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3 years ago

How to divide 20 by radical 3

Mathematics

1 answer:

Alex Ar [27]3 years ago

5 0

Answer:

"How do you divide a whole number by a radical?

Method 1 Dividing Radicands

Set up a fraction. If your expression is not already set up like a fraction, rewrite it this way. ...

Use one radical sign. If your problem has a square root in the numerator and denominator, you can place both radicands under one radical sign. ...

Divide the radicands. ...

Simplify, if necessary.

4 Simple and Easy Ways to Divide Square Roots - wikiHow

https://m.wikihow.com/Divide-Square-Roots "

Step-by-step explanation:

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Differential cos^2x dy/dx =e^y-tanx

gulaghasi [49]

Answer:

y=t−1+ce

−t

where t=tanx.

Given, cos

+y=tanx

⇒

+ysec

x=tanxsec

x ....(1)

Here P=sec

x⇒∫PdP=∫sec

xdx=tanx

∴I.F.=e

tanx

Multiplying (1) by I.F. we get

tanx

ysec

x=e

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tanxsec

Integrating both sides, we get

tanx

=∫e

tanx

.tanxsec

xdx

Put tanx=t⇒sec

xdx=dt

∴ye

=∫te

dt=e

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⇒y=t−1+ce

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2 years ago

Linda tossed a fair coin 6 times, and the result was heads each time. Which statement

Ivanshal [37]

Answer:

Step-by-step explanation:

Each toss is independent so the probability of getting a tail is the same thE OF GETTING A HEAD - 50%^.

6 0

1 year ago

There are 40 runners in a race. How many ways can the runners finish first, second, and third?

k0ka [10]

Answer:

117600

Step-by-step explanation:

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3 years ago

What does it mean that stocks and bonds are relatively liquid?

solong [7]

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3 years ago

Read 2 more answers

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

stealth61 [152]

Answer:

Proof in the explanation

Step-by-step explanation:

Trigonometric Equalities

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

5 0

3 years ago