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The simplest way is by factoring, and quadratics can be solved with the Quadratic Formula. One of the simpler ways for factoring polynomials would be synthetic division. Of course, you can graph the equation, but the roots are sometimes unclear on graphs.

This site should explain it more in depth: http://www.mathsisfun.com/algebra/polynomials-solving.html

-3^2=9

9*x=9x (obvious)

To find the area of a trapezoid, use this formula:  $${\mathtt{A}} = {\mathtt{h}}{\mathtt{\,\times\,}}\left({\frac{\left({\mathtt{b1}}{\mathtt{\,\small\textbf+\,}}{\mathtt{b2}}\right)}{{\mathtt{2}}}}\right)$$

H is the height of the trapezoid, and b₁ and b₂ are the lengths of the top and bottom bases.

you mean "not a terminating number"

The point of intersection is (-19/24 , -11/12)

81*300=24300m

$${{\mathtt{xi}}}^{{\mathtt{2}}} = {\mathtt{\,-\,}}\left({{\mathtt{x}}}^{{\mathtt{2}}}\right) \Rightarrow \left\{ \begin{array}{l}{\mathtt{xi}} = {\mathtt{\,-\,}}\left({i}{\mathtt{\,\times\,}}{\mathtt{x}}\right)\\
{\mathtt{xi}} = {i}{\mathtt{\,\times\,}}{\mathtt{x}}\\
\end{array} \right\}$$

When you first learn about trigonometry it will used to find the side lengths and angles in right angled triangles.  

The sine ratio of one of the acute angles it it opposite side length / hypotenuse length.

$$Sin(x)=\dfrac{opposite}{hypotenuse}$$

it is an expression, not an equation. if you need an answer, substitute x for any number, then solve. square x then add x-6.

So I assume you have this....

log₇ (1/343)

First...look at something

Note......(7)^-1 = (1/7¹) = (1/7)

And    (7)^-2= (1/7²)  = (1/49)

And    (1/7)^-3  = (1/7³) = (1/343)

So .... log₇ (1/343) is just asking us..."What exponent (power) do we need to put on 7 to get (1/343) ??"     And the answer is....    -3   

(Don't get confused....we're not "multiplying" anything.....!!)

Hope that helps.....

$${\frac{{\mathtt{1.4}}}{{\mathtt{5.082}}}} = {\frac{{\mathtt{100}}}{{\mathtt{363}}}} = {\mathtt{0.275\: \!482\: \!093\: \!663\: \!911\: \!8}}$$

$$x^2-16<0$$

$$x^2<16$$

$$-4

assuming everyone has the same chance it's

p=12/900 = 1/75

This is so unfair GoldenLeaf.  You will get your points back it may just take a while.

Sometimes people/programmers can be so mean!

I'm am sorry that you didn't get you place in the symphony.  You will have better luck next time I am sure. Maybe your age was against you?

If you be rude to me Zegroes it might be like having a unicorn for an Avitar.

THAT'S WHAT I KNOW!

It measures the size of an angle or the temperature.  Depending on context.

It calculates nothing!

I want you to look at this post.

$$16^{1/2}=\sqrt{16}=4\\\\

16^{-1/2}=\frac{16^{-1/2}}{1}=\frac{1}{1\times 16^{+1/2}}=\frac{1}{\sqrt{16}}=\frac{1}{4}$$

NOTE: The 1 that was on the bottom stayed on the bottom.  I put the one on the top because there was nothing else there. when there is nothing else you put a one. does that make sense? 

Whenever anything is raised to a negative power you swap it to the otherside of the fraction line and change the negative power to a positive power.

No look at your problem again you would use arctan and subtract!

$${\mathtt{\pi}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}^{{\mathtt{2}}}{\mathtt{\,\times\,}}{\mathtt{4.5}} = {\mathtt{31.808\: \!625\: \!617\: \!596\: \!656\: \!5}}$$

I just did this on the web2 calc - I used the math(inpu=Result) key above.

Maybe you are putting the brackets in the wrong place?

$$6C4(0.60)^4(0.40)^2$$

$${\left({\frac{{\mathtt{6}}{!}}{{\mathtt{4}}{!}{\mathtt{\,\times\,}}({\mathtt{6}}{\mathtt{\,-\,}}{\mathtt{4}}){!}}}\right)}{\mathtt{\,\times\,}}{{\mathtt{0.6}}}^{{\mathtt{4}}}{\mathtt{\,\times\,}}{{\mathtt{0.4}}}^{{\mathtt{2}}} = {\frac{{\mathtt{972}}}{{\mathtt{3\,125}}}} = {\mathtt{0.311\: \!04}}$$

10P10

$${\left({\frac{{\mathtt{10}}{!}}{({\mathtt{10}}{\mathtt{\,-\,}}{\mathtt{10}}){!}}}\right)} = {\mathtt{3\,628\,800}}$$

or just 10!

What do you want me to do with this?  discuss it?

I can see straight away that 

$$x^3-432\ge 0\\
x^3\ge 432\\
\mbox{Hence}\\
x\ge \sqrt[3]{432}$$

I can also see that as x approaches infinity y plus or minus infinity.

I can 'see' that the cuve will be symmetrical about the x axis. (This is the line y=0)

Maybe that is enough discussion.

Here is the graph.

I assume you might mean this.....(5x^2-3+5) - (3x-4x-7) =

(5x^2 +2) - (-1x -7) =

5x^2 +2 + 1x +7 =

5x^2 +x + 9

Using the "change of base" rule, we can write:

log1944/log2n = log 486√2/logn      ...and rearranging, we have

log2n/logn = log 1944/log 486√2     .....simplify the right side

log2n/logn = 1.1591543597738549   .......multiplying both sides by logn and using a multiplicative log property to expand the left hand side, we  have

log2 + logn = (1.1591543597738549)logn     ......subtract logn from both sides

log2 = (.1591543597738549)logn      ........isolate logn on the right by dividing by                                                                (.159......49)

log2/.1591543597738549 = logn      ....simplify the left side

1.8914341780628552767 = logn   .......writing in exponential form to find "n," we have:

10^{1.8914341780628552767} = n = 77.8814767965999254121447 ≈ 77.88