All Answers 352713 Answers

Suppose we have n points equally spaced around a circle. Then there are exactly

(n-2) different possible values for the angle PQR and they are 

i*(180/n) for 1<=i<=n-2

Like so:

Thanks Ginger  

Need some help in summing up the following sequence:
4/1.2.3 + 4/2.3.4 + 4/3.4.5 +..............+ 4/998.999.1000

Calculate:
\(\begin{array}{|lcll|} \hline \mathbf{ S_n = 4\cdot \left( \dfrac{1}{1 \cdot 2 \cdot 3} + \dfrac{1}{2 \cdot 3 \cdot 4} + \dfrac{1}{3 \cdot 4 \cdot 5} + \dfrac{1}{4 \cdot 5 \cdot 6} + \cdots \ + \dfrac{1}{n \cdot (n+1) \cdot (n+2)} \right) } \\ \mathbf{ S_n = 4\cdot \sum \limits_{k=1}^n \dfrac 1{k(k+1)(k+2)} } \\ \hline \end{array} \)

\(\text{Calculate $ S = \sum \limits_{k=1}^n \dfrac 1{k(k+1)(k+2)} $} \)

Formula:

\(\begin{array}{|lcll|} \hline \text{in general}:\ \dfrac{1}{n(n+d)} = \dfrac{1}{d}\left(\dfrac{1}{n}- \dfrac{1}{n+d} \right) \\ \hline \\ \begin{array}{lrcll} \text{we need}: & \dfrac{1}{(n+1)(n+2)} &=& \dfrac{1}{n+1}-\dfrac{1}{n+2} \\\\ & \dfrac{1}{n(n+1)} &=& \dfrac{1}{n}-\dfrac{1}{n+1} \\\\ & \dfrac{1}{n(n+2)} &=& \dfrac{1}{2} \left( \dfrac{1}{n}-\dfrac{1}{n+2} \right) \\ \end{array} \\ \hline \end{array}\)

We rearrange:

\(\begin{array}{|rcll|} \hline \dfrac{1}{n \cdot (n+1) \cdot (n+2)} \\\\ &=& \dfrac{1}{n}\times \dfrac{1}{(n+1) \cdot (n+2)} \\\\ &=& \dfrac{1}{n}\times \left( \dfrac{1}{n+1}-\dfrac{1}{n+2} \right) \\\\ &=& \dfrac{1}{n}\times \dfrac{1}{n+1} - \dfrac{1}{n}\times \dfrac{1}{n+2} \\\\ &=& \left(\dfrac{1}{n}-\dfrac{1}{n+1} \right)- \dfrac{1}{2} \times \left(\dfrac{1}{n} -\dfrac{1}{n+2} \right) \\\\ &=& \dfrac{1}{n} - \dfrac{1}{n+1} -\dfrac{1}{2n} + \dfrac{1}{2(n+2)} \\\\ \mathbf{\dfrac{1}{n \cdot (n+1) \cdot (n+2)} } & \mathbf{=} & \mathbf{ \dfrac{1}{2n} - \dfrac{1}{n+1} + \dfrac{1}{2(n+2)} } \\ \hline \end{array}\)

Telescoping series:

\(\begin{array}{|rcll|} \hline S_n &=& \mathbf{\dfrac{1}{2}} &\mathbf{-}& \mathbf{\dfrac{1}{2}} &\color{red}+& \color{red}\dfrac{1}{6} \\\\ &\mathbf{+}& \mathbf{\dfrac{1}{4}} &\color{red}-& \color{red}\dfrac{1}{3} &\color{blue}+& \color{blue}\dfrac{1}{8} \\\\ &\color{red}+& \color{red}\dfrac{1}{6} &\color{blue}-& \color{blue}\dfrac{1}{4} &\color{red}+& \color{red}\dfrac{1}{10} \\\\ &\color{blue}+& \color{blue}\dfrac{1}{8} &\color{red}-& \color{red}\dfrac{1}{5} &\color{green}+& \color{green}\dfrac{1}{12} \\\\ && \ldots \\\\ &+\color{red}& \color{red}\dfrac{1}{2(n-2)} &\color{green}-& \color{green}\dfrac{1}{n-1} &\color{red}+& \color{red}\dfrac{1}{2n} \\\\ &\color{green}+& \color{green}\dfrac{1}{2(n-1)} &\color{red}-& \color{red}\dfrac{1}{n} &\mathbf{+}& \mathbf{\dfrac{1}{2(n+1)}} \\\\ &\color{red}+& \color{red}\dfrac{1}{2n} &\mathbf{-}& \mathbf{\dfrac{1}{n+1}} &\mathbf{+}& \mathbf{\dfrac{1}{2(n+2)}} \\ \hline \end{array}\)

The part of each term cancelling with part of the next two diagonal terms:
Example:

