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I think that the answer is 19C5*14C5*9C3*6C3, whatever that works out to be.

This business of dividing by 2 and 3! when you have one group of five then the other group of five is determined, or for the groups of three it doesn't matter in what order they are chosen, is wrong.

There is a difference between splitting the 19 into two groups of five and three groups of three without distinction, and splitting them into groups where each group studies a different topic.

Consider, for example, a group of four students, call them A,B,C and D, split into two twos.

4C2 = 6, so we have 6 pairs, AB, AC, AD, BC, BD, CD.

However, when the first pairing is chosen, in effect the second pairing is also chosen, so we have just three (= 6/2) possible groupings, {AB, CD}, {AC, BD} and {AD, BC}. That's the situation if it doesn't matter which pairing comes first, that is, for example, {AB, CD} is the same as {CD, AB}.

Now suppose that the students are going into two different language classes, say the first pairing study French and the second pairing study German. Now, the order of the pairings makes a difference, {AB, CD} is different from {CD, AB}.

In the first case, A and B study French and in the second case they study German, (and the other way round for C and D).

If this is the situation, the original calculation of 4C2 = 6 stands.

(Notice btw that, for the 19 student case, as far as the calculation of the final number is concerned, it doesn't matter which group is chosen first, that is, for example, 19C5*14C5*9C3*6C3 = 19C3*16C3*13C5*8C5 = ... .)

BobP

physics problem help. find t

-100=150sin(37.1)t-1/2(9.81)t^2

Hello Guest!

\({\color{BrickRed}-100=150\cdot sin(37.1°)\cdot t-\frac{1}{2}\cdot 9.81\cdot t^2}\\ \frac{1}{2}\cdot 9.81\cdot t^2 -150\cdot sin(37.1°)\cdot t-100=0\\ 4.905t^2-90.4812t-100=0\)

   a                   b                c

\(t= {-b \pm \sqrt{b^2-4ac} \over 2a}\)

\(\large t = {90.4812 \pm \sqrt{8186.8472-4\times 4.905\times (-100)} \over 2\times 4.905}\)

\(\large t = {90.4812 \pm \sqrt{10148.8472} \over 9.81}\)

\(\large t=\frac{90.4812\pm 100.7415}{9.81}\)    \((-100.7415\ entf\ddot allt)\)

\(t= 19.493\)

  !

\(x = {1+\dfrac{\sqrt{2}}{1+\dfrac{\sqrt{2}}{1+...}}}\).

\(x=1+\frac{\sqrt2}{x}\\ x^2=x+\sqrt2\\ x^2-x=\sqrt2\\ \)

    \(\frac{1}{(x+1)(x-2)}\\ =\frac{1}{(x+1)(x-2)}\\ =\frac{1}{x^2-x-2}\\ =\frac{1}{\sqrt2-2}\\ =\frac{-1}{2-\sqrt2}\\ =\frac{-1}{2-\sqrt2}\times \frac{2+\sqrt2}{2+\sqrt2}\\ =\frac{-(2+\sqrt2)}{4-2}\\ =\frac{-2-\sqrt2}{2}\\ =\frac{2+\sqrt2}{-2}\\ \)  .

When your answer is in the form   \(\frac{A+\sqrt{B}}{C}\)   ,

A=+2,   B=+2,   C= -2

where A, B, and C are integers, what is |A|+|B|+|C|?

|A|+|B|+|C| = 2+2+2 = 6

Algebra

Compute the sum \(\mathbf{\frac{2}{1 \cdot 2 \cdot 3} + \frac{2}{2 \cdot 3 \cdot 4} + \frac{2}{3 \cdot 4 \cdot 5} + \cdots}\)

\(\begin{array}{lcll} \mathbf{ \dfrac{2}{1 \cdot 2 \cdot 3} + \dfrac{2}{2 \cdot 3 \cdot 4} + \dfrac{2}{3 \cdot 4 \cdot 5} + \dfrac{2}{4 \cdot 5 \cdot 6} + \cdots \ + \dfrac{2}{n \cdot (n+1) \cdot (n+2)} + \cdots =\ \mathbf{ ? } } \\\\ \begin{array}{|lcll|} \hline s_n = \dfrac{2}{1 \cdot 2 \cdot 3} + \dfrac{2}{2 \cdot 3 \cdot 4} + \dfrac{2}{3 \cdot 4 \cdot 5} + \dfrac{2}{4 \cdot 5 \cdot 6} + \cdots \ + \dfrac{2}{n \cdot (n+1) \cdot (n+2)} \\ \hline \end{array} \\ \end{array}\\\)

Formula:

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

we rearrange:

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

telescoping series

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

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

Example:

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

So \(s_n\) is, we have all black terms left :

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

 \(\lim \limits_{n\to \infty} { \dfrac{1}{n+1}} = 0 \quad \text{ and } \quad \lim \limits_{n\to \infty} { \dfrac{1}{n+2} } = 0 \)

\( \begin{array}{|rcll|} \hline \lim \limits_{n\to \infty} s_n &=& \dfrac{1}{2} - 0 + 0 \\ &=& \dfrac{1}{2} \\ \hline \end{array} \)

\(\begin{array}{lcll} \mathbf{ \dfrac{2}{1 \cdot 2 \cdot 3} + \dfrac{2}{2 \cdot 3 \cdot 4} + \dfrac{2}{3 \cdot 4 \cdot 5} + \dfrac{2}{4 \cdot 5 \cdot 6} + \cdots \ + \dfrac{2}{n \cdot (n+1) \cdot (n+2)} + \cdots =\ \mathbf{ \dfrac{1}{2} } } \\ \end{array}\\\)

I solved it on my own...

inital angle= 30.0

initial velocity= 29.4 m/s

total time= 3.00s

cos30(29.4m/s)= 25.46m/s

vx=25.46m/s

vx=dx/t

vx(t)=dx

25.46m/s(3.00s)=dx

76.38m=dx

76.4m=dx

When you multiply the "last" terms, you get.....

