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Like so:

.

In how many different ways can 2/15 be represented as 1/A + 1/B,

if A and B are positive integers with A less than or equal to B

\(\text{For odd $n>2$ there is always at least one decomposition into exactly two unit fractions: $\dfrac{2}{n} = \dfrac{1}{A} + \dfrac{1}{B}$ } \\ \text{Finding all possibilities.}\\ \text{The prime factorization of $n^2$ results in all possible decompositions into two unit fractions.} \)

\(n=15\ \text{is odd} \\ n^2 =225 \\ \text{All divisors of $n^2=225$ are: $1,\ 3,\ 5,\ 9,\ 15,\ 25,\ 45,\ 75,\ 225$}\\ \text{Let $n^2=p\times q$ }\)

\(\begin{array}{|r|r|r|c|c|c|c|c|c|c|c|c| } \hline & p & q & n^2 & s & t & r & k & A & B & A\le B & \\ & & & =p\cdot q & = \frac{p+q}{2} & = \frac{p-q}{2} & =\frac{t}{2} & = \frac{15+\sqrt{15^2+t^2} }{2} & =k-r & =k+r & \\ \hline 1. & 225 & 1 & 225=225 \cdot 1 & 113 & 112 & 56 & 64 & 8 & 120 & \checkmark & \mathbf{\dfrac{2}{15} = \dfrac{1}{8} + \dfrac{1}{120}} \\ \hline 2. & 75 & 3 & 225= 73 \cdot 3 & 39 & 36 & 18 & 27 & 9 & 45 & \checkmark & \mathbf{\dfrac{2}{15} = \dfrac{1}{9} + \dfrac{1}{45}} \\ \hline 3. & 45 & 5 & 225= 45 \cdot 5 & 25 & 20 & 10 & 20 & 10 & 30 & \checkmark & \mathbf{\dfrac{2}{15} = \dfrac{1}{10} + \dfrac{1}{30}} \\ \hline 4. & 25 & 9 & 225= 25 \cdot 9 & 17 & 8 & 4 & 16 & 12 & 20 & \checkmark & \mathbf{\dfrac{2}{15} = \dfrac{1}{12} + \dfrac{1}{20}} \\ \hline 5. & 15 & 15 & 225= 15 \cdot 15 & 15 & 0 & 0 & 15 & 15 & 15 & \checkmark & \mathbf{\dfrac{2}{15} = \dfrac{1}{15} + \dfrac{1}{15}} \\ \hline \end{array}\)

Yes it would have been a misprint.  :)

or a poor print  ;)

Hello Melody!!!!,

I am going to be VERY honest about this....the answer I actually got, was exactly the same as what you got...I promise you that. this morning when I posted the problem, i did not have my paper that i worked on, with me, so I did the sum afresh...and got lost. I figured that it was'nt a train smash if i did not put the sum exactly as I actually had it....the answer is wrong any case...so I left it as such

Yes I also did the 3 "i"'s etc...

Melody, yes, I checked and re-checked that answer....the "i" looks like it is within the square root sign...maybe they did a mis-print?

Hi Juriemagic :)

\(\sqrt{-5}=\sqrt{5*-1}=\sqrt5*\sqrt{-1}=\sqrt5\;i\)

So your mistakes strat in the third row.

\(\sqrt{-5}(\sqrt{-3}*\sqrt{-4})\\ \sqrt5*\sqrt3*\sqrt4*i*i*i\\ 2\sqrt{15}*-1*i\\ -2\sqrt{15}\;i\)

This is different from your answer but different from your book too.  

Are you sure you copied in the book answer correctly?

Suppose that

\(ABC_4+200_{10}=ABC_9\)
ABC_4+200_{10}=ABC_9,
where A, B, and C are valid digits in base 4 and 9.
What is the sum when you add all possible values of A, all possible values of B, and all possible values of C?

1.

A, B, and C are valid digits in base 4 and 9

\(A = \{ 0,1,2,3 \} \\ B = \{ 0,1,2,3 \} \\ C = \{ 0,1,2,3 \} \)

2.

