heureka

Tom has 6 red marbles, 4 blue marbles, and 3 green marbles.

How many distinct ways can he line them up in a row?

\(\begin{array}{|rcll|} \hline && \mathbf{\dfrac{(6+4+3)!}{6!4!3!}} \\\\ &=& \dfrac{13!}{6!4!3!} \\\\ &=& \dfrac{6!\cdot 7\cdot 8\cdot 9\cdot 10\cdot 11\cdot 12\cdot 13}{6!4!3!} \\\\ &=& \dfrac{7\cdot 8\cdot 9\cdot 10\cdot 11\cdot 12\cdot 13}{4!3!} \\\\ &=& \dfrac{7\cdot 8\cdot 9\cdot 10\cdot 11\cdot 12\cdot 13}{1\cdot 2\cdot 3\cdot 4\cdot 1\cdot 2 \cdot 3} \\\\ &=& 7\cdot 3\cdot 5\cdot 11\cdot 4\cdot 13 \\ &=& \mathbf{60060} \\ \hline \end{array}\)

1)
Suppose the function \(f(x,y,z)=xyz\)
is defined for
\(x+y+z=7, x,y,z\geq 0\).
What is the range of \(f(x,y,z)\)?

\(\begin{array}{|rcll|} \hline f(x,y,z) &=& xyz \quad | \quad x+y+z=7 \text{ or } z = 7-x-y \\ &=& xy(7-x-y) \\\\ f(x,y) &=& xy(7-x-y) \\ f(x,y) &=& 7xy-x^2y-xy^2 \\ \hline f_x = \dfrac{\partial f(x,y)}{\partial x} &=& 7y-2xy-y^2 \\ f_y = \dfrac{\partial f(x,y)}{\partial y} &=& 7x-x^2-2xy \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline (1) & f_x=0 &=& 7y-2xy-y^2 \\ (2) & f_y=0 &=& 7x-x^2-2xy \\ \hline (1) & y(7-2x-y) &=& 0 \\ & \mathbf{y} &=& \mathbf{0} \\\\ & 7-2x-y &=& 0 \\ & \mathbf{y} &=& \mathbf{7-2x} \\ \hline (2) & x(7-x-2y) &=& 0 \\ & \mathbf{x} &=& \mathbf{0} \\\\ & 7-x-2y &=& 0 \\ & 7-x-2(y) &=& 0 \\ & 7-x-2(0) &=& 0 \\ & 7-x &=& 0 \\ & \mathbf{x} &=& \mathbf{7} \quad | \quad y=0 \\\\ & 7-x-2(y) &=& 0 \\ & 7-x-2(7-2x) &=& 0 \\ & 7-x-14+4x &=& 0 \\ & 3x &=& 7 \\ & \mathbf{x} &=& \mathbf{\dfrac{7}{3} } \quad | \quad y=7-2x \\\\ \hline & y &=& 7-2x \\ & y &=& 7-2(0) \\ & \mathbf{y} &=& \mathbf{7} \quad | \quad x=0 \\\\ & y &=& 7-2x \\ & y &=& 7-2(7) \\ & \mathbf{y} &=& \mathbf{-7} \quad | \quad x=7 \\\\ & y &=& 7-2x \\ & y &=& 7-2(\dfrac{7}{3}) \\ & \mathbf{y} &=& \mathbf{\dfrac{7}{3} }\quad | \quad x=\dfrac{7}{3} \\ \hline \end{array}\)

\(\begin{array}{|c|c|c|c|c|r|} \hline x&y&z=7-x-y & \text{solution} & f(x,y,z) & \\ \hline 0 &7 &0&&0&\text{minimum} \\ 7 &0 &0&&0&\text{minimum} \\ 7 &-7 &0& \text{no }(y\geq 0!)&& \\ \dfrac{7}{3} & \dfrac{7}{3} & \dfrac{7}{3} && \dfrac{343}{27} & \text{maximum} \\ \hline \end{array}\)

Range of \(f(x,y,z):\ 0\ldots \dfrac{343}{27}\)

2)
The arithmetic mean, geometric mean, and harmonic mean of a, b, c are 8, 5, 3 respectively.
What is the value of \(a^2+b^2+c^2\)?