-4√13 * -4√13   =   16 * √13 * √13   =   16 * √13²   =   16 * 13   =   208

how'd you get 208 ?

Here is one, by GingerAle and Hectictar, that Hectictar just asked me to include.  Thanks Girls :)

This is the original thread but I will includ most of the salient coding underneath.

x^0  =  1  , when  x ≠ 0   

(a)  angle(A)  =  38^o  and  angle(C)  =  85^o   --->   angle(B)  =  180^o - angle(A) - angle(B)  =  180^o - 38^o - 85^o  =  57^o

       Since you know side(c)  =  32 cm, you can use the Law of Sines to find sides a and b:

            side(a) / sin( angle(A) )  =   side(c) / sin( angle(C) )

            side(a) / sin( 38^o )  =  32 / sin( 85^o ) 

            side(a)  =  19.77 cm

            side(b) / sin( angle(B) )  =   side(c) / sin( angle(C) )

            side(b) / sin( 57^o )  =  32 / sin( 85^o ) 

            side(a)  =  26.94 cm

(b)  angle(A)  =  24^o  and  side(b)  =  12.5 m  side(c)  =  13.2 m

      This describes a S-A-S situation, so you can use the Law of Cosines:

            a²  =  b² + c² - 2·b·c·cos(A)

            a²  =  (12.5)² + (13.2)² - 2·(12.5)·(13.2)·cos(24)

            a  =  5.39 m

       You can use the Law of Sines to find angle(B)

            [Warning:  Don't use the Law of Sines to find the largest angle in a triangle unless you have to!]

        sin(B) / side(b)  =  sin(A) / side(a)

        sin(B) / 12.5  =  sin(24^o) / 5.39

        angle(B)  =  70.6^o

        By subtracting angle(C)  =  180^o - angle(A) - angle(B)  =  180^o - 24^o - 70.6^o   =  85.4^o

Wait isn't x^0 always 1? That is how I was taught at least? :P

Mean is the 40 because (25+55+40+65+29+45+59+35+25+37+52+30+8+40+55)/15=40

Standered deviation is 14.9086104875 because (25-40)^2=225 (Then do this for all the numbers you have) and then add them together, divide by 15 and take the square root.

Here's a helpful guide for next time:

The number of elements in set A is found by the formula:  largest - smallest + 1

n(A)  =  127 - 101 + 1  =  27

The number of elements in set B is found by the formula:  (largest - smallest) / 2 + 1

n(B)  =  (301 - 101) / 2 + 1  =  101

He gets a discount of 9% off

(845*.09)=76.05 (Trade disocunt)

845-76.05=768.95 (Checking to make sure problem works out)

Hi Chris,

This question confused me but what you have done is certainly correct. :)

I have included the graph here.

For the benefit of others.

The inverse of a function is the reflection of that function in the line y=x.

That is why I included y=x on my graph.

N / -7  = 8      multiply both sides by -7

N = -56

3N - 10  = 23     add 10 to both sides

3N  = 33            divide both sides by 3

N = 11

-5N + 11  =  -29       subtract 11 from both sides

-5N  = -40     dibide both sides by -5

N = 8

 1 gal  = 128 oz

So  12.5 gallons  =  12.5 * 128  =  1600 oz

So....the ratio  of   8 oz to 1600 oz  = 

8 / 1600  =      [8 * 1] / [8 *  200]  =  1 / 200

( 6∛2 - 4√13)^2  =

( 6∛2 - 4√13) ( 6∛2 - 4√13)  =

36∛4 - 24*√13*∛2 - 24*√13*∛2 + 16*13 ) =

36∛4 - 48*√13*∛2 + 208

No.....note that

32√5 + √5    is the same as

32√5 + 1√5  =

33√5

y =  2 + ( x - 2)^2

y - 2  =  (x - 2)^2       take both roots 

±√[ y - 2 ] = x - 2      add 2 to both sides

±√[ y - 2 ] + 2  =  x    exchange x and y

±√[ x - 2 ] + 2 =  y  =  possible choices for k(x)

Since  +√[ x - 2 ] + 2   > 0    for all x  >2, then  -√[ x - 2 ] + 2  is the correct  function

[ To see this, note that   (-1,11)  is a point on  2 + ( x - 2)^2......but ( 11, -1) is not on  +√[ x - 2 ] + 2 .....however,  (11 , -1)  is on  -√[ x - 2 ] + 2  ]

So....k(x)  =  -√[ x - 2 ] + 2 

See  the  inverses here : https://www.desmos.com/calculator/iiz2jtxdok

wouldn't it be √ [ 15j ] + 32√5 ?

Sounds good to me!