CPhill

I actually don't know how to answer this, but I think this is what you're looking for:

Well...if we put in any real number for n...don't we get a real number back?? And since we can return any real number we want...that's what I was getting at.....the "y" is implied..

It's just nonsense, rosala.....

It's a derivative of......"How much wood could a woodchuck chuck if a woodchuck could chuck wood??"

I wouldn't worry about that one too much, either.....

So I assume you want to find the perimeter of a trianle with sides two sides of (8+5/16) and one side of (3 + 3/8) ??

Let's just add the fractions, first...we'll go back and "tack on" the whole numbers at the end...this makes it easier !!

So   .        (5/16) + (5/16) + (3/8) =  (5/16) + (5/16) + (6/16) = 16/16 = 1

So..... 1 + 8 + 8 + 3 = 20     .....pretty easy, huh??

And that's the perimeter...

Where 2-secx = 0

So secx = 2  or  cosx = (1/2)

So we have (pi/3 + 2n*pi)   and (5/3)pi + 2n*pi).....where n is an integer

Rs 51

cos^2 x-sinx-sin^2 x=0      writing cos^2x as 1 - sin^2x, we have

1-sin^x - sinx - sin^2x = 0

-2sin^x - sinx + 1  =  0        ....multiply through by -1

2sin^x + sinx - 1 = 0            .....factor

(2sinx - 1) (sinx + 1) = 0      ....set each factor to 0

2sinx - 1 = 0            sinx + 1 - 0

sinx = 1/2                sinx = -1

x = 30° , 150°  and 270°

So....x = 30°

Great job, NinjaDevo......well presented, too...Thumbs Up and Points......!!!!

2log₄(16xy²)

2 [ log₄16 + log₄x + log₄y² ]

2 [ 2 + log₄x + 2log₄y ]

4 + 2log₄x + 4log₄y

And that's in "linear" form..... (no exponents)

y = 4log₅(3-x) + 2   .........set y =0

0 = 4log₅(3-x) + 2   .........subtract 2 from both sides

-2 = 4log₅(3-x)       ..........divide by  4

-1/2 = log₅(3-x)        ....in exponential form, we have

5 ^-1/2 = 3-x               .....or........

1/√5 = 3 -x              .......add x to both sides, subtract 1/√5 from both sides

x = 3 - 1/√5  =  [3√5 - 1]/√5 

So we have.......  ( [3√5 - 1]/√5 , 0 )