What is this I don’t know lol help me

2 tan 30°<br>II<br>1 + tan- 300

shusha [124]

Question:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$

Answer:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= sin(60^{\circ})$

Step-by-step explanation:

Given

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$

Required

Simplify

In trigonometry:

$tan(30^{\circ}) = \frac{1}{\sqrt{3}}$

So, the expression becomes:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2 * \frac{1}{\sqrt{3}}}{1 + (\frac{1}{\sqrt{3}})^2}$

Simplify the denominator

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2 * \frac{1}{\sqrt{3}}}{1 + \frac{1}{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\frac{2}{\sqrt{3}}}{1 + \frac{1}{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\frac{2}{\sqrt{3}}}{ \frac{3+1}{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\frac{2}{\sqrt{3}}}{ \frac{4}{3}}$

Express the fraction as:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2}{\sqrt 3} / \frac{4}{3}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2}{\sqrt 3} * \frac{3}{4}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{1}{\sqrt 3} * \frac{3}{2}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{3}{2\sqrt 3}$

Rationalize

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{3}{2\sqrt 3} * \frac{\sqrt{3}}{\sqrt{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{3\sqrt{3}}{2* 3}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\sqrt{3}}{2}$

In trigonometry:

$sin(60^{\circ}) = \frac{\sqrt{3}}{2}$

Hence:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= sin(60^{\circ})$

3 0

3 years ago

If the Circumference of a circle is 12 cm, what is the circles Diameter? Please help thankss!

BARSIC [14]

Answer:

Step-by-step explanation:

Circumference = πd = 12

d = 12/π cm

5 0

3 years ago

F(a) = -4a + 3<br> gla)=3a +4<br> Find f(3) · g(3)

sladkih [1.3K]

F(3)=-4(3)+3=-9
g(3)=3(3)+4=13
f(3).g(3)=-9*13=-117

8 0

3 years ago

Pls help!!!!!!!!!!!!

nikklg [1K]

The measure of angle R is 50° because x=3.

7 0

3 years ago

at first the average mass some durians was 3.4kg.when 2 durians were added the aevrage mass of the 2 durians was 1.3kg was added

LekaFEV [45]

We conclude that the initial number of durians is 12.

<h3>How many durians were there at first?</h3>

Let's say that initially, we had N durians, such that these N durians have an average mass of 3.4kg.

When 2 durians (with an average mass of 1.3kg) more are added, the new average mass is given by:

(N*3.4kg + 2*1.3kg)/(N + 2)

And we know that the above amount is equal to 3.1kg, then we can write:

(N*3.4kg + 2*1.3kg)/(N + 2) = 3.1kg

Now we can solve this equation for N:

(N*3.4kg + 2.6kg) = 3.1kg*(N + 2)

N*(3.4kg - 3.1kg) = 6.2kg - 2.6kg

N*(0.3kg) = 3.6kg

N = (3.6kg)/(0.3kg) = 12

We conclude that the initial number of durians is 12.

If you want to learn more about mass:

brainly.com/question/1354972

#SPJ1

4 0

2 years ago