jagatuba
  • jagatuba
Can anyone show me how to prove two equations are equivalent?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
which equations?
jagatuba
  • jagatuba
L*[c/(1- (1 + c) ^ -n) ] and L*[c(1 + c)^n]/[(1 + c)^n - 1]
anonymous
  • anonymous
???

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

jagatuba
  • jagatuba
Bot of those expressions should equal P
jagatuba
  • jagatuba
In other words: P = L*[c/(1- (1 + c) ^ -n) ] and P = L*[c(1 + c)n]/[(1 + c)n - 1]
anonymous
  • anonymous
L , c, n are variables ???
jagatuba
  • jagatuba
Yes I get as far as eliminating L by division, but then I get stuck.
anonymous
  • anonymous
Could you use the equation button to express it in a better way....I m pretty gud at solving equations...but its just written weirdly...
jagatuba
  • jagatuba
Yeah I know what you mean. I'll try. I was messing with the equation button last night but was not getting the equations to look right. lol Give me a few
anonymous
  • anonymous
ok
jagatuba
  • jagatuba
\[P=L \times \left[ c \div \left( 1-\left( 1+c \right)^{-n} \right) \right]\] \[P=L \times \left[ \left( c \times \left( 1+c \right)^{n} \right) \div \left( \left( 1+c \right)^{n} \right)-1\right]\]
jagatuba
  • jagatuba
Sorry it took so long
anonymous
  • anonymous
no prob...let me try
anonymous
  • anonymous
heyyyyyyyy
anonymous
  • anonymous
so simple ...it was...
jagatuba
  • jagatuba
I don't know if this will help you at all but these are two equations that are used to determine the monthly payment on a mortgage loan (P=monthly payment). I know they do come up with the same results I just have to prove mathematically
anonymous
  • anonymous
in the first equation, make (1+c)^ -n = 1/ (1+c)^ n and then take the lcm
jagatuba
  • jagatuba
(1=c)?
jagatuba
  • jagatuba
'Scuse me (1+c)?
anonymous
  • anonymous
yes in the first equation...
jagatuba
  • jagatuba
I could write the right side as 1/(c+1)(c+1) but how would I write the left side with it's negative exponent?
anonymous
  • anonymous
but that was (1+c)^n ...wasnt it?
jagatuba
  • jagatuba
-n is the exponent
jagatuba
  • jagatuba
Wait now I'm getting confused. lol
anonymous
  • anonymous
OK...let me make sure again....we are trying to prove that first equation equals to second equation...right??
anonymous
  • anonymous
yes...I was right...
anonymous
  • anonymous
in the first equation...take the lcm in the denominator
jagatuba
  • jagatuba
Okay give me a minute to work this out on paper and I'll post again.
anonymous
  • anonymous
which will eventually go to numerator
jagatuba
  • jagatuba
SO after dividing both sides of the equation by L I get: \[\left( c \div \left( 1-\left( 1+c \right)^{-n} \right) \right)=\left( \left( c \times \left( 1+c \right)^{n} \right)\div \left( \left( 1+c \right)^{n} \right)-1 \right)\] Next I take care of the negative exponent like this: \[c-\left( 1+c \right)^{n }=\left( c \times \left( 1+c \right)^{n} \right)\div \left( \left( 1+c \right)^{n}-1 \right)\] Is this right so far?
jagatuba
  • jagatuba
Actually that right side should be \[c \times -\left( 1+c \right)^{n}\]
jagatuba
  • jagatuba
But now I'm lost again. Dang it.
anonymous
  • anonymous
u have got it !!!
jagatuba
  • jagatuba
But if I divide each side by \[-\left( 1+c \right)^{n}\] I still end up with: \[c =c \div \left( \left( 1+c \right)^{n}-1 \right)\]
jagatuba
  • jagatuba
That can't be right can it?

Looking for something else?

Not the answer you are looking for? Search for more explanations.