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Thanks Heureka,

I didn't even see the tan(2x)      

Find sin 2x, cos 2x, and tan 2x from the given information. tan x = −1/2 , cos x > 0

In trigonometry, the tangent half-angle formulas relate the tangent of one half of an angle to trigonometric functions of the entire angle. They are as follows:

\(\begin{array}{rcl} \boxed{~ \begin{array}{rcl} \sin{(\alpha)} &=& \dfrac{ 2\cdot \tan{\frac{\alpha}{2} } } { 1 + \tan^2{\frac{\alpha}{2} } } \\\\ \cos{(\alpha)} &=& \dfrac{ 1 - \tan^2{\frac{\alpha}{2} } } { 1 + \tan^2{\frac{\alpha}{2} } }\\\\ \tan{(\alpha)} &=& \dfrac{ 2\cdot \tan{\frac{\alpha}{2} } } { 1 - \tan^2{\frac{\alpha}{2} } } \end{array} ~} \quad \text{ or } \quad \boxed{~ \begin{array}{rcl} \sin{(2\alpha)} &=& \dfrac{ 2\cdot \tan{\alpha } } { 1 + \tan^2{\alpha } } \\\\ \cos{(2\alpha)} &=& \dfrac{ 1 - \tan^2{\alpha } } { 1 + \tan^2{\alpha } }\\\\ \tan{(2\alpha)} &=& \dfrac{ 2\cdot \tan{\alpha } } { 1 - \tan^2{\alpha } } \end{array} ~} \end{array}\)

\(\begin{array}{rcl} \sin{(2x)} &=& \dfrac{ 2\cdot \tan{x } } { 1 + \tan^2{x } } \\ \sin{(2x)} &=& \dfrac{ 2\cdot ( -\frac12 ) } { 1 + ( -\frac12 )^2 } \\ \sin{(2x)} &=& \dfrac{ -1 } { 1 + \frac14 } \\ \mathbf{\sin{(2x)}} &\mathbf{=}& \mathbf{-\frac45}\\ \\ \cos{(2x)} &=& \dfrac{ 1 - \tan^2{x } } { 1 + \tan^2{x } }\\ \cos{(2x)} &=& \dfrac{ 1 - \frac14 } { 1 + \frac14 }\\ \cos{(2x)} &=& \frac34 \cdot \frac45 \\ \mathbf{\cos{(2x)}} &\mathbf{=}& \mathbf{\frac35} \\ \\ \tan{(2x)} &=& \dfrac{ 2\cdot \tan{x } } { 1 - \tan^2{x } } \\ \tan{(2x)} &=& \dfrac{ 2\cdot ( -\frac12 ) } { 1 - ( -\frac12 )^2 }\\ \tan{(2x)} &=& \dfrac{ -1 } { 1 - \frac14 }\\ \mathbf{\tan{(2x)}} &\mathbf{=}& \mathbf{- \dfrac43}\\ \end{array}\)

g/h^2÷g^2/h^3

Are there meant to be any brackets here?

I will read your mind and put some in :)

\((g/h^2)÷(g^2/h^3)\\ =\frac{g}{h^2}\div\frac{g^2}{h^3}\\ =\frac{g}{h^2}\times \frac{h^3}{g^2}\\ =\frac{h}{g}\)

Find sin 2x, cos 2x, and tan 2x from the given information. tan x = −1/2 , cos x > 0

If a right angles triangle has the two short sides of 1 and 2 then the hypotenuse must be sqrt5

since tan is neg the angle is in the 2nd or 4th quads and since cos is pos, x must be in the 4th quad.

tan x = −1/2        cosx= +2/sqrt5         sinx= -1/sqrt5

sin2x= 2sinxcosx = 2* -1/sqrt5 * 2/sqrt5 = -4/5

cos2x= cos^2x-sin^2x =  4/5-1/5 = 3/5

I did it in my head so I hope I didn't do anything stupid.  You are capable of checking yourself. 

Any question, just ask. :)

Simplify by removing factors of 1.

6y^2-24/9y^2-36

\(\begin{array}{rcl} \dfrac{ 6y^2-24 } { 9y^2-36 } &=& \dfrac{ 2\cdot ( 3y^2-12 ) } { 3\cdot ( 3y^2-12 ) } \\ &=& \dfrac{ 2 } { 3 } \end{array}\)

6y^2-24/9y^2-36 - I'm simply reading your mind as to what actually want or intend, because no bracketing involved. If this is not your answer, then re-post it accurately, with proper brackets.

Simplify the following:
(6 y^2-24)/(9 y^2-36)

Factor 9 out of 9 y^2-36:
(6 y^2-24)/9 (y^2-4)

y^2-4 = y^2-2^2:
(6 y^2-24)/(9 (y^2-2^2))

Factor the difference of two squares. y^2-2^2 = (y-2) (y+2):
(6 y^2-24)/(9 (y-2) (y+2))

Factor 6 out of 6 y^2-24:
6 (y^2-4)/(9 (y-2) (y+2))

y^2-4 = y^2-2^2:
(6 (y^2-2^2))/(9 (y-2) (y+2))

Factor the difference of two squares. y^2-2^2 = (y-2) (y+2):
(6 (y-2) (y+2))/(9 (y-2) (y+2))

