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what is 1000(1+0.09/12)^4.5=$1,034.20. Are you sure the 4.5 isn't 4.5 X 12?

A CHALLENGE TO YOU

I have just been playing around with a harder version of this question.

Say you have all that info EXCEPT you do not know how many people attended.

 In PTA fundraising activity, couples who came were charged $400. However if only one parent came, he was charged $250.00.

59 People attended the activity and $12,150 was raised. How many couples attended?

You want to know all the possibilities of how many singles and how many couples attended.

I mean all the possible pairs of numbers in order for EXACTLY  $12,150 to be raised.

Can you do that ?

I found exactly 6 posibilities.  Can you find ANY or all of those?

GOOD LUCK :))

The car depreciates by 1/1.012=0.98814229, then:

Log(3,000/12,500)=nLog(0.98814229)

-0.61978875=n X -0.005180513

n=119.64 months. Anything > this will depreciate the car just under $3,000. In fact, at 120 months will be worth=$2,987.09, exactly.

Just multiply 73 by  1.7699999999...........and that will be your answer

(2x + 7)^2  - 49   =  (2x + 7) (2x + 7 )   - 49   =  

           2x + 7 

           2x + 7

           --------

           14x  + 49

4x^2 + 14x

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

4x^2 + 28x + 49

And subtracting 49 from this     gives us    4x^2 + 28x + 49 - 49

If it depreciates every month at 1.2%...then each successive month, it must be worth only   98.8%  = .988  of what it was the prior month......so we have

3000  =  12,500 (.988)^m        divide both sides by 12500

30/125  = (.988)^m     take the log of both sides

log (30/ 125)  = log .988 ^m      and we can write

log(30/ 125)   = m log .988      divide both sides by log.988

log(30 / 125)  / log(.988) = m  = about 118.2 months.....actually....  m > 118.2 months since,at any time after this, it will be worth less than $3000

You purchase a new vehicle for 12, 500 dollars. This particular vehicle depreciates at a rate of 1.2%. How many months will it take for your new vehicle to be worth LESS THAN $3,000?

Is it 1.2% per month? If yes, then the car will depreciate to just under $3,000 in 120 months or 10 years. Use: 12,500/ (1.012)^n=< $3,000. Solve for n=120.

1200 nCr 4 =85,968,659,700=1200! / [(1200 - 4)!.4!]

a 50.0 kilogram cheetah has a kinetic energy of 18000 joules. How fast is the cheetah running

KE=1/2.m.v^2, where m=mass, v=velocity

18000=1/2.50.v^2

v^2=720 take the sqrt of sides

v=26.83m/second=96.588Km/hour.

Hi, Hayley. My name is Olivia.

OK, jc: I think I know what you did wrong. Look below:

1-year CD that pays 3% compounded semi-annually.     $22,811.55 X (1.015)^2=$23,501.03

2-year CD that pays 4% .............................................    $21,711.32  X (1.02)^4  =same amount

3-year CD that pays 5%..............................................    $20,264.86  X (1.025)6  =same amount

4-year CD that pays 6%..............................................    $18,551.93  X (1.03)^8  =same amount

5-year CD that pays 7%..............................................    $16,660.32  X (1.035)^10=same amount

Hi Hayley, hi Riddler    

Just beat me to it cphill

jc: I don't think you understand the problem!. Each CD that he/she purchased at the amounts calculated by Melody will mature with exactly the same amount, which is $23,501.03. That is what an "annuity" is.

This person could have easily gone to an Insurance Company and purchased a 5-year annuity, with equal payments being sent to him/her. If the Insurance Co. agreed and told him/her that they would receive 5 payments each for $23,501.03, then that would be exactly like this investment.

Hello :)!! i'd call you sir or mam.. but i dont no your age so, i wouldnt want to be rude. :P

Assuming C=90  A=50 b/c all angls of a traingle are equal to 180 degrees so 90+40=130. 180-130=50

I don't know about b and c though

Well.....I'm assuming a is a leg.... we have....with the Law of Sines

6/ sin 50   = b/ sin 40      [ note if B = 40, A must  = 50]

b =  6sin40 / sin50  = about 5.035

And the   hypotenuse  = c =   sqrt [6^2 + 5.035^2]  = about  7.833

Nice to meet you Hayley!

Thanks, Guest.......here's a sightly easier appproach  using one variable

Call the number of 4 kilos he sold, x       and the number of 3 kilos,  30 - x     and we have

4x + 3(30 - x)  = 100   simplify

x + 90   = 100     subtract 90 from both sides

x = 10     = the number of 4 kilos sold          30 - x   = 30  - 10   = 20    =   the number of 3 kilos sold 

Solve for x:
x^4-4 x^2 = 0

The left hand side factors into a product with three terms:
x^2 (x-2) (x+2) = 0

Split into three equations:
x-2 = 0 or x^2 = 0 or x+2 = 0

Add 2 to both sides:
x = 2 or x^2 = 0 or x+2 = 0

Take the square root of both sides:
x = 2 or x = 0 or x+2 = 0

Subtract 2 from both sides:
Answer: | | x = 2    or x = 0    or x = -2
 

What's what about?

What's what about?

radium has a half life of 1600 years. what percentage of a sample would be left after 2.07*10^3 years ?

2070/1600=1.29375 half-lives, therefore,

2^-1.29375=0.40789.....X 100=40.789% of the original amount left.

Geno3141: Please, always bear in mind Occam's razor!.