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6C3 * (0.5)^3 * (1-0.5)^3 =

The first derivative gives the gradient of the tangent.

At stationary points the gradient must be 0.

So find the first derivative (differentiate it) and put that equal to 0.

Solve this equaton and you will have x value. 

Sub that into the initial f(x) equation to get the y value.     In this case y is called f(x).

Do as much as you can of that and then I can help more.

(a)

Two candles of equal length are lit at the same time.

One candle takes 6 hours to burn out, and the other takes 9 hours to burn out.

After how much time will the slower burning candle be exactly twice as long as the faster burning one?

$\begin{array}{|rcll|} \hline \text{Rest fast candle}&=& 1Candle - \dfrac{1candle}{6h}\times t \\ \text{Rest slow candle}&=& 1Candle - \dfrac{1candle}{9h}\times t \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline 2\times \text{Rest fast candle } &=& \text{Rest slow candle} \\\\ 2\times \left( 1Candle - \dfrac{1candle}{6h}\times t\right) &=& 1Candle - \dfrac{1candle}{9h}\times t \\\\ 2 - \dfrac{2}{6}\times t &=& 1 - \dfrac{1}{9}\times t \\\\ 2 - \dfrac{1}{3}\times t &=& 1 - \dfrac{1}{9}\times t \\\\ \dfrac{1}{3}\times t- \dfrac{1}{9}\times t &=& 1 \\\\ \dfrac{(9-3)}{3*9}\times t &=& 1 \\\\ \dfrac{6}{3*9}\times t &=& 1 \\\\ \dfrac{2}{9}\times t &=& 1 \\\\ t &=& \dfrac{9}{2} \\\\ \mathbf{t} &=& \mathbf{4.5~ hours} \\ \hline \end{array}$

heureka: Isn't this the way it is supposed to work?

1,2,3,4 =3,5,7 =8,1, 2 =9,3 =12  ??

My not-so-short computer code agrees with Melody's number:

a=0;b=0;c=0;d=0;e=0;f=0;g=0;p=0; cycle:n=a*1000000+b*100000+c*10000+d*1000+e*100+f*10+g; if(a*b*c*d*e*f*g==10000, goto loop, goto next);loop:printn,", ",;p=p+1; next:g++;if(g<10, goto cycle, 0);g=1;f++;if(f<10, goto cycle, 0);g=1;f=1;e++;if(e<10, goto cycle,0);g=1;f=1;e=1;d++;if(d<10, goto cycle,0);g=1;f=1;e=1;d=1;c++;if(c<10, goto cycle,0);g=1;f=1;e=1;d=1;c=1;b++;if(b<10, goto cycle,0);g=1;f=1;e=1;d=1;c=1;b=1;a++;if(a<10, goto cycle,0);print"Total = ",p

OUTPUT: 

 1255558 , 1255585 , 1255855 , 1258555 , 1285555 , 1445555 , 1454555 , 1455455 , 1455545 , 1455554 , 1525558,.............................  8525551 , 8551255 , 8551525 , 8551552 , 8552155 , 8552515 , 8552551 , 8555125 , 8555152 , 8555215 , 8555251 , 8555512 , 8555521 , Total =  420

I have a question.

How many times do people need to ask the same question in the space of 1 day?

Obviously the answer must be some degree of infinity.     

The light just never quite comes on.

5,1,1 

Means 1 person got 5 and the other 2 people got 1.

There are 3 ways that this can happen becasue the 5 can go to person A,B or C.

then the other two get the same number each.

4,2,1

Means 1 person got 4, one got 2 and 1 got only one.

The 4 can go to person AB or C  that is 3 possible ways, then there are two people left. Either of those can get the single card, so that is 2 ways, the last person gets the others. So that is 3*2=6 ways.

So far I have not paid attention to the fact that the cards are different from each other. I dealt with that afterwards.

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Try digesting that and then if you need me to I can continue to explain the next bit.

Are you saying that this first answer is wrong?  Why do you believe this to be the case?

You must have a reason. You should fully state it.

Mack the bug starts at (0,0) at noon and each minute moves one unit right or one unit up. He is trying to get to the point (5,7). However, at (4,3) there is a spider that will eat him if he goes through that point. In how many ways can Mack reach (5,7)?

I get 617 ways. Can you please confirm if you think this is correct or not.

If you believe you know the answer then please state what it is.

Here is an outline of my logic.

There are 35 ways to get from (0,0) to (4,3)

There are 5 ways to get from (4,3) to (5,7)

So there are 35*3=175 ways to get from (0,0) to (5,7) passing through (4,3)

There are 792 ways to get from (0,0) to (5,7)

So there must be 792-175=617 ways to get from (0,0) to (5,7) without passing through (4,3)

My logic seems ok to me.

Kevin will start with the integers 1, 2, 3 and 4 each used exactly once
and written in a row in any order.
Then he will find the sum of the adjacent pairs of integers in each row
to make a new row, until one integer is left.
For example, if he starts with 3, 2, 1, 4, then he takes sums to get 5, 3, 5,
followed by 8, 8, and he ends with the final sum 16.
Including all of Kevin's possible starting arrangements of the integers 1, 2, 3 and 4,
how many possible final sums are there?

I assume  5 possible final sums $\{16,~ 18,~ 20,~ 22,~ 24 \}$

$\begin{array}{|cccc|} \hline \text{Start} & a & & b & & c & & d \\ \text{First sum} & & a+b & & b+c & & c+d \\ \text{Second sum} & & & a+2b+c & & b+2c+d & & \\ \text{Final sum} & & & &a+b+c+d+2(b+c)& & & \\\hline \end{array}$

$\begin{array}{|rcll|} \hline \text{Final sum} &=& a+b+c+d+2(b+c) \quad | \quad a+b+c+d=1+2+3+4= 10\\ \mathbf{\text{Final sum}} &=& \mathbf{10+2*(b+c)} \\ \hline \end{array}$

$\mathbf{b+c=\ ?}$

$\begin{array}{|c|c|c|c|c|c|} \hline \mathbf{b+c} & c= & 1 & 2 & 3 & 4 \\ \hline b=& & & & & \\ \hline 1 && - &3 & 4 & 5 \\ \hline 2 && 3 &- & 5 & 6 \\ \hline 3 && 4 &5 & - & 7 \\ \hline 4 && 5 &6 & 7 & - \\ \hline \end{array}$

$\begin{array}{|c|c|c|c|} \hline & b+c & 2x & +10 \\ \hline 1 & 3 & 6 & 16 \\ 2 & 4 & 8 & 18 \\ 3 & 5 & 10 & 20 \\ 4 & 6 & 12 & 22 \\ 5 & 7 & 14 & 24 \\ \hline \end{array}$

How many ways are there to form a 7-digit code where each digit can be from 0 to 9 and the product of all the digits in the code is 10,000?

Thanks Geno and Mathmathj28,

Since there is disagreement I will add my 10cents worth.

factor(10000) = 2^4*5^4

So I can only use 2,4,8,5,1  and 5 must be used 4 times

10000/5^4 = 16

5 5 5 5 

I need 3 digits that multiply to 16       8,2,1       4,4,1     4,2,2,  and that is it.

So the digits can be

5 5 5 5 8 2 1         7!/4!         =210

5 5 5 5 4 4 1          7!/(4!2!)  = 105

5 5 5 5 4 2 2          7!/(4!2!)  =  105

210+105+105 = 420 ways

I got more than 10.  Draw it and count them for yourself.  

Pooja is the hypotenuse of a right triangle with 40 foot and 9 foot legs.

By Pythagoras' Theorem:   hypotenuse²  =  40² + 9²    

                                           Pooja's height  =  square root of 1681       

                                           Pooja's height  =  41 feet 

_.

Actually, what I meant is, how did you find the ways?

51 coins

x  = #dimes            amount   = .10 x

51-x = # quarters   amount = .25 (51-x)          added together   = 10.35

.10x    +  .25(51-x)   = 10.35      solve for x    which is the number of dimes. then #quarters = 51-x

When you start walking, your friend has been walking 5 minutes (1/12 hr)  at 3 m/hr

Rate  x time = distance

3 m/hr  x 1/12/hr = 3/12 mile = 1/4 mile

Total solution will be 500 ml

  500 ml of 60%   =  500 (.6) = 300 ml of 100%

 75 % solution is   x           .75 x   is amount

25 % solution                     .25 (300-x)

.75x   +   .25 (300 - x)     = 300    Solve for 'x' = amount of 75% solution to use

The area of a hexagon  ABCDEF = 39.84 u²

Wouldn't 5 x 5 x 5 x 2 x 2 x 2 x 1 be the prime factorization for 1,000?

The combinations of digits that I found were 5 x 5 x 5 x 5 x 2 x 2 x 4, 5 x 5 x 5 x 5 x 4 x 4 x 1, and 5 x 5 x 5 x 5 x 8 x 2 x 1

I'm assuming that this: $\frac79$ and $\frac45$, means: 7/9 and 4/5 ??

If so, the nth term =First term + [common difference * Term number - 1]

The 13th term =7/9 + [D * (13 - 1)]

4/5 =7/9 + 12D, solve for D

D = 1 / 540

7th term =7/9 + 1 /540 * 6

7th term = 71 / 90

The number of ways is 18*17*16 - C(20,2) = 4706.

The prime factorization of 10,000  =  5 x 5 x 5 x 2 x 2 x 2 x 1      (using 7 digits).

So the 7 digit possibilities are  5 x 5 x 5 x 2 x 2 x 2 x 1

a)  5 x 5 x 5 x 2 x 2 x 2 x 1

b)  5 x 5 x 5 x 4 x 2 x 1 x 1

c)  5 x  5 x 5 x 8 x 1 x 1 x 1

There are  7! / (3! x 3!)  =  140 ways to do part a.

There are  7! / (3! x 2!)  =  420 ways to do part b.

There are  7! / (3! x 3!)  =  140 ways to do part c.

Adding, there are 700 ways.

Does this mean: $7\cdot24\cdot48$ =7 * 24 * 48 ??.If so, then you have:

=8064 =2^7 * 3^2 * 7

So, y =2^2 * 3 * 7^2 =588, so that:

8064 * 588 =4,741,632^(1/3) = 168 - and is a perfect cube.

That makes sense, thank you so much for explaining it!!