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On a certain college faculity, 4/7 of the professors are male, and the ratio of the professors older than 50 to the professors less than or equal to 50 years is 2:5. if 1/5 of the male professors are older than 50 years, what fraction of the female professors are less than or equal to 50 years ?

\(\begin{array}{rcl} m_{>50} &=& \frac47\cdot \frac15 = \frac{4}{35} \\ m_{\le50} &=& \frac47\cdot \frac45 = \frac{16}{35} \\ m_{>50} + m_{\le50} &=& \frac{4}{35} + \frac{16}{35} = \frac{20}{35} \\\\ f_{>50} + f_{\le50} &=& 1- ( m_{>50} + m_{\le50} ) = 1- \frac{20}{35} = \frac{15}{35} \\ f_{>50} &=& \frac{15}{35} - f_{\le50} \\\\ \frac{m_{>50}+f_{>50}}{m_{\le50}+f_{\le50}} &=& \frac25 \\ 5\cdot ( m_{>50}+f_{>50} ) &=& 2\cdot ( m_{\le50}+f_{\le50} ) \qquad | \qquad f_{>50} = \frac{15}{35} - f_{\le50} \\ 5\cdot ( m_{>50}+\frac{15}{35} - f_{\le50} ) &=& 2\cdot ( m_{\le50}+f_{\le50} ) \\ 5\cdot ( \frac{4}{35}+\frac{15}{35} - f_{\le50} ) &=& 2\cdot ( \frac{16}{35}+f_{\le50} ) \\ 5\cdot ( \frac{19}{35} - f_{\le50} ) &=& 2\cdot ( \frac{16}{35}+f_{\le50} ) \\ \frac{95}{35} - 5\cdot f_{\le50} &=& \frac{32}{35}+2\cdot f_{\le50} \\ 7\cdot f_{\le50} &=& \frac{95}{35} - \frac{32}{35} \\ 7\cdot f_{\le50} &=& \frac{63}{35} \\ f_{\le50} &=& \frac{63}{35\cdot 7 } \\ \text{female}_{\le50} &=& \frac{9}{35} \\ \end{array}\)

i found 9/35

Bertie

I don't have a complete solution either but this is what I have so far.

\(\displaystyle f(n) = 9^{n+1}-8n-9\),   

so    

\(\displaystyle f(n+1) = 9^{n+2}-8(n+1)-9\) ,

in which case

\(\displaystyle \frac{f(n+1)}{f(n)}=\frac{9-\{8(n+1)+9\}/9^{n+1}}{1-\{8n+9\}/9^{n+1}}\),

having divided top and bottom by \(\displaystyle 9^{(n+1)}\).

That means that for increasing n, \(\displaystyle f(n+1) \approx 9f(n) \text{ or } f(n+1) \approx 2.08^{3}f(n)\) ,

and that implies that for each increase of 1 in n, m will increase by 3.

That effect seems to kick in early,

f(1) = 64 = 2^6, so m = 6

f(2) = 704 = 2^9 + 192, so m = 9

f(3) = 6528 = 2^12 + 2432 so m = 12

f(4) = 59008 = 2^15 + 26420 so m = 15

That in turn suggests the rule

m = 3n+3.

I've not yet convinced myself whether or not that this is true for all n.

-Bertie

Here's my attempt (basically the same as heureka's):

I tried to upload this from my phone yesterday (I was away from my computer), but with no success!

what is 111...1(total of one hundred ones) divided by 1,111,111 ( answer with remainder, not fraction or decimal )

You can divide the number in  arbitrarily sections.

Devide all parts by 1 111 111, but attach the remainder of the previous calculation left.

