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Well it is a really good thing that there is an excellent calculator on the home page.   

Let x be the tank's capacity.......And the difference between being 5/8ths full and 1/6th full is 180 (gallons??)

So

(5/8)x - (1/6)x = 180

(30 - 8)x /48 = 180

(22/48)x = 180     multiply both sides by 48/22

x = (48.22)180 = about 392.72 (gallons??)

Thanks Chris  

Very nice, Melody.

the formula is 

$$\\\boxed{\int f(x)g'(x)\;dx=f(x)g(x)-\int f'(x)g(x)\;dx}\\\\
or\\\\
\boxed{\int uv'\;dx=uv-\int u'v\;dx}\\\\$$

$$\\\int sin^{-1}\;dx=\int sin^{-1}*1\;dx\\
let\;u=sin^{-1}\;\;and\;\;v'=1\\\\
u=sin^{-1}x\\
x=sin\;u\\
\frac{dx}{du}=cos\;u\\
\frac{du}{dx}=\frac{1}{cos\;u}\\
\frac{du}{dx}=\frac{1}{\sqrt{cos^2 u}}\\
\frac{du}{dx}=\frac{1}{\sqrt{1-sin^2 u}}\\
\frac{du}{dx}=\frac{1}{\sqrt{1-x^2}}\\$$

$$so\\
u=sin^{-1}x\\
u'=\frac{1}{\sqrt{1-x^2}}\\
v'=1\\
v=x\\\\
\int sin^{-1}x\;dx\\
=\int sin^{-1}*1\;dx\\
=uv-\int u'v\;dx\\
=sin^{-1}x*x-\int \frac{x}{\sqrt{1-x^2}}\;dx\\
=xsin^{-1}x-\int \frac{x}{\sqrt{1-x^2}}\;dx\\\\
let\;\;t=1-x^2\\\\
\frac{dt}{dx}=-2x\\\\
dx=\frac{dt}{-2x}=\frac{dt}{-2\sqrt{1-t}}\\\\
\sqrt{1-t}=x\\\\$$

$$so\\\\
xsin^{-1}x-\int \frac{x}{\sqrt{1-x^2}}\;dx\\\\
=xsin^{-1}x-\int \frac{\sqrt{1-t}}{\sqrt{t}}\;\frac{dt}{-2\sqrt{1-t}}\\\\
=xsin^{-1}x-\int \frac{\sqrt{1-t}}{\sqrt{t}(-2\sqrt{1-t})}\;dt\\\\
=xsin^{-1}x-\int \frac{\sqrt{1-t}}{-2\sqrt{t}\sqrt{1-t}}\;dt\\\\
=xsin^{-1}x-\int \frac{1}{-2\sqrt{t}}\;dt\\\\
=xsin^{-1}x+\frac{1}{2}\int t^{-0.5}\;dt\\\\
=xsin^{-1}x+\frac{1}{2}\times \frac{t^{0.5}}{0.5}+c\\\\
=xsin^{-1}x+ \sqrt{t}+c\\\\
=xsin^{-1}x+ \sqrt{1-x^2}+c\\\\$$

987654321 divided by 987654321234567890987654321

$$2532=2^2*3*211$$

 cos(2a) = -√2 /2

2a must be in the 2nd or 3rd quadrant

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

The first thing that I notice is that

  $$\frac{\sqrt2}{2}\\\\
=\frac{\sqrt2\times\sqrt2}{2\sqrt2}\\\\
=\frac{2}{2\sqrt2}\\\\
=\frac{1}{\sqrt2}\\\\\\
so \\\\
cos(2a)=\frac{1}{\sqrt2}\\\\
$Now you should recognise that $ \\\\
cos45^0=\frac{1}{\sqrt2}\\\\$$

So 

2a=180-45, 180+45, 180-45+360, 180+45+360 degrees

2a=135, 225, 495, 585 degrees

a=67°30', 112°30', 247°30', 292°30'

That is all the answers  for  $$0\le a\le360^0$$

I know i have used pre-knowledge that you may not have - if you want me to explain differently/more then let me know.  

$$\\x+7-3x=16-5x+3\\
7-2x=19-5x\\$$

now add 5x to bot sides.

then

subtract 7 from both sides.

