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An integer is like a whole number, but it also includes negative numbers. They DO NOT contain fractions. Examples of integers would be: -3, -2, -1, 0, 1, 2, 3 (and so on)

Use the foil method and you will get: 4x^2 - 12x + 9

sin(x)=cos(2x)

sin(x)-cos(2x)=0                 | Move x to one side of equation

sin(x)-(1-sin²(x))=0             | Double angle identity

sin²(x)+sin(x)-1=0              | simplify terms

(2sin(x)-1)(sin(x)+1)=0       |

x=π/6 , 3π/2                       |   if you need degrees, x=30º ,270º

How did you know to use those numbers inside of the parenthesis? I understand the 3, but I don't know how you knew to use -2 and 1. The way I was taught was to find two numbers that added to equal b, (-5) and multiplied to equal c, (2). And if you couldn't do that, refer to the formula.

That's an interesting tecnique, Alan and Bertie......

I'm with Alan...........a calculator seems vital......

Did you mean (64/125)^(-2/3) ??

If so, we have  

(125/64)^(2/3) =

[(125/64)^(1/3)]^(2) =

(5/4)^(2) =

(25/16)

This factors as

(3x-2) (x-1)

If you're tryig to find the "zeroes," just set each factor to 0. You will obtain x = 2/3 and x = 1.

g(4)=(8*4)+3=32+3=35

13.798±0.037 billion years old

50*6= 300 

300/10=30

30-6=24

24+36=60

60 is your answer.

This is an example of a quadratic equation. 

Since you can't factor easily, you should use the quadratic formula.

Your two solutions should be:

x=1 & x=0.6666666667

What you have here is an example of a quadratic equation. So x is going to have 2 values.

Since you can't factor this, use the quadratic formula.

b + or - = the square root of -b(4ac)/2

Your solutions should be:

x= -3.236  & x= 1.236

______________________

5^A=2

ln(5^A)=ln(2)

A*ln(5)=ln(2)

A=ln(2)/ln(5)

A=0.4306765580733931 #

______________________

The iterative method Bertie mentioned, to find the cube root of "a", say, is:

$$$$x_{n+1}=\frac{1}{3}(\frac{a}{x_n^2}+2x_n)$$

This is derived using Newton-Raphson.

However, in general, I wouldn't want to use this without a calculator of some sort! 

That's a great answer!!!

Rom's method is correct; he just used a value for r1 of 1 instead of the requested 4. 

1/R = 1/4 + 1/3  = 3/12 + 4/12 = 7/12

Hence R = 12/7 = 1.714 ohms

i.e. the same as DavidQD

$${\mathtt{1}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{\left(-{\mathtt{14}}\right)}{\mathtt{\,-\,}}{\mathtt{7.24}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{\left(-{\mathtt{4}}\right)} = -{\mathtt{0.000\: \!723\: \!999\: \!999\: \!99}}$$

And all this time I thought the answer was nine million divided by two and a quarter million.

How could i get it so wrong?  

Thanks rosala,

That was some really good advice.

This the exact  address that Rosala is sending you to

Good one anonymous!

Hi anonymous,

I have no idea which is correct.  As the question asker it is up to you to have enough knowledge to know which answer to accept and which to reject.

Maybe Rom or someone else will come along and help you with this task.

Well, I have checked using a hand held casio calculator and Wolfram|Alpha calculator and

Alan has checked using Mathcad.

It is correct to 12 significant figures.  The last five digits have some rounding error.  For most people's purposes this error would be insignificant.

$$1 \mbox{ billion }=10^9$$

$$(10^9)^{(10^9)} = 10^{9\cdot 10^9}=10^{(9 \mbox{ billion })}$$

that's a 1 with 9 billion zeros after it.

might take a while to write down