Example:

\(\underbrace{1111111}_{1.\ \text{partition}}\ \underbrace{1111111}_{2.\ \text{partition}}\ \underbrace{1111111}_{3.\ \text{partition}}\ \underbrace{1111111}_{4.\ \text{partition}}\ \underbrace{1111111}_{5.\ \text{partition}}\ \underbrace{1111111}_{6.\ \text{partition}}\ \underbrace{1111111}_{7.\ \text{partition}}\ \\ \underbrace{1111111}_{8.\ \text{partition}}\ \underbrace{1111111}_{9.\ \text{partition}}\ \underbrace{1111111}_{10.\ \text{partition}}\ \underbrace{1111111}_{11.\ \text{partition}}\ \underbrace{1111111}_{12.\ \text{partition}}\ \underbrace{1111111}_{13.\ \text{partition}}\ \underbrace{1111111}_{14.\ \text{partition}}\ \underbrace{11}_{15.\ \text{partition}}\ \)

\(\begin{array}{lcl} \underbrace{1111111}_{1.\ \text{partition}} : 1\ 111\ 111 = 1 \quad \text{ Remainder } = \color{red}{0} \\ \text{attach Remainder}\\ {\color{red}0}\underbrace{1111111}_{2.\ \text{partition}}: 1\ 111\ 111 = 1 \quad \text{ Remainder } = \color{green}{0} \\ \text{attach Remainder}\\ {\color{green}0}\underbrace{1111111}_{3.\ \text{partition}}: 1\ 111\ 111 = 1 \quad \text{ Remainder } = 0 \\ \text{attach Remainder}\\ 0\underbrace{1111111}_{4.\ \text{partition}}: 1\ 111\ 111 = 1 \quad \text{ Remainder } = 0 \\ \text{attach Remainder}\\ 0\underbrace{1111111}_{5.\ \text{partition}}: 1\ 111\ 111 = 1 \quad \text{ Remainder } = 0 \\ \text{attach Remainder}\\ 0\underbrace{1111111}_{6.\ \text{partition}}: 1\ 111\ 111 = 1 \quad \text{ Remainder } = 0 \\ \text{attach Remainder}\\ 0\underbrace{1111111}_{7.\ \text{partition}}: 1\ 111\ 111 = 1 \quad \text{ Remainder } = 0 \\ \text{attach Remainder}\\ 0\underbrace{1111111}_{8.\ \text{partition}}: 1\ 111\ 111 = 1 \quad \text{ Remainder } = 0 \\ \text{attach Remainder}\\ 0\underbrace{1111111}_{9.\ \text{partition}}: 1\ 111\ 111 = 1 \quad \text{ Remainder } = 0 \\ \text{attach Remainder}\\ 0\underbrace{1111111}_{10.\ \text{partition}}: 1\ 111\ 111 = 1 \quad \text{ Remainder } = 0 \\ \text{attach Remainder}\\ 0\underbrace{1111111}_{11.\ \text{partition}}: 1\ 111\ 111 = 1 \quad \text{ Remainder } = 0 \\ \text{attach Remainder}\\ 0\underbrace{1111111}_{12.\ \text{partition}}: 1\ 111\ 111 = 1 \quad \text{ Remainder } = 0 \\ \text{attach Remainder}\\ 0\underbrace{1111111}_{13.\ \text{partition}}: 1\ 111\ 111 = 1 \quad \text{ Remainder } = 0 \\ \text{attach Remainder}\\ 0\underbrace{1111111}_{14.\ \text{partition}}: 1\ 111\ 111 = 1 \quad \text{ Remainder } = 0 \\ \text{attach Remainder}\\ 0\underbrace{11}_{15.\ \text{partition}}: 1\ 111\ 111 = 0 \quad \text{ Remainder } = \mathbf{11} \\ \end{array}\)

Thanks guest #9.    I have edited my answer.     

press the slash button. for ex. 5/7

10^(0.3333(log22+log6.25+log4.2-log7))

\(10^{(0.3333(log22+log6.25+log4.2-log7))}\\~\\ =10^{0.3333log22+0.3333log6.25+0.3333log4.2-0.3333log7)}\\~\\ =10^{0.3333log22}*10^{0.3333log6.25}*10^{0.3333log4.2}*10^{0.3333log7}\\~\\ =10^{log22^{0.3333}}*10^{log6.25^{0.3333}}*10^{log4.2^{0.3333}}*10^{log7^{0.3333}}\\~\\ =22^{0.3333}*6.25^{0.3333}*4.2^{0.3333}*7^{0.3333}\\~\\ =(22*6.25*4.2*7)^{0.3333}\\~\\ =(4042.5)^{0.3333}\\~\\ \approx 15.92562388\)

（x − 1/y）/ （x/y） = x（y-1）/x

\(\frac{（x − 1/y）}{ （x/y）} =\frac{ x（y-1）}{x}\\ Note: x\ne0 \qquad y\ne 0\\ \frac{xy − 1}{y}\div\frac{x}{y} =\frac{ x（y-1）}{x}\\ \frac{xy − 1}{y}\times \frac{y}{x} =\frac{ x（y-1）}{x}\\ \frac{xy − 1}{1}\times \frac{1}{x} =\frac{ 1（y-1）}{1}\\ \frac{xy − 1}{x} =y-1\\ xy − 1=xy-x\\ -1=-x\\ x=1\\ \mbox{So this is the line x=1 except it has a 'hole' at y=0} \)

That was interesting :)

3^x = 12       take the log of both sides

log 3^x  = log 12       and we can write

x * log 3   = log 12      divide both sides by log 3

x =   log 12  / log 3  =  about  2.262

If the growth rate doubled every week, it would grow by 100% per week.

Since the growth rate triples every week, it grows by 200% per week.

This is easy to see.....

If 1 person is sick at the start.......by the end of the first week, 2 more would be sick. And 2 is 200% of 1. So, 3 people are sick at the end of the first week- triple that of the start.

At the end of the second week,  6 more would be sick , and 6 is 200% of 3. So, 9 people total are sick at the end of the second week - triple that of the beginning of the second week.......etc.

Hallo aditya@calc.com,

sorry i have no closed solution. Here my way so far.

\(\small{ \begin{array}{rcl} 3^{2n+2} -8n -9 &=& 3^{2n}\cdot 3^2-8n-9 \\ &=& 3^{2n}\cdot 9-8n-9 \\ &=& (3^2)^n\cdot 9-8n-9 \\ &=& 9^n\cdot 9-8n-9 \\ &=& 9^{n+1}-8n-9 \\ &=& (1+8)^{n+1}-8n-9 \\ &=& -8n-9 + (1+8)^{n+1}\\ &=& -8n-9 + \underbrace{\binom{n+1}{0}+ \binom{n+1}{1}\cdot 8}_{=8n+9} + \binom{n+1}{2}\cdot 8^2 + \binom{n+1}{3}\cdot 8^3 + \binom{n+1}{4}\cdot 8^4 +\cdots \\ && + \binom{n+1}{n-1}\cdot 8^{n-1} + \binom{n+1}{n}\cdot 8^{n} + \binom{n+1}{n+1}\cdot 8^{n+1} \\\\ &=& \binom{n+1}{2}\cdot 8^2 + \binom{n+1}{3}\cdot 8^3 + \binom{n+1}{4}\cdot 8^4 +\cdots + \binom{n+1}{n-1}\cdot 8^{n-1} + \binom{n+1}{n}\cdot 8^{n} + \binom{n+1}{n+1}\cdot 8^{n+1} \\\\ &=& \binom{n+1}{2}\cdot 2^6 + \binom{n+1}{3}\cdot 2^9 + \binom{n+1}{4}\cdot 2^{12} +\cdots + \binom{n+1}{n-1}\cdot 2^{3n-3} + \binom{n+1}{n}\cdot 2^{3n} + \binom{n+1}{n+1}\cdot 2^{3n+3} \\ \end{array} }\)

It is not easy to factorize the binoms \(\binom{n+1}{2},~\binom{n+1}{3},~\binom{n+1}{3}\cdots\)  to get the exponent of the prim number 2 for every n

but m mimimum is 6, because we have \(2^6\)

It seems that (no proof !!!!)

  \(\begin{array}{rcl} \binom{n+1}{2}\cdot 2^6 &=& 2^m\cdot p \\\\ (n+1)\cdot n \cdot 2^5 &=& 2^m\cdot p \\ \end{array}\)

Example:

\(\begin{array}{|r|c|l|} \hline n & m & p \\ \hline 1 & 6 & 1 \\ 2 & 6 & 3 \\ \hline 3 & 7 & 3 \\ 4 & 7 & 5 \\ \hline 5 & 6 & 15 \\ 6 & 6 & 21 \\ \hline 7 & 8 & 7 \\ 8 & 8 & 9 \\ \hline 9 & 6 & 45 \\ 10 & 6 & 55 \\ \hline 11 & 7 & 33 \\ 12 & 7 & 39 \\ \hline 13 & 6 & 91 \\ 14 & 6 & 105 \\ \hline 15 & 9 & 15 \\ 16 & 9 & 17 \\ \hline 17 & 6 & 153 \\ 18 & 6 & 171 \\ \hline \cdots \\ \hline \end{array}\)

LOL Eistein, M=infinity duh!

2sqrt(5)   =   5sqrt(2)      ????

Note that....if we squared both things, and they were equal to begin with, they should still be equal....so we have

4 * 5   = 25 * 2      ????

20  =   50    ????  ....  nope.....       thus.........2sqrt(5)   <  2sqrt(2)........!!!!!

F = 75 + 5m     ......   m in this case  = 50- 30   = 20........so we have

F = 75 + 5(20)  =  75 + 100   = $175

Notice the pattern:  [you may have to shrink your screen to see the complete results]

A string  of 20 1's   = 10000001000000* 1,111,111 +  111,111  

A string of 40 1s = 1000000100000010000001000000100000 * 1,111,111 +  11,111  

A string of 60 1's =

100000010000001000000100000010000001000000100000010000 * 1111111 + 1111  

80 1's    10 groups of 1000000 append 1000 * 1,111,111  +  111   

100 1's =    13 groups of 1000000 append100 * 1,111,111 + 11  

So, 100 1's   divided by 1,111,111  =

1000000100000010000001000000100000010000001000000100000010000001000000100000010000001000000100 +11/1,111,111

So....the remainder is 11

Maybe Heureka knows an easier method to calculate this.....????

Simplify the following:
1/(sqrt(2)+sqrt(3))

Multiply numerator and denominator of 1/(sqrt(2)+sqrt(3)) by sqrt(2)-sqrt(3):
(sqrt(2)-sqrt(3))/((sqrt(2)+sqrt(3)) (sqrt(2)-sqrt(3)))

(sqrt(2)+sqrt(3)) (sqrt(2)-sqrt(3)) = sqrt(2) sqrt(2)+sqrt(2) (-sqrt(3))+sqrt(3) sqrt(2)+sqrt(3) (-sqrt(3)) = 2-sqrt(6)+sqrt(6)-3 = -1:
(sqrt(2)-sqrt(3))/(-1)

Multiply numerator and denominator of (sqrt(2)-sqrt(3))/(-1) by -1:
-(sqrt(2)-sqrt(3))

-(sqrt(2)-sqrt(3)) = sqrt(3)-sqrt(2):
Answer: | sqrt(3)-sqrt(2)

some money was put away at 5% interest annualy. in the end he has $375 what did he start of with?

He started with $357.14

You use this formula to calculate this:

PV=FV(1 + R)^-N

PV=375(1.05)^-1

PV=$357.14

-2, -8, -32, -96, -384

You have to set up an equation to solve this problem. 

He burned 200 calories jogging + 5 calories for every minute swam = 11 calories for every minute jogged

200 + 5x = 11x 

Subtract 5x from both sides.

200 + 5x = 11x

200 + 5x - 5x = 11x - 5x

200 = 6x 

Divide both sides by 6.

200 = 6x

200 / 6 = 6x /6 

33.3 = x 

Round to the nearest whole minute. 

33.3 rounds up to 34, because he has to do at least 33.3 minutes, so you will round up to 34. 

So, he must jog at least 34 minutes. 

what's the question?

1.7 x 10⁷ mi

= 17000000 mi 

17000000 / 1420

= 11971.8 hrs  (rounded up) 

11971.8 / 24 

= 498.8 days (rounded up) 

or 

498 days and 19 hours (approximately) 

8.18535277187

I don't like the grapher in this calculator.......I use Desmos......easy to use and it will handle exponential functions........here's the website address  :  https://www.desmos.com/calculator