Then think about how to finish it :)

What's 60-5?3(squared)+(4?2)

What are the question marks for?

@@ End of Day Wrap            Sat 18/10/14            Sydney, Australia           Time 8:05pm     ♬

Hi everyone,

Great answers today from Heureka, CPhill, Alan, alvarez, GoldenLeaf, Genop3141 and TheLoneWolf.  Thank you, you are all wonderful.

We had a visit from NinjaDevo and GoldenLeaf today.  That was great.  Thanks guys.  

Now for some interest posts:

1) A lovely visit from NinjaDevo and GoldenLeaf.   

Sun 19/10/14

1) Integration by parts - It was fun :)

Which game are you talking about Pyramid?

Australians say maths too.

Most Australian students would not know what you were talking about if you said you were going to your math class.  It would sound really weird.  

When we say maths it often sounds more like mass.   We are a lazy bunch.    LOL

Wait! This game is is a bit like the Chinese card game number 24 except the magic number doesn't change!

Thanks Alan.  

Puzzle 1 answer: I am confused. Don't get it.

you can enter it into the calc like this

(18+3/4)

or you can look here

http://web2.0calc.com/questions/how-do-i-type-in-fractions

I'll just do the first one

if x=-2 then

2*(-2)+y=8     See how I replaced the x in the equaton with -2    

-4+y=8

y=12   

(-2,12)

now you can copy the technique and do the others yourself.  

A man borrows $700 at 4% per annum. He repays $200 per year. How much still owes after the 3rd repayment.

At the end of the first year he owes     1.04*700-200=$528

At the end of the second year he owes     1.04*528-200=$349.12

At the end of the second year he owes     1.04*349.12-200=$163.48

Hi Alan,

I understand at least some of what you have done.

I am wondering about your derivation.

I can follow it alright but how did you do the simplification.     (Where it actually says simplify)

Can you do that by hand or do you need to use a calculator such as Mathcad or Wolfram|Alpha?

Thank you.:)

$$opposite*1.5$$

or

$$adjacent*1.5$$

I think

(Old amount - New amount) / Old amount

For instance, percent decrease from 10 to 8 =

(10 - 8) / 10   =    2 / 10  = 20%

Doesn't that make sense.....???    ...... 2 is 20% of 10...and if we take 2 from 10, it has "decreased" by 20% !!!

Notice the last two points........"x" and "y" change by 1 from (6,4) to (7,5)...And the "slope" is defined by (change in y) /(change in x) = (1 / 1) = 1

So using the slope and one of the points ....we'll use ( 4, 2)....we can use the point-slope formula to write the equation..we have....

y - y₁  = m(x - x₁)           where x₁ = 4, y₁ = 2 and the slope, "m" = 1....so we have...

y - 2 = 1 (x - 4)

y - 2 = x - 4                   add 2 to both sides

y =  x - 2

Check to see that the other two points "work" in this equation !!!

Let me add something, Tori ....rotating a point  in a quadrant either 90 degrees clockwise or counter-clockwise simply puts the point in the next quadrant (for 90 degrees clockwise) or the prior quadrant (for 90 degrees counter-clockwise). As long as you remember the signs on x and y in each quadrant, you might not even have to do much "math' in these kinds of problems. For instance, "A" is the correct choice here, because it's the only one that makes logical sense.  "E" is a first-quadrant point whose signs are both positive. And when we rotate it 90 degrees clockwise, it ends up in the 4th quadrant where x is positive and y is negative. Look at the rest of the answers......the signs are incorrect for each point......we don't even have to look at the values in this case !!!!