Math Is Fun Forum

hi yaz!

4)  30coins = quarters + dimes
we are given that: quarters > $5.20
one quarter is .25, so 20 quarters is: 20(.25) = $5.  however, 20 quarters < $5.20, so we must have an additional quarter to make that statement true giving us a total of 21 quarters.  knowing this, we can now solve for dimes:
30 = 21 + dimes
30 - 21 = dimes
9 = dimes

8) let n be the small integer
    let (n+1) be the consequitive larger integer
we are given:
5n < 4(n+1)
solving for n:
5n - 4n < 4
n < 4

11)
A = $4(maintenance fee) + .10(check cashed)
B = $6(maintenance fee) + .06(check cashed)
so we wanna find out when is A > B:
$4 + .10x > $6 + .06x     solve for x
4 - 6 > .06x - .10x
-2 > -(1/25)x
-2(-25) < x
50 < x

griever is easy to defeat only if you are prepared.  stupid me never bothered to learn the junction system so none of my characters were properly equipped.  even worse, i fought all my battles using the GFs exclusively.  As soon as battle began i would whip them out and hope for the best.  not once did i attack, used an item, nor cast magic – the GFs did all the fighting.  i never upgraded my weapons nor stocked up on potions as a result.

my avatar is also a game character.  you know what game he's from??  zach seems to know a lot about games he's never played so may be he might just know.

hey john!

there is a similar problem like this here:
http://www.mathsisfun.com/forum/viewtopic.php?id=2310

i'll work it out for ya!

y = x^x                   apply log to both sides

lny= lnx^x

lny = xlnx                now differentiate both sides

y'/y = lnx + x/x

y' = y(lnx + 1)         remember that y=x^x

y' = (x^x)(lnx + 1)

y' = (x^x)lnx + x^x

y' = (x^x)lnx + x*x^(x-1)

that last part is typical calculator manipulation.  they always spit out some funky looking solutions like x*x^(x-1) when it could just be x^x.

siva:  that is a very interesting approach and one that i'm not too sure if it's true, however, it looks like you found the x-intercept and not the summation.

it's not that difficult really.  just apply log to both sides and differentiate.

y = (cos x )^0.7x

lny = ln(cos x )^0.7x

lny = .7xln(cox)

give it a try!  oh and happy holidays everyone!

hooray!!!  i solved it!!!

(e^x - e^-x)/2 = 7

e^x - e^-x = 14                             multiply everything by e^x

e^x(e^x - e^-x) = 14e^x

e^2x - 1 = 14e^x

(e^x)² - 14e^x - 1 = 0                   let w=e^x

w² - 14w - 1 = 0                            using the quadratic formula, i obtain

     14±(200)^1/2)
w= ------------------                        which simplifies to
            2

w= 7±(50)^(1/2)                           plug w back in w=e^x

7±(50)^(1/2) = e^x

ln(7±(50)^(1/2)) = x                     but lnx > 0

ln(7+(50)^(1/2)) = x

what is "Maths at A2 Level, (C3)"???

hi yaz chemist!!

  (1+x)^(1/2)
∫---------------dx          and you are trying to solve this integral using x=cosθ ⇒ dx=-sinθdθ
  (1-x)^(1/2)

  (-sinθ)(1+cosθ)^(1/2)
∫-------------------------dθ     multiply everything by (1+cosθ)^(1/2)
  (1-cosθ)^(1/2)

  (-sinθ)(1+cosθ)^(1/2)*(1+cosθ)^(1/2)
∫--------------------------------------------dθ
  (1-cosθ)^(1/2)*(1+cosθ)^(1/2)

  (-sinθ)(1+cosθ)
∫--------------------------dθ
  (1-cos²θ)^(1/2)

  (-sinθ)(1+cosθ)
∫------------------dθ
        sinθ

i guess you can take it from here!!  i may have done something wrong so please √√

Rod: hello in russian is "привет"

и не, я не говорю по-русски, но я могу притвориться!

http://www.freetranslation.com/

uh oh!  you are not allowed to post your email.  you gotta send in your birth certificate to rod proving you are of a certain age first

and as for your query.... i guess it would be (4!)^(6!)^(8!)^(9!)

my calculator can't even handel 2^9!

one use is in optimization problems.  if you want to find the minimum cost of constructing, oh i don't don't, say a can, you would probably use partial derivaties.

i'm sure someone in the field can give you a better use for it.