\(\begin{array}{|lcll|} \hline \dfrac{1}{6}-\dfrac{1}{3}+\dfrac{1}{6} = 0 \\\\ \dfrac{1}{8}-\dfrac{1}{4}+\dfrac{1}{8} = 0 \\\\ \dfrac{1}{10}-\dfrac{1}{5}+\dfrac{1}{10} = 0 \\\\ \ldots \\\\ \dfrac{1}{2n}-\dfrac{1}{n} + \dfrac{1}{2n} = 0 \\ \hline \end{array}\)

So S is, we have all black terms left :

\(\begin{array}{|rcll|} \hline S &=& \dfrac{1}{2}-\dfrac{1}{2}+\dfrac{1}{4} + \dfrac{1}{2(n+1)} - \dfrac{1}{n+1} + \dfrac{1}{2(n+2)} \\\\ &=& \dfrac{1}{4} - \dfrac{1}{2(n+1)} + \dfrac{1}{2(n+2)} \\\\ &=& \dfrac{1}{4} - \dfrac{1}{2}\left( \dfrac{1}{n+1} - \dfrac{1}{n+2} \right) \\\\ &=& \dfrac{1}{4} - \dfrac{1}{2}\left( \dfrac{1}{(n+1)(n+2)} \right) \\\\ &=& \dfrac{1}{4} - \dfrac{1}{2(n+1)(n+2)} \\\\ &=& \dfrac{1}{4} \left(1- \dfrac{2}{(n+1)(n+2)} \right) \\\\ &=& \dfrac{(n+1)(n+2)-2}{4(n+1)(n+2)} \\\\ &=& \dfrac{n(n+2)+n+2-2}{4(n+1)(n+2)} \\\\ &=& \dfrac{n(n+2)+n}{4(n+1)(n+2)} \\\\ &=& \dfrac{n(n+2+1)}{4(n+1)(n+2)} \\\\ \mathbf{ S }& \mathbf{=} & \mathbf{ \dfrac{n(n+3)}{4(n+1)(n+2)} } \\ \hline \end{array}\)

\(\large{ \mathbf{ S = \sum \limits_{k=1 }^{n} \dfrac 1{k(k+1)(k+2)} = \dfrac{n(n+3)}{4(n+1)(n+2)} } }\)

\(\begin{array}{|lcll|} \hline \mathbf{ S_n} & \mathbf{ =} & \mathbf{ 4\cdot \sum \limits_{k=1}^n \dfrac 1{k(k+1)(k+2)} } \\\\ S_n &=& 4\cdot \dfrac{n(n+3)}{4(n+1)(n+2)} \\\\ \mathbf{ S_n} &\mathbf{=} & \mathbf{ \dfrac{n(n+3)}{(n+1)(n+2)} } \quad & | \quad n = 998 \\\\ S_{998} & {=} & \dfrac{998\cdot(998+3)}{(998+1)(998+2)} \\\\ S_{998} & {=} & \dfrac{998\cdot 1001}{999\cdot 1000} \\\\ S_{998} & {=} & \dfrac{499499}{499500} \\\\ S_{998} & {=} & 0.999\overline{997} \\\\ \hline \end{array}\)

For the trail mix portions to be identical, they need to have the same number of bags in each portion, and we can quickly figure out this by doing GCM (16,8) and getting 8. So the highest is 8 portions

Your answer is 8.

First, we have to see what the least difference between any could be. We find the least distance to be possible, for example, when there is 64 and 60, because the difference is there 4 and otherwise, it would be a higher number. 

I will list out the numbers for anyone who wants to try and prove me wrong (not that I'm saying you're a bad community, this is a great one) :

Sequence A: 2, 4, 8, 16, 32, 64, 128, 256, 512

Sequence B: 20, 40, 60, 80, 100, 120, 140, 160, 180, 200, 220, 240, 260, 280, 300, 320

The answer is 4

First we can try solving the first part: -2(2)^4

We do 2^4 = 16,  then 16(-2) = -32

Then we solve the (4)(-3), which equals -12.

Then we subtract -12 from -32 and get:

-32 - (-12) = -20

-20 is the answer!

It should be C because:

the boxed portion is not the lower and upper extremes

it is not the lower and upper extremes: that is 4 and 10

it is not the median of entire data set: that is 8

it is not the mode of entire set: we don't know that.

Another hand scrawled answer:

Thank you! That makes a big banana-boat load of sense!

The probability of their occurrence is higher than the final observable count.  

While this seems to the case for elements selected from a single (dependent) set, when elements are selected from independent sets, such as “pick 3” lottery based questions, the probability directly reflects the enumeration, when a player “boxes” (permutes) a selection where (2) of the numbers are the same.   

Dissection of the statistical processes clearly shows why duplicate numbers require division to calculate the probability of success.

Your analysis of a dependent set shows why duplicate numbers require multiplication to calculate the probability of success.

 GA

Base area = Pi x r^2
81Pi =Pi x r^2
r =Sqrt(81) = 9 mm - the radius of the cone.
Lateral Surface Area=Pi x r x s
135Pi=Pi x 9 x s
s =135 / 9
s = 15 mm - slant height of the cone.

Ooops.......

