Ask question

Ask question

All categories

Y_Kistochka [10]

2 years ago

8

Tan 45°. tan² 30° tan 60° slove it

1 answer:

makvit [3.9K]2 years ago

4 0

Answer:

$\frac{\sqrt{3} }{3} or \frac{1}{\sqrt{3} }$

Step-by-step explanation:

We can break the problem into three parts,

tan45 = sin(45)/cos(45)= $\frac{\frac{\sqrt{2}}{2} }{\frac{\sqrt{2}}{2} } =1$

$tan^{2}30 =(tan30))^{2}=(sin30/cos30)^2=((1/2)/(\sqrt{3}/2))^2 =(1/\sqrt{3})^2$

$tan60=sin60/cos60=\frac{\sqrt{3}/2 }{1/2} =\sqrt{3}$

Then, $tan45*tan^230*tan60=1*(1/sqrt(3))^2*sqrt(3)=1/sqrt(3)=sqrt(3)/3$

Therefore, the solution is sqrt(3)/3

You might be interested in

Which shows the expression x^2-1/x^2-x in simplest form

____ [38]

Answer:

$\large\boxed{\dfrac{x+1}{x}=1+\dfrac{1}{x}}$

Step-by-step explanation:

$\dfrac{x^2-1}{x^2-x}=(*)\\\\x^2-1=x^2-1^2\qquad\text{use}\ a^2-b^2=(a-b)(a+b)\\=(x-1)(x+1)\\\\x^2-x=(x)(x)-(x)(1)\qquad\text{use the distributive property}\\=x(x-1)\\\\(*)=\dfrac{(x-1)(x+1)}{x(x-1)}\qquad\text{cancel}\ (x-1)\\\\=\dfrac{x+1}{x}=\dfrac{x}{x}+\dfrac{1}{x}=1+\dfrac{1}{x}$

4 0

3 years ago

Read 2 more answers

___________________

strojnjashka [21]

Answer:

The answer is A

Step-by-step explanation:

6 0

3 years ago

Write down the explicit solution for each of the following: a) x’=t–sin(t); x(0)=1

Kay [80]

Answer:

a) x=(t^2)/2+cos(t), b) x=2+3e^(-2t), c) x=(1/2)sin(2t)

Step-by-step explanation:

Let's solve by separating variables:

$x'=\frac{dx}{dt}$

a) x’=t–sin(t), x(0)=1

$dx=(t-sint)dt$

Apply integral both sides:

$\int {} \, dx=\int {(t-sint)} \, dt\\\\x=\frac{t^2}{2}+cost +k$

where k is a constant due to integration. With x(0)=1, substitute:

$1=0+cos0+k\\\\1=1+k\\k=0$

Finally:

$x=\frac{t^2}{2} +cos(t)$

b) x’+2x=4; x(0)=5

$dx=(4-2x)dt\\\\\frac{dx}{4-2x}=dt \\\\\int {\frac{dx}{4-2x}}= \int {dt}\\$

Completing the integral:

$-\frac{1}{2} \int{\frac{(-2)dx}{4-2x}}= \int {dt}$

Solving the operator:

$-\frac{1}{2}ln(4-2x)=t+k$

Using algebra, it becomes explicit:

$x=2+ke^{-2t}$

With x(0)=5, substitute:

$5=2+ke^{-2(0)}=2+k(1)\\\\k=3$

Finally:

$x=2+3e^{-2t}$

c) x’’+4x=0; x(0)=0; x’(0)=1

Let $x=e^{mt}$ be the solution for the equation, then:

$x'=me^{mt}\\x''=m^{2}e^{mt}$

Substituting these equations in <em>c)</em>

$m^{2}e^{mt}+4(e^{mt})=0\\\\m^{2}+4=0\\\\m^{2}=-4\\\\m=2i$

This becomes the solution <em>m=α±βi</em> where <em>α=0</em> and <em>β=2</em>

$x=e^{\alpha t}[Asin\beta t+Bcos\beta t]\\\\x=e^{0}[Asin((2)t)+Bcos((2)t)]\\\\x=Asin((2)t)+Bcos((2)t)$

Where <em>A</em> and <em>B</em> are constants. With x(0)=0; x’(0)=1:

$x=Asin(2t)+Bcos(2t)\\\\x'=2Acos(2t)-2Bsin(2t)\\\\0=Asin(2(0))+Bcos(2(0))\\\\0=0+B(1)\\\\B=0\\\\1=2Acos(2(0))\\\\1=2A\\\\A=\frac{1}{2}$

Finally:

$x=\frac{1}{2} sin(2t)$

7 0

3 years ago

Mulitply (2-2)(3x+4) using distributive property

Sati [7]

8x.......................

3 0

3 years ago

Which expression is equivalent to the expression below. 3x+5(x-2).

worty [1.4K]

It would be B) 8x-10. You distribute the 5 to the (x-2) to get 5x-10 and then simplify the expression.

4 0

3 years ago

Read 2 more answers