\(\begin{array}{|rcll|} \hline ABC_4+200_{10} &=& ABC_9 \\ \overbrace{A\cdot 4^2 + B\cdot 4 + C}^{ABC_4=} + 200 &=& \overbrace{ A\cdot 9^2 + B\cdot 9 + C}^{ABC_9=} \\ 16A+4B+200 &=& 81A+9B \\ 65A+5B &=& 200 \quad & | \quad :5\\ \mathbf{13A+B} &\mathbf{=}&\mathbf{40} \\ \hline \end{array}\)

3.

Possible values of A and B

\(\begin{array}{|c|c|r|r|} \hline A & B & 13A+B & =40\ ? \\ \hline 0 & 0 & 0 \\ & 1 & 1 \\ & 2 & 2 \\ & 3 & 3 \\ \hline 1 & 0 & 13 \\ & 1 & 14 \\ & 2 & 15 \\ & 3 & 16 \\ \hline 2 & 0 & 26 \\ & 1 & 27 \\ & 2 & 28 \\ & 3 & 29 \\ \hline 3 & 0 & 39 \\ & 1 & 40 & \checkmark \\ & 2 & 41 \\ & 3 & 42 \\ \hline \end{array}\\ A=3,\ B=1,\ C=0,1,2,3 \)

\(\text{sum} = 3+1+0+1+2+3 = \mathbf{10}\)

The sum when you add all possible values of A, all possible values of B, and all possible values of C is 10

check:

\(310_4+200_{10} = 310_9 =252_{10} \\ 311_4+200_{10} = 311_9 =253_{10} \\ 312_4+200_{10} = 312_9 =254_{10} \\ 313_4+200_{10} = 313_9 =255_{10} \)

For how many ordered pairs of positive integers (x,y)  does the equation  \(\dfrac{1}{x} +\dfrac{2}{y}=\dfrac{1}{3}\\ \) hold?

\(\dfrac{1}{x} +\dfrac{2}{y}=\dfrac{1}{3}\\ \)

x=0 and y = zero are both asymptotes.

as x tends to infinity    y tends to 6

as y tends to infinity     x tends to 3

\(\dfrac{1}{x} +\dfrac{2}{y}=\dfrac{1}{3}\\ \frac{2}{y}=\frac{1}{3}-\frac{1}{x}\\ \frac{2}{y}=\frac{x-3}{3x}\\ \frac{y}{2}=\frac{3x}{x-3}\\ y=\frac{6x}{x-3}\\ \)

for y to be positive x>3

For x to be positive y>6

I found these points using  EXCEL.   I found 6 ordered pairs.

I'll give you a hint.

The asympotes for f(x) is where    x^2-2x-3=0

This is because you cannot divide by 0.

it would look like this:

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This question was answered by Cphill before

This is what he answered

What real value of t produces the smallest value of the quadratic t^2 -9t - 36

Since this is  an upwards-turning parabola, the "t" that produces the smallest possible value  is given by :

-b/ [2a] =  -[9] / [2 (1) ] =   9/2 = 4.5

Is this what you mean?

1/8  + 1 / 120    = 2/15
1/9  +  1 / 45     = 2/15
1/10  +  1 / 30   = 2/15
1/12  +  1 / 20   = 2/15

1/15  +  1/15     = 2/15

slope = y1-y2   /   x1-x2

    = 6- -2  /   -2-  -4

    =8 / 2  =4

y=mx+b form

y = 4x + b      Sub in one of the points to find b

6 = 4(-2) + b

b = 14

y = 4x+14      when x = 0     b is the y intercept = 14

First one:

Take square root of both sides of the equation to get

x^2 - 82 = = 18  or  -18

x^2 =  100    or   64

x= + or - 10    or  + - 8        -10,-8,8,10

Second one

x^2+6x-16=0  factor

(x-2)(x+8) = 0                   so x = 2 or -8       -8 is the smallest

Third one

 (x+6)^2 = -25

x+6 =    -5i     5i 

x = -6-5i     -6+5i

Just sustitute in and do the math.....

P = 2(7.9) + 2(12.4)      and round to tenth...

       15.8   +  24.8

         40.6 cm

11! days   = 11! = 39916800 days

That is about 109,000 years. 

So since cows, or people do not live that long he can essentially have a different order for every day.

5C + 4B = 216     and

5C + 2B = 138     Subtract 2nd equation from first

        2B = 78    B=39

5C + 2(39) = 138

5C= 60        C = 12