\(\text{$(1)\ $Arithmetic mean: $\dfrac{a+b+c}{3}=8$ } \\ \text{$(2)\ $Geometric mean: $ \sqrt[3]{abc}=5$ } \\ \text{$(3)\ $Harmonic mean: $ \dfrac{3}{\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}} = 3 $ }\)

\(\begin{array}{|lrcll|} \hline (1) & \dfrac{a+b+c}{3} &=&8 \\ & a+b+c &=& 3\cdot 8 \\ &\mathbf{a+b+c} &=& \mathbf{24} \\ \hline \end{array} \)

\(\begin{array}{|lrcll|} \hline (2) & \sqrt[3]{abc} &=& 5 \\ & abc &=& 5^3 \\ &\mathbf{abc} &=& \mathbf{125} \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline (3) & \dfrac{3}{\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}} &=& 3 \\\\ & \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}&=& \dfrac{3}{3} \\\\ &\mathbf{\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}} &=& \mathbf{1} \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline (4) & \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} &=& 1 \quad | \quad \times abc \\ & \dfrac{abc}{a}+\dfrac{abc}{b}+\dfrac{abc}{c} &=& abc \\ & \mathbf{bc+ac+ab} &=& \mathbf{abc} \\\\ (5) & (a+b+c)^2 &=& a^2+b^2+c^2 + 2(\mathbf{bc+ac+ab}) \\ & (a+b+c)^2 &=& a^2+b^2+c^2 + 2abc \quad | \quad a+b+c = 24,\ abc = 125 \\ & 24^2 &=& a^2+b^2+c^2 + 2\times 125 \\ & 576 &=& a^2+b^2+c^2 + 250\\ & a^2+b^2+c^2 &=& 576-250 \\ & \mathbf{a^2+b^2+c^2} &=& \mathbf{326} \\ \hline \end{array}\)

Let a,b and c be nonzero real numbers such that a/b + b/c + c/a =7 and b/a + c/b + a/c = 9. 

Find a^3/b^3 + b^3/c^3 + c^3/a^3.

\(\begin{array}{|rcll|} \hline && \mathbf{\left( \dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{a} \right)^3 -3 \left(\dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{a} \right) \left(\dfrac{b}{a} + \dfrac{c}{b} + \dfrac{a}{c}\right )} \\\\ &=& \dfrac{a^3}{b^3} + \dfrac{c^3}{a^3} + \dfrac{3a^2}{bc} + \dfrac{3bc}{a^2} + \dfrac{3ac}{b^2} + \dfrac{3b^2}{ac} + \dfrac{3ab}{c^2} + \dfrac{3c^2}{ab} + \dfrac{b^3}{c^3} + 6 \\ && - \left(\dfrac{3a^2}{b c} + \dfrac{3bc}{a^2} + \dfrac{3ac}{b^2} + \dfrac{3b^2}{ac} + \dfrac{3ab}{c^2} + \dfrac{3c^2}{a b} + 9\right) \\\\ &=& \mathbf{\dfrac{a^3}{b^3} + \dfrac{b^3}{c^3} + \dfrac{c^3}{a^3} - 3} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{\dfrac{a^3}{b^3} + \dfrac{b^3}{c^3} + \dfrac{c^3}{a^3} - 3 } &=& \mathbf{\left( \dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{a} \right)^3 -3 \left(\dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{a} \right) \left(\dfrac{b}{a} + \dfrac{c}{b} + \dfrac{a}{c}\right )} \\\\ \dfrac{a^3}{b^3} + \dfrac{b^3}{c^3} + \dfrac{c^3}{a^3} &=& 3+ \left( \dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{a} \right)^3 -3 \left(\dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{a} \right) \left(\dfrac{b}{a} + \dfrac{c}{b} + \dfrac{a}{c}\right ) \\\\ \dfrac{a^3}{b^3} + \dfrac{b^3}{c^3} + \dfrac{c^3}{a^3} &=& 3+ \left( 7 \right)^3 -3 \left(7 \right) \left(9\right ) \\\\ &=& 3+ 343 - 189 \\\\ \mathbf{\dfrac{a^3}{b^3} + \dfrac{b^3}{c^3} + \dfrac{c^3}{a^3} } &=& \mathbf{157} \\ \hline \end{array}\)

Two circles have the same center O.

