are expressions involving trigonometric functions which must be proven, generally working on only one side of the equality

For this particular equality, we'll use the following equation

$\displaystyle cos(x-y)=cos\ x\ cos\ y+sin\ x\ sin\ y$

The equality we want to prove is

$\displaystyle arccos\ \sqrt{\frac{2}{3}}-arccos\left(\frac{1+\sqrt{6}}{2\sqrt{3}}\right)=\frac{\pi}{6}$

Let's set the following variables:

$\displaystyle x=arccos\ \sqrt{\frac{2}{3}},\ y=arccos(\frac{1+\sqrt{6}}{2\sqrt{3}})$

And modify the first variable:

$\displaystyle x=arccos\ \frac{\sqrt{6}}{3}}=>\ cos\ x= \frac{\sqrt{6}}{3}}$

Now with the second variable

$\displaystyle y=arccos\ \frac{1+\sqrt{6}}{2\sqrt{3}}=>cos\ y=\frac{1+\sqrt{6}}{2\sqrt{3}}=\frac{\sqrt{3}+3\sqrt{2}}{6}$

Knowing that

$sin^2x+cos^2x=1$

We compute the other two trigonometric functions of X and Y

$\displaystyle sin \ x=\sqrt{1-cos^2\ x}=\sqrt{1-(\frac{\sqrt{6}}{3})^2}=\sqrt{1-\frac{6}{9}}=\frac{\sqrt{3}}{3}$

$\displaystyle sin\ y=\sqrt{1-cos^2y}=\sqrt{1-\frac{(\sqrt{3}+3\sqrt{2})^2}{36}}}$

$\displaystyle sin\ y=\sqrt{\frac{36-(3+6\sqrt{6}+18)}{36}}=\sqrt{\frac{15-6\sqrt{6}}{36}}$

Computing

$15-6\sqrt{6}=(3-\sqrt{6})^2$

Then

$\displaystyle sin\ y=\frac{3-\sqrt{6}}{6}$

Now we replace all in the first equality:

$\displaystyle cos(x-y)=\frac{\sqrt{6}}{3}.\frac{\sqrt{3}+3\sqrt{2}}{6}+\frac{\sqrt{3}}{3}.\frac{3-\sqrt{6}}{6}$

$\displaystyle cos(x-y)=\frac{3\sqrt{2}+6\sqrt{3}}{18}+\frac{3\sqrt{3}-3\sqrt{2}}{18}$

$\displaystyle cos(x-y)=\frac{9\sqrt{3}}{18}=\frac{\sqrt{3}}{2}=cos\ \pi/6$

Thus, proven

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