(6 (y-2) (y+2))/(9 (y-2) (y+2)) = ((y-2) (y+2))/((y-2) (y+2))×6/9 = 6/9:
6/9

The gcd of 6 and 9 is 3, so 6/9 = (3×2)/(3×3) = 3/3×2/3 = 2/3:
Answer: | 
| 2/3

Find the line perpendicular to 8x-2y=4 that goes through (-12,-1)

Find the line perpendicular to 8x-2y=4 that goes through (-12,-1)

\(\begin{array}{rcl} 8x-2y &=& 4 \qquad | \qquad \cdot (-1) \\ -8x+2y &=& -4 \\ 2y &=& 8x - 4 \\ y &=& \frac{8x-4}{2} \\ y &=& \frac{8x}{2} - \frac{4}{2}\\ y &=& 4x - 2 \\ \end{array}\)

\(\begin{array}{rcl} \text{Formula } \boxed{~ \begin{array}{lrcl} y = mx+b \\\\ \dfrac{y-y_p}{x-x_p} = m_{\text{perpendicular}} \\ m_{\text{perpendicular}} = -\frac{1}{m} \end{array} ~}\\\\ \end{array}\\ \begin{array}{rcl} P(x_p,y_p) &=& (-12,-1) \\\\ y &=& 4x - 2 \qquad m = 4 \\ m_{\text{perpendicular}} &=& -\frac{1}{ 4 } \\ \dfrac{y-y_p}{x-x_p} = \dfrac{y-(-1)}{x-(-12)} &=& -\frac{1}{ 4 } \\ y+1 &=& -\frac{1}{ 4 } \cdot (x+12) \\ y+1 &=& -\frac{1}{ 4 } x -\frac{12}{ 4 } \\ y+1 &=& -\frac{1}{ 4 } x -3 \\ y &=& -\frac{1}{ 4 } x -3 -1 \\ y &=& -\frac{1}{ 4 } x -4 \\ \end{array}\)

csc

\(cosec(x)_=\frac{1}{sin(x)} = \frac{hyp }{ opp}\qquad \mbox{in a right angled triangle}\)

150

Let the distance from the Port to the Island=d, then we have:

d/13 + d/14=27

Solve for d:
(27 d)/182 = 27

Multiply both sides of (27 d)/182 = 27 by 182/27:
((182×27 d)/(27))/(182) = 182/27×27

182/27×27 = (182×27)/27:
(182×27 d)/(27×182) = (182×27)/27

(182×27 d)/(27×182) = (27×182)/(27×182)×d = d:
d = (182×27)/27

(182×27)/27 = 27/27×182 = 182:
Answer: | 
| d = 182, Miles-the distance from the Port to the Island.

I will start you off  :)

Let t1 be the time to get to the island and t2 be the time to get back.

Let d be the distance to the island.

t1+t2=27    so     t1=27-t2

speed = distanc/time

so

13= d/t1      and 14=d/t2

13t1 = d      and    14t2=d

so

13t1 = 14t2

13(27-t2)=14t2

etc

After you get t2 you can sub back into to get d.

Trig

\(\frac{cosx(secx-cosx)}{sinx}\\ =cosx(\frac{1}{cosx}-\frac{cos^2x}{cosx})\div sinx\\ =(1-cos^2x)\div sinx\\ =sin^2x\div sinx\\ =sinx\\ \)

6.626 x 10exponent-34 x 6.5 x 10 exponent 14

(6.626*10^-34)*(6.5*10^14)=

(6.626*10^-34)*(6.5*10^14) = 0.00000000000000000043069

If you enter it like this on the home page and then press = again, it will change it to scientific notation. :)

Unfortunately that does not work from inside the forum. :(

I am going to try it our guests way. :)

6.626e-34*6.5e14 = 0.00000000000000000043069

0.00000000000000000043069 = 4.3069e-19

It will disply in scientific notation in the forum that way too.

Thanks guest you have shown me something really good.     

Are you talking about the calculator here? If yes, this is what you do: Enter your first number, 6.626, then press the letter "e" in the top left-hand, then enter 34 followed by +,- key to the right of zero. It should look like this:6.626e-34. "e" stands for "exponent of 10". Now, you hit one the four operations, +,-,x,/. Next you enter your second number exactly the same as the first number. Finally, you hit the = key twice. That should get you what you want. Good luck. 

Well, it is NOT "no solution" because there was nothing to solve in t he first place.

You can just say that the statement is true :)

The cube route negative 27

\(\sqrt[3]{-27}=-3\)

because -3*-3*-3 = -27

There are 2 other complex roots as well but you probably do not need to worry about those.

Those would  be     \(3e^{\frac{\pm\pi i}{3}}\)

So it's a no solution?

Use this online calculator:http://www.calculatorsoup.com/calculators/math/fractions.php

The cube route negative 27= -27^1/3= -3, some people will quibble with this. OR,

(-27)^1/3=3(-1)^(1/3)-This is the "complex" root.

Yes it is not an equation because there are no unknowns.

LHS=15*-1-3((4*-1)-6) = 

15*-1-3((4*-1)-6) = 15

RHS =

3*-1+9-9*-1 = 15

So you NUMBER STATEMENT  is true. :)

Your "equation" is meaningless!.

I am not really sure what your question is but if it is a right angled triangle and you want the 3rd side you can use the pythagorean theorem. :)

6lnx=1

lnx = 1/6

x = e^(1/6)