Point X is the midpoint of segment OP.

What is the ratio of the area of the circle with radius OX to the area of the circle with radius OP?

\(\begin{array}{|rcll|} \hline A_1 &=& \pi \times OX^2 \\ A_2 &=& \pi \times OP^2 \\ \hline \dfrac{A_1}{A_2} &=& \dfrac{\pi\times OX^2} {\pi\times OP^2} \\\\ \dfrac{A_1}{A_2} &=& \dfrac{ OX^2} { OP^2} \quad | \quad OP = 2\times OX \\\\ \dfrac{A_1}{A_2} &=& \dfrac{ OX^2} { \left(2\times OX \right)^2} \\\\ \dfrac{A_1}{A_2} &=& \dfrac{ OX^2} { 4\times OX^2} \\\\ \mathbf{\dfrac{A_1}{A_2}} &=& \mathbf{ \dfrac{1} { 4 } } \quad \text{ or } \quad \mathbf{\dfrac{A_1}{A_2}} = \mathbf{ \left(\dfrac{1}{2}\right)^2 }\\ \hline \end{array}\)

Determine the value of  \(\dfrac{2016}{1} + \dfrac{2015}{2} + \dfrac{2014}{3} + \dots + \dfrac{1}{2016} \above 1pt \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} + \dots + \dfrac{1}{2017} \)

\(\begin{array}{|rcll|} \hline && \mathbf{ \dfrac{2016}{1} + \dfrac{2015}{2} + \dfrac{2014}{3} + \dots + \dfrac{1}{2016} \above 1pt \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} + \dots + \dfrac{1}{2017} } \\\\ &=& \dfrac{\dfrac{2017-1}{1} + \dfrac{2017-2}{2} + \dfrac{2017-3}{3} + \dots + \dfrac{2017-2016}{2016}} {\dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} + \dots + \dfrac{1}{2016}+ \dfrac{1}{2017}} \\\\ &=& \dfrac{\dfrac{2017}{1} + \dfrac{2017}{2} + \dfrac{2017}{3} + \dots + \dfrac{2017}{2016}-2016\cdot 1} {\dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} + \dots + \dfrac{1}{2016}+ \dfrac{1}{2017}} \\\\ &=& \dfrac{\dfrac{2017}{2} + \dfrac{2017}{3} + \dots + \dfrac{2017}{2016}+2017-2016 } {\dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} + \dots + \dfrac{1}{2016}+ \dfrac{1}{2017}} \\\\ &=& \dfrac{\dfrac{2017}{2} + \dfrac{2017}{3} + \dots + \dfrac{2017}{2016}+1 } {\dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} + \dots + \dfrac{1}{2016}+ \dfrac{1}{2017}} \\\\ &=& \dfrac{\dfrac{2017}{2} + \dfrac{2017}{3} + \dots + \dfrac{2017}{2016} +1 } {\left( \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} + \dots + \dfrac{1}{2016}+\dfrac{1}{2017} \right)\times \dfrac{2017}{2017} } \\\\ &=& \dfrac{\dfrac{2017}{2} + \dfrac{2017}{3} + \dots + \dfrac{2017}{2016} +1 } {\left( \dfrac{2017}{2} + \dfrac{2017}{3} + \dfrac{2017}{4} + \dots + \dfrac{2017}{2016}+\dfrac{2017}{2017} \right)\times \dfrac{1}{2017} } \\\\ &=& \dfrac{\dfrac{2017}{2} + \dfrac{2017}{3} + \dots + \dfrac{2017}{2016} +1 } {\left( \dfrac{2017}{2} + \dfrac{2017}{3} + \dfrac{2017}{4} + \dots + \dfrac{2017}{2016}+1 \right)\times \dfrac{1}{2017} } \\\\ &=& