All Answers 358108 Answers

thank you very much alan for aswering my question(im sorry i didnt notice you had another approach on the 1st problem and yeah that is the right one) , but i got confused on how you wrote the second problem is y= squaroot of 1+ tan^2 (theta) , is it just equal to your y= squaroot of 1+ tan (theta) ^2?

|5-2| + |-6+4-3| - |2-5|

Complete the equations inside first.

|3| + |-5| - |-3|

Now bring them out. Make sure to change them to positive if negative.

3 + 5 - 3

And complete the whole thing

5

Woops, thanks for correcting me. I'll try to avoid those mistakes in future.

And in honesty, my method was quite makeshift there. Kinda fiddled with the numbers way more than really was needed xD (And I completely forgot about how fractions worked. Gj me). I Much prefer Heureka's working (Particularly as "5^x+7" more likely means "(5^x)+7").

I wanted a recap of what i learnt, i kinda did learn something.

Since you been teaching me that way i've been presenting my work like that.

Before i didn't add the divide bit with the fraction when dividing x to be just x. Usually i do it in my head.

Continuing from the last post.

I'm doing good in Angles so far i wasn't really getting the protractor bit because i didn't know where to start i get it now.

In school we're doing expanding brackets, but i totally get it.

Sorry but i had to comment this.

Well said Nauseated.

I dont want to experience death

if i were to die, id rather die in my sleep like mary lincoln (she died in her sleep) 

Use Pythagoras Theory.

Like Mellie said

$${{\mathtt{a}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{b}}}^{{\mathtt{2}}} = {{\mathtt{c}}}^{{\mathtt{2}}}$$

The hypotenuse is the longest side.

If you square the adjacent side and add it to the Opposite side squared you get the Hypotenuse squared.

Square root the answer to find the hypotenuse.

or Hachary???!

Lol

It doesn't look too bad.

Don't worry, listen to your doctor.

*Presses Challenge Button

If we accept Sir-Emo-etc's interpretation of the question we have:  5^2x = 5^x+7 so 2x = x+7 and x = 7.

You went awry after 1 + 7/x = log(25)/log(5) Sir-Emo..., you wrote 7/x = log(25)/log(5) + 1, but should have had 7/x = log(25)/log(5) - 1 

Also, you need to be careful when inverting both sides of an equation that you invert the whole of the side, not the individual terms!

.

@@ End of Day Wrap  Tues 23/6/15   Sydney, Australia Time   12:40am   (Yes it is Wed)  ♪ ♫

Hi all, 

Today our great answers were provided by Radix, Nauseated, Alan, Geno3141, Heureka, Mellie, MathsGod1, TitaniumRome, Eloise, Sir-Emo-Chappington, BrittanyJ and DarkBlaze347.  Thank you  

I think Nauseated predicted the future

Remember when my good friend Heather joined?

This is what Nauseated told Heather Feather her very first day besides welcome:

This site is devoted to the cutting edge developments of theoretical and applied mathematics. You will find amazing ideas and concepts on here. In fact, some of the concepts are so advanced it requires a unique universe for them to function.

-Nauseated

I think this is what Nauseated meant. Not only of my number thingys or Beyonce rules, but also Heureka's unique universe

i think Nauseated may be a fortune teller, a witch at witchcraft, or the oracle of something from Greece

Oh, btw, 3 is an evil number thingy

Thurs 25/6/15

If you would like to comment on other site issues please do so on the Lantern Thread.  Thank you.    

Interest Posts: 

FTJ means 'For the juniors' 

I do not understand what you are doing with these questions Heureka  

I don't understand your answer Heureka.  

I think what you are just being asked is:

How many numbers less then 12 are relatively prime to 12, that is, that is the only common factor of the two numbers is 1

so the numbers will be  5,7 and 11.   That is 3

So      $$\phi(12)=3$$

Maybe I just don't understand the question.   

[ I was supposed to include 1 as well]

25^x=5^x+7

First we need to bring those x's to a more usable form.

Using logarithms, I can turn the x and x+7 into factors.

Log(A^B) = B * Log(A)

x * log(25) = (x + 7) * log(5)

Now re-arrange so all instances of x are on the same side of the equation.

^x/_(x+7) = ^log(5)/_log(25)

Now we got the annoying situation of x+7 being the denominator. I'm going to quickly flip the fractions around because we do not want an equation as the denominator.

^(x+7)/_x = ^log(25)/_log(5)

Let's quickly expand out our fraction to seperate the x's.

⁽⁷⁾/_x + ^(x)/_x = ^log(25)/_log(5)

Simplify...

1 + ⁷/_x = ^log(25)/_log(5)

Again, let's re-arrange to keep x seperate from everything else.

⁷/_x = ^log(25)/_log(5) + 1

Now, x is back on the denominator! Let's flip these fractions again to fix that (Note the "reciprocal" of 1 is 1, so that's why it won't change)

^x/₇ = ^log(5)/_log(25) + 1

Now we can remove the 7 by multiplying both sides by it.

x = (^log(5)/_log(25) + 1) * 7

And finally, let's work out these numbers.

x = ^{(log(5) * 7)}/_log(25) + 7 = 3.5 + 7 = 10.5

(Of course by the time I finish typing this, I'm a bit late at answering :P)

///////////////////////////////

ok I'm in :)

 2k-13-3= 2k-16

I accept the challenge.

I guess in a few hours, i will have the first ever math challenge for you all

prepare to be challenged!

its a real life situation

Very nice Heureka :)

Thank you really much, Melody