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I've have constructed the triangle to show you that there are exactly 2 solutions solutions.

This triangle is drawn to scale.

We need to find angle B first.

You can do this using the sine rule but remember, since sin is positive in the first and second quadrant there will be an accute angle B and an obtuse angle B - Two possible solutions for B

Once you have angle B you can find Angle C by using angle sum of a triangle.  Once agian there will be 2 solutions.

You try finding this these angles.

If you get stuck again put another post after this one to ask for more help.    

If this number is divisible by 2 and 5 it must be a multiple of 10.

And the only thee digit number which fits the bill is 290. It's not divisible by 3 and the sum of its digits  = 11.

Maybe this page will help you.  

I think that you mean

  $$If $sin\theta = 0.588 \qquad $ then how do I find the angle $\theta$\\$$

When you are looking for an angle you have to use inverse sine  (arc sine is the same thing)

So on the web2 site calc (on the home page) just type in

asin(0.588) and that will give you the acute angle answer - which is probably the answer that you want.

(there are other answers as well but don't worry about those unless you have to)

$$(-0.4) + (-7.9) = -0.4\; +\; - \;7.9$$

I think of negative numbers as cold numbers and positive numbers as hot numbers.

Here you are adding cold to a room.   Does it get hotter or does it get colder?  It gets colder ofcourse!

So 

$$-0.4\; +\; - \;7.9 = -0.4-7.9$$

Now to do this you start at -0.4 in the number line and then move 7.9 units in the negative direction.

This is a bit trickier than subtraction with whole numbers  but if you think about this picture you may realise that  

$$-0.4-7.9=-(0.4+7.9)=-(8.3)=-8.3$$

I'll take your word for it Takahiro lol.    

You can not solve it because there is no equal sign.

But, you can expand and simplify it.  

$$4(a-1)^2\\
=4*(a-1)(a-1)\\
=4*[a(a-1)-1(a-1)]\\
=4*[a^2-a-a+1]\\
=4*[a^2-2a+1]\\$$

Can you take it from there?

$$\sqrt{96}$$

In order to simplify a surd you need to find a square number that goes into the number under the root.

If there is no square number factor then the surd cannot be simplified.

square numbers are 1,4,9,16,25,36,49,64,81....

Now 4 goes into 96 so lets use that one.

$$\sqrt{96}\\
=\sqrt{4*24}\\
=\sqrt4 * \sqrt{24}\\
=2*\sqrt{24}\\
=2\sqrt{24}\\$$

Now can this be simplified further?  4 is a square number and it is a factor of 24.  So we can do more!

$$2 *\sqrt{24}\\
=2*\sqrt{4*6}\\
=2*\sqrt{4} * \sqrt{6}\\
=2*2*\sqrt{6}\\
=4*\sqrt{6}\\
=4\sqrt{6}$$

Now can this be simplified further?  There are no square factors of 6 so this is finished!

NOTE:  If you had noticed that 16, which is a square number, is also a factor of 96 this could have been done more quickly.

$$\sqrt{96}\\
=\sqrt{16*6}\\
=\sqrt{16}*\sqrt{6}\\
=4*\sqrt{6}\\
=4\sqrt{6}\\$$

so if you find the Biggest square number to start with then it is quicker but if you find a smaller one you will still get there eventually.   

2cm * 4 = 8cm

so

12km * 4 = 48km

$$\left(\frac{8}{27}\right)^{(2/3)}\\\\
=\frac{8^{(2/3)}}{27^{(2/3)}}\\\\
now\\\\
8^{2/3}=\left(8^{1/3}\right)^2=\left(\sqrt[3]{8}\right)^2=2^2=4\\\\
and\\\\
27^{2/3}=\left(27^{1/3}\right)^2=\left(\sqrt[3]{27}\right)^2=3^2=9\\\\
\mbox{So the answer is }\\\\
\dfrac{4}{9}$$

The only method that I can see uses trig.

I called each of the four small angles at C, $$\alpha$$, and after five (easy) lines arrived at

$$\sin2\alpha\cos4\alpha=0,$$

implying that $$4\alpha=90$$, as required.

14:8

Which simplifies to

7:4

Because I want to know how many sheep I own so that I know if the one is stolen.

32, if I can remember correctly :P

To aid life.

$$1/2x^{-1/2} *((2xe^x)/ln(3x^2 +2))\\\\
=\frac{x^{-1/2}}{\not{2}}*\frac{\not{2}xe^x}{ln(3x^2 +2)}\\\\
=\frac{1}{1}*\frac{x^{(1-1/2)}e^x}{ln(3x^2 +2)}\\\\
=\frac{x^{(1/2)}e^x}{ln(3x^2 +2)}\\\\
=\frac{\sqrt{x}e^x}{ln(3x^2 +2)}\\\\$$

I don't think that there is anything else to be done with this.  

Has anyone got any further/different thoughts?

$$1/3(8y-2)+1/9(12y+10)\\\\
=\frac{(8y-2)}{3}+\frac{12y+10}{9}\\\\
=\frac{(8y-2)*3}{3*3}+\frac{12y+10}{9}$$

Can you take it from here?

$$\left(\dfrac{-2a^4}{b}\right)^2\\\\\\
=\dfrac{(-2)^2(a^4)^2}{b^2}\\\\
=\dfrac{4a^8}{b^2}\\\\$$

for question 20 ... 

I think that they should say in the question that the weight is in Equilibrium   right ~ ! 

thank you heureka

 f(x)=x^2-15x

What are the zeros of the quadratic function?

$$\left
\begin{array}{rcl}
x^2-15x&=&0\\
x(x-15) &=&0
\end{array}
\right\}
\quad x_1=0\qquad (x_2-15) = 0 \quad \Rightarrow \; x_2=15$$

Solution: B

$$\texttyle{2\sin^2{(x)}+sin(x)-1 = 0 \quad \cos{2x} = \cos^2{(x)}-\sin^2{(x)} \quad \cos^2{(x)} = 1 - \sin^2{(x)} }$$

$$\Rightarrow \sin{(x)} = \cos{(2x)}$$

(3 - 2i) /(4 - 5i)

Multiplyimg by the conjugate of (4 - 5i) = (4 + 5i)  in the numerator and denominator, we have

(15 + 7i + 10) / (16 + 25) =

(22 + 7i) / 41 =

(22 /41) + (7i / 41) =

(1/41)(22 + 7i)

(z₁)^2 = (1 - i√3)^2 = (1 - 2i√3 - 3) = (- 2 - 2i√3) = -2(1 + 2i√3)

(z₂)^2 = (√3 + i)^2 = (3 +2i√3 - 1) =   ( 2 + 2i√3) = 2( 1 + 2i√3)

So ......

[(z₁)^2] / [(z₂)^2] = [ -2(1 + 2i√3)] / [2(1 + 2i√3)]  = -2/2 = -1  =  -1 + 0i  = z₃

So

tan Θ = (b/a)  →  tan^-1 (b/a) = Θ  = arg z₃

And

tan^-1 (0/-1) = pi

So

arg z₃ = pi