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Similar problem, use same method:

https://web2.0calc.com/questions/algebra_9719

symbolab.com is a really good online calculator!

https://web2.0calc.com/questions/help_90174

https://web2.0calc.com/questions/counting-problem_32
or
https://web2.0calc.com/questions/help-counting_19

I am just looking at your solution.   

Now I have finished looking and I really like it.  Thanks.

from 1 and 2

\((f'(0))^2=f(0)f''(0)\\ (f'(0))^2=f''(0)\\~\\ so\\ f(0)=1,\qquad f''''(0)=9\qquad f''(0)=(f'(0))^2 \qquad find\;\; f'(0) \\\)

\((f'(x))^2=f(x)f''(x)\\ \text{diff wrt x}\\ 2(f'(x))(f''(x)) =f(x)f'''(x) +f'(x)f''(x)\\ \text{diff again wrt x}\\ 2(f'(x)(f'''(x)) + 2(f''(x)(f''(x)) =f(x)f''''(x) +f'(x)f'''(x) +f'(x)f'''(x) +f''(x)f''(x)\\ f''(x)f''(x) =f(x)f''''(x) \\ sub\;\;x=0\\ f''(0)f''(0) =f(0)f''''(0) \\ (f'(0))^4 =1*9 \\ f'(0)=\pm \sqrt[4]{9}\\ f'(0)=\pm \sqrt{3}\\\)

LaTex:

(f'(0))^2=f(0)f''(0)\\
(f'(0))^2=f''(0)\\~\\
so\\
f(0)=1,\qquad f''''(0)=9\qquad f''(0)=(f'(0))^2 \qquad find\;\; f'(0) \\

(f'(x))^2=f(x)f''(x)\\
\text{diff  wrt x}\\
2(f'(x))(f''(x))    =f(x)f'''(x)    +f'(x)f''(x)\\

\text{diff  again wrt x}\\
2(f'(x)(f'''(x))  + 2(f''(x)(f''(x))  =f(x)f''''(x) +f'(x)f'''(x)   +f'(x)f'''(x)  +f''(x)f''(x)\\
 f''(x)f''(x)  =f(x)f''''(x)    \\
sub\;\;x=0\\
f''(0)f''(0)  =f(0)f''''(0)    \\
(f'(0))^4 =1*9   \\
f'(0)=\pm \sqrt[4]{9}\\
f'(0)=\pm \sqrt{3}\\

Hi Guest,

I do believe that Alan's answer is the correct one but I would like to understand what you have tried to do.

I think you have accidentally omitted an x in the second row but that is no bigf deal

When you say 'and with more bounding'  I have no idea what you mean.

sumfor(n, 1, 12, 1/(2^n)) ==4095 / 4096

That is a great solution Alan, which means I can easily understand it.  Thanks 

30 miles / 45 m/hr =  2/3 hr = 40 minutes

15 minutes earlier would be 25 minutes to get there  ( 5/12 hr)

30 miles / x m/hr = 5/12 hr 

x =    30 * 12/5 = 72 m/hr

5/18  * 27 = 7.5 minutes

10 oclock hour twice a day = 120 minutes

00 - 10      twice  a day for each of 22 hours     484 mins

20 30 40 50 twice a day for 22 hours  176 minutes

780 minutes out of 1440      =   13/24    (I think!)

Yep....thought it said ninth when I was answering.... thanx for the notice !

Check your arithmetic.  That would be the 9^th term. 

Consider the geometric sequence  3, 9, 27, 81, .... 

Find the eighth term of the sequence.

Express your answer as a common fraction. 

Let's write that same sequence as 3¹, 3², 3³, 3⁴, ...  

so the eighth term would be 3⁸ which is 6561. 

_.

I tried comparing this to choosing \(n\) diameters for a circle. That clearly forms subsets of the circle (i.e. segments between the \(2n\) points of intersection between the diameters and the circumference)...

Maybe that can help answer the original question?

The other person's answer is incorrect 

Joe spent 50 worth = $72.50

Number of 50 cent coins spent = 72.50/0.50 = 145
Number of 20 cent coins he spent = 5/7 * 7/8x = 5/8x

(7/8x - 5/8x) + (1/8x - 145) = 2/7x

x = 1624

Number of 20 cent coins left = (7/8x - 5/8x) = 2/8x = 406

Number of 59 cent coins left = (1/8x - 145) = (1/8 * 1624 - 145) = 58

Amount Joe had left = 406 * 0.20 + 58 * 0.50 = $110.20

This is the correct answer

We know that all the 4 digit palindromes are divisible by 11 so that can be our 2 digit prime number. Since the other factor is a 3 digit, we can only look be looking at palindromes between 1100 and 2000. With that information we can easily say thatthe last digit of the 3 digit number is 1. The 3 digit numbers: 101, 111...,181, we find that there are 4 3 digit prime numbers. We know this because when we multiply by 11 we get palindromes.
 

Okay beware that im here it willjust take some time to type it

didn't you post this here?
https://web2.0calc.com/questions/paper-squares

Just asking

Hmmm Thx for your time, but how??

Search your question up before asking. Many users have good explanations

https://web2.0calc.com/questions/suppose-that-a-b-and-c-are-non-zero-distinct-digits

It should be 4

https://web2.0calc.com/questions/help-need_2

Here is a good explanation!

Whats the question here? You just posted an equation

464 is 2/7 of 1624