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Yes where are the questions
Solve this equation for x. Round your answer to the nearest hundredth. 0.59 = log x
assuming log is base 10
10^.59 = x
i got 3.890
that is correct... or so the calculator tells me :)
6 = In(x+4)
e^6 = x+4 e^6 -4 = x
now.. here is a type i'm stuck on
You invest $4200 in an account that pays an interest rate of 6.5%, compounded continuously. Calculate the balance of your account in 10 years. round your answer to the nearest hundredth
to compound continuosly means we use pert for our answer.. 4200(e^(.065)(10))
8045.271 is what i get... but i have no 'e' button
e generally equal 2.71828
for that i got 8045.268
you got a choice of answers to go from?
no. i have to type in an answer
does the material you are working from give you an approximation for 'e' to use?
no but i searched it & was also told that almost every time e = 2.71828
4200 * (e^.65) = 8045.27148 when i type it into google, the calc pops this out...
id go with: 8045.27 or .28
i went with .27
nextt question: holly wants to save money for an emergency. holly invests $1200 in an account that pays an interest rate of 4%. how many years will it take for the account to reach $15,400? round your answer to the nearest hundredth
Remember to give amistre a medal if they helped you out!
was already planning on doing that
4% interest is SIMPLE INTEREST or COMPOUND INTEREST???
i really don't know. it doesn't say. some have said compounded continuously. but this one doesn't
they can also forget to give me a medal.... i got plenty to spare ;)
if its over a year, then compund interest is applied usually
Using Simple Interest, the time comes out way too long, must be Compound interest
savings accounts are usualy compounded yearly...
i do believe it is compounded
Pls check and reconfirm the numbers in the question. The given ones don't seem to b right
those are correct
The given numbers do not work for Compound interest So, for Simple interest, the answer is 295.83 years (Way too long!!??) This one beats me:( Anyone else take a shot at it??
i'll post the question again as another question & see if anyone will answer it. i'm confused on these ones.