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With 3 children the probabilities are:

1 boy and 2 girls, or 1 girl and 2 boys =37.50%=N!/(B!.G! x 2^N) x 100

2 boys and 1 girl or 2 girls and 1 boy =37.50% =,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,

3 boys and no girls or 3 girls and no boys=12.50%=,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,

1 acre =43,560 sq.ft.

1.5 x 43,560 =65,340

65,340 / 160 =408.375 ft. the length of the Lot.

Given that 

1) x - b2 = a2

2)  + b = a

3) a > b > 0

Find x : y when a : b = 5 : 3 

24 mod 5 =4

64ft/s is correct.

How does this work ? 11/9-2/3+5/18

\(\begin{array}{|rcll|} \hline \frac{11}{9} - \frac{2}{3} + \frac{5}{18} &=& x \qquad &|\qquad \cdot 9 \\ 9\cdot \frac{11}{9} - 9\cdot \frac{2}{3} + 9\cdot \frac{5}{18} &=& 9x \\ 11 - 6 + \frac{5}{2} &=& 9x \\ 5 + \frac{5}{2} &=& 9x \\ \frac{10}{2} + \frac{5}{2} &=& 9x \\ \frac{15}{2} &=& 9x \qquad &|\qquad : 9\\ \frac{15}{2\cdot 9} &=& x \\ \frac{5}{2\cdot 3} &=& x \\ \frac{5}{6} &=& x \\ \frac{5}{6} &=& \frac{11}{9} - \frac{2}{3} + \frac{5}{18} \\ \hline \end{array}\)

4ab           2a             2b          2a

---------- - ---------- - ---------- = -----

a^2-b^2      a-b          a+b        a+b        is this correct?

\(\begin{array}{|rcll|} \hline && \frac{4ab}{a^2-b^2} -\frac{2a}{a-b} -\frac{2b}{a+b} \qquad &|\qquad a^2-b^2 = (a-b)(a+b)\\ &=& \frac{4ab}{(a-b)(a+b)} -\frac{2a}{a-b} -\frac{2b}{a+b} \\ &=& \frac{4ab}{(a-b)(a+b)} -\frac{2a}{a-b}\cdot \left( \frac{a+b}{a+b} \right) -\frac{2b}{a+b}\cdot \left( \frac{a-b}{a-b} \right)\\ &=& \frac{4ab-2a(a+b)-2b(a-b)}{(a-b)(a+b)} \qquad &|\qquad (a-b)(a+b)= a^2-b^2 \\ &=& \frac{4ab-2a(a+b)-2b(a-b)}{a^2-b^2} \\ &=& \frac{4ab-2a^2-2ab-2ab+2b^2}{a^2-b^2} \\ &=& \frac{4ab-4ab-2a^2+2b^2}{a^2-b^2} \\ &=& \frac{-2a^2+2b^2}{a^2-b^2} \\ &=& \frac{-2\cdot(a^2-b^2)}{a^2-b^2} \\ &=& -2 \\ \hline \end{array}\)

What  is  the determinant of  C?
\(C=\begin{bmatrix} 1 && 2 &&-2\\ 3 && 0 && -1\\ 5 && 3 && 2 \end{bmatrix}\)

\(\begin{array}{|rcll|} \hline C_{determinant} &=& \begin{vmatrix}1 && 2 &&-2\\ 3 && 0 && -1\\ 5 && 3 && 2 \end{vmatrix} \\ &=& 1\cdot 0 \cdot 2 + 2\cdot(-1)\cdot 5 + (-2)\cdot 3\cdot 3 \\ && - 5\cdot 0 \cdot (-2) -2\cdot 3 \cdot 2 -1\cdot 3 \cdot (-1) \\ &=& 2\cdot(-1)\cdot 5 + (-2)\cdot 3\cdot 3 -2\cdot 3 \cdot 2 -1\cdot 3 \cdot (-1) \\ &=& -10 -18 -12 +3 \\ &=& -37 \\ \hline \end{array} \)

Btw the extension is just repeating the first 2 columns of the matrix.

Btw the extension is just repeating the first 2 columns of the matrix.

Extend it like this:\(\begin{bmatrix} 1 && 2 &&-2 &&1&&2\\ 3 && 0 && -1&&3&&0\\ 5 && 3 && 2&&5&&3 \end{bmatrix}\)

Multiply the numbers if they can be linked top-to-bottom diagonally with 3 numbers: 

Example: 1, 0, and 2 is in the same diagonal and they are 3 numbers.

Then add up the products obtained from multiplying the number of top-left to bottom-right. Call that 'answer 1'

(1)(0)(2) + (2)(-1)(5) + (-2)(3)(3) = 0 - 10 - 18 = -28 -----(1)

Then add up the products obtained from multiplying the number of top-right to bottom-left. Call that 'answer 2'

(-2)(0)(5) + (1)(-1)(3) + (2)(3)(2) = -10 - 3 + 12 = -1 -----(2)

Add up answer 1 and answer 2: -28 - 1 = -29.

Therefore det(C) = -29. Best method ever , I don't have a proof of it though but I got through the determinant questions correctly when I use this method.

So my answer is 64ft/second, is that correct?

I got 96ft/second for Part 1. Is that correct? because I never did word equation....... 

Note:  In part 1 my hint contains an error.  \(\frac{du}{dt}=\frac{s}{\sqrt{s^2+1800^2}}\frac{ds}{dt}\) is the correct expression for the rate of change of u.  The 2 shouldn't be there!

.

Hi Max, your question on the integral of 1/x^2 -9.

Use partial fractions instead of substition .   X^2 -9   = (x+3)(x-3)

So 1/(x+3)(x-3) = 1/6{ (1/(x+3) - (1/x-3)} Integrate that and you'll get the answer given in the book.

Why all of them sounded so funny I laughed in front of my computer LOL.

Btw, use the tag [off-topic] when the question is not about Maths, or you can you [Physics] when the question is about Physics.

Title example:  [Off-topic] Spanish

1.  Here's a hint for part 1 Max.

I'm sure you can take it from here.

2.  Here's part 2.

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I think you mean : \(\dfrac{y^2}{y^7}\)

And that equals \(y^{2-7} = y^{-5}\)OR \(\dfrac{1}{y^5}\)

43/50 = 86/100 = 86 x 1/100 = 86 x 0.01 = 0.86

Look here..... Exact same question......

I saw you answering when I just started to answer...... And I use LaTeX too. Just practice more and you will be as fast as I am :)

There is nothing wrong with taking the square root, it is afterall where, after completion of the square, the formula for the solution of a quadratic comes from.

Taking the square root, 8y - 11 = 13 from which y = 3, or, 8y - 11 = -13 from which y = -1/4.

Sorry I took so long because it is hard to write in LaTeX.