1 3/8 = 11/8   

11/8 -(-7/8) =

11/8 + 7/8 = 18/8 = 2 2/8 = 2 1/4

Here is my hand  scrawled answer:

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

Rationalize the denominator. 1/sqrt(2) = 1/sqrt(2)×(2^(1 - 1/2))/(2^(1 - 1/2)) = (2^(1 - 1/2))/2:
sqrt(2) + (2^(1 - 1/2))/2 + sqrt(3) + 1/sqrt(3)

Combine powers. (2^(1 - 1/2))/2 = 2^((1 - 1/2) - 1):
sqrt(2) + 2^((1 - 1/2) - 1) + sqrt(3) + 1/sqrt(3)

Put 1 - 1/2 over the common denominator 2. 1 - 1/2 = 2/2 - 1/2:
sqrt(2) + 2^(2/2 - 1/2 - 1) + sqrt(3) + 1/sqrt(3)

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

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

Put 1/2 - 1 over the common denominator 2. 1/2 - 1 = 1/2 - 2/2:
sqrt(2) + 2^(1/2 - 2/2) + sqrt(3) + 1/sqrt(3)

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

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

Rationalize the denominator. 1/sqrt(2) = 1/sqrt(2)×(sqrt(2))/(sqrt(2)) = (sqrt(2))/2:
sqrt(2) + (sqrt(2))/2 + sqrt(3) + 1/sqrt(3)

Rationalize the denominator. 1/sqrt(3) = 1/sqrt(3)×(sqrt(3))/(sqrt(3)) = (sqrt(3))/3:
sqrt(2) + (sqrt(2))/2 + sqrt(3) + (sqrt(3))/3

Put each term in sqrt(2) + (sqrt(2))/2 + sqrt(3) + (sqrt(3))/3 over the common denominator 6: sqrt(2) + (sqrt(2))/2 + sqrt(3) + (sqrt(3))/3 = (6 sqrt(2))/6 + (3 sqrt(2))/6 + (6 sqrt(3))/6 + (2 sqrt(3))/6:
(6 sqrt(2))/6 + (3 sqrt(2))/6 + (6 sqrt(3))/6 + (2 sqrt(3))/6

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

Add like terms. 6 sqrt(2) + 3 sqrt(2) + 6 sqrt(3) + 2 sqrt(3) = 9 sqrt(2) + 8 sqrt(3):

(9 sqrt(2) + 8 sqrt(3))/6 = a =9, b=8, c=6: a + b + c =9 + 8 + 6 = 23

Pythag theorem

6^2 + b^2 = c^2

Given    6 + b + c = 18

      rearranged:  b=12-c      sub into pythag

6^2 + (12-c)^2 = c^2      expand

36 + 144 -24c + c^2 = c^2     simplify

180 =24c    solve

c = 7.5          then b = 18-7.5-6=4.5

Check   6^2 + 4.5^2 = 7.5^2  ??

              36 + 20.25 = 56.25           yes!

Make the denominator the same:

1/5 = (2)(1)/(2)(5) = 2/10

Then add the numerators, or subtract in this case, and then put the denominator on:

(2-9)/10 = -7/10

The answer is -7/10

There is no max value, since it is 1/2 * x^2, and 1/2 is positive.

The minimum value is going to be (6,7)

The axis of symmetry is: x=6

The domain is (-infinity,infinity)

The range is (7, infinity)

It means the same thing as: "How many times can you put 24 into 37 without going over?" Unless your looking for the remainder as well...

If you don't want any remainder, then you can easily find the answer because: 37-24=13, but you can't subtract any more 24's without going negative. Your answer in this case would be 1.

If you want a remainder, then you can do again the same process: 37-24=13, and then you'd know your answer would be 1 13/24, or as an improper fraction would be 37/24.

Hope this helps!

First, make the denominators the same for both fractions:

-1/2 = (-1)(4)/(2(4) = -4/8

Then simplify the fractions because now you have same denominator for both fractions:

-3/8 - 4/8 = -7/8

-7/8 is your answer!

You first make the fractions improper ones:

-1 2/3 = -5/3

Make the denominators of the fractions the same, since you would want the calculations to later be easier:

-5/3 = (-5)(2)/(3)(2) = -10/6

Now do as you would simplify like denominators:

((-10) - (-5))/6 = -5/6

Your answer is -5/6

First, make that whole number an improper fraction:

1 3/8 = 11/8

Then, do the subtraction from your numerator and add on the denominator:

(11 - (-7))/8 = (18)/8 = 18/8

Your answer is 18/8, unless you want to simplify to a whole number. That would be 2 1/4.

Done!

(p.s. thanks electricpavlov for pointing my mistake!)

Keep denominator same and do the numerator almost as if the denom. wasn't there; treat it like regular integers. Then add on the denominator to the number once you're done simplifying:

- 4/5 - 4/5 = (-4-4) / 5 = -8/5

-8/5 is your answer

To subtract fractions with like denominators, you have to keep the denominator the same and then do the numerator as required, so:

-6/11 - (-5/11) =  (-6 - (-5))/11 = (-6 + 5)/11 = -1/11

Hmmmm..... Using Quadratic Formula   I get   (4+- 2sqrt10)/3