Melody

the square of a number is the second power of the number.      $$x^2$$

the cube of a number is the third power of the number . explain why these names are reasonable     $$x^3$$

Because they are the same thing. 

There is no answer.  This is the equation of a line.

OR

5^5

^ is the computer symbol for power of

Just start  form the most inside bracket and work outwards from there.  

YOu will get the same answer as the calculator. 

$$6561^{1/4}$$

6561^(1/4)  will give you the answer.  

DragonSlayer554    gave a really good answer to a question like this just a day or two ago.

Where is THE Dragon when you need him.   Off fighting for Camelot I expect!  I should not complain, we need our valiant defender!

Hi   TheAnswerGuy

It stands for sine.

It is a part of trigonometry that you obviously have not studied yet.

Here is an introduction if you are really serious.     

Mmmm

There is no equal zero which is what you have all solve for.

In fact there is NO question here.

Perhaps we were supposedto factorise it in which case the answer is

$$\\F(x)=x^2-8x+15\\\\
\mbox{You need to find to numbers that mult to 15 and add to -8 (-3 and -5)}\\\\
F(x)=(x-3)(x-5)\\\\$$

And that is all that you can do.  

This will be the same answer no matter how far it is so to make life easy I am going to say it is 200km

At 50km/h it will take 4 hours

At 40km/hour it will take 5 hours

That is 9 hours altogether to travel 400km

So the average speed it 400/9  km/hour =    $$44\frac{4}{9}\;\;km/hour$$  

I always use the units to help me do rates.

My method is quite unique I believe but when you get the hang of it it makes working out any rates related questions really easy - even really complicated looking ones.

I have explained this one in detail so it looks long but I normally would do it in 3 easy lines.  

I think it is worth puting the effort in and working out what I have done.     

$$\\\frac{30000kb}{1sec}\qquad \mbox{this can also be expressed as }\qquad \frac{1sec}{30000kb}\\\\
\mbox{Alex wants to download 4.24GB }\\
\mbox{(I have to assume this because it is not written into the question)}\\\\
\mbox{There are } \frac{8.6\times 10^6kb}{1GB}\quad \mbox {and Alex want to know seconds/download}\\\\\\
\mbox{Start with the ratio that has seconds on the top}\\\\
\frac{1sec}{30000kb}\\\\
\mbox{Now look for the one with kb on the top because we want to cancel those out}\\\\
\frac{1sec}{30000kb} \times \frac{8.6\times 10^6kb}{1GB}\\\\
\mbox{Now look for the one with GB on the top because we want to cancel those out}\\\\
\frac{1sec}{30000kb} \times \frac{8.6\times 10^6kb}{1GB}\times \frac{4.24GB}{1} \\\\
\mbox{So this will leave us with } \\\\$$

$$\\\frac{8.6\times10^6 \times 4.24 \;seconds}{30000} = 1215\;\; seconds\qquad \mbox{(All the other units cancelled out)}\\\\
\frac{1min}{60sec}\times \frac{1215 sec}{1} = \frac{1215 min}{60} = 20.25\;minutes\\\\
\mbox{Therefore download time is approximately 20 minutes}$$