All Answers 358514 Answers

Is that a fact Alan? You definitely should know this.  Thank you.

You question does not make a lot of sense.  Perhaps you could re-word it.

u r questioning and answering urself ! is all this for our knowledge or what ! 

It is a beaut photo.

I love it!

$${\mathtt{9}}{m}{\mathtt{\,\times\,}}{\mathtt{10}}{m}{\mathtt{\,\times\,}}{\mathtt{30}}{m}{\mathtt{\,\times\,}}{\mathtt{9}}{m} = {\mathtt{24\,300}}\left[{m}{\mathtt{\,\times\,}}{m}{\mathtt{\,\times\,}}{m}{\mathtt{\,\times\,}}{m}\right]$$ = $$24300[m^4]$$

Yes, all the superscripts/subscripts were a bit of a pain, but no more so than typing \frac{}{} all the time in LaTeX!

Incidentally, I like your photo.  You look younger than I had expected!!

lets do it like this;

3 times - 5 times means 3x-5x  and -2 to the power of 3 means -2³

so ;3x-5x-2³

=-2x-8

thats the answer i believe ! 

Thanks, alan...I'll look at your way again.......I always like seeing how someone else might approach the same thing differently !!

(You didn't break the Format "superscript" thingy typing all those exponents, did you??)

Virtually all car speedometers are deliberately overclocked, usually by between 5% and 10% (to judge by comparisons with roadside speed warning indicators), to make you think you are going slightly faster than you are!

this nes right CPhill , i had just made a slight mistake ! thumbs up !

As Anonymous has noted, this is an expression not an equation.  However, notice that it factors nicely into (x+3)(x-2), so if you wanted to find the values of x that make the expression equal zero, you can see immediately that those values would be x = -3 and x = 2.

There is another way of approaching this (you have to think in terms of factorizing 486 =  2*3⁵ and 1944 = 2³3⁵ for this):

Let a = log_n(486√2)  ...(1)

Then it's also true that a = log_2n(1944)   ...(2)

It's useful to write (1) as   a = log_n(2*3⁵√2) = log_n(3⁵2^3/2)   so that  n^a = 3⁵2^3/2  ...(3)

Similarly, with (2) we have  a = log_2n(2³3⁵) so that (2n)^a = 2³3⁵   or 2^an^a = 2³3⁵  ...(4)

Substituting from (3) into (4) we get  2^a3⁵2^3/2 = 2³3⁵ from which  2^a = 2^3/2 so that a = 3/2.

Putting this back into (3) we have n^3/2 = 3⁵2^3/2 so n = (3⁵2^3/2)^2/3  or

n = 2(3⁵)^2/3 = 2*3^10/3 = 2*(3⁹3¹)^1/3

n = 2*3³*3^1/3

Finally:  n = 54*3^1/3

Sorry...Anonymous.....I didn't look at your answer closely enough...it is arctan(4/15)

My mistake.....!!!

Just put this into the on-site calculator as is.......make sure that you're in the correct mode (probably degrees??)

Well Sir/Madam, you are a wealth of knowledge!

so it would be arcsin=4/15 not arctan?

-y since log(5)=y , log(1/5)=-log5=-y

Thanks, kitty<3....I hadn't seen that "Bounds on Zero" thing since Pre-Cal, and I forgot how to do it !!!.......it's presented in an easy manner on those pages......

Thanks Kitty.

I hadn't heard that terminology before.

It is this: (- 2 - 4i) ^ 2

Is the result but not be as it is the procedure for the exercise! Ayudaa

by Anonymous (3 hours ago)

This is just a perfect square  

$$(a+b)^2=a^2+2ab+b^2$$

--------------------------------

$$\begin{array}{rll}
(-2-4i)^2&=&(-2)^2+2*(-2)*(-4i)+(-4i)^2\\
&=& 4+8i+16i^2\\
&=&4+8i-16\\
&=&-12+8i
\end{array}$$

Or you can use your keyboard and type in  ^2

In computing, ^ means 'power of'  It is above the 6 on your keyboard.

^ is a caret symbol that in this instance can be referred to as a 'hat'

All of you forgot...

34-12=22

OK, Anonymous......the opposite side is 4  and the adjacent side is 15....what relates these?? Set this up....and use an inverse function to find your angle.....

BTW.....you have the right inverse function.......your numbers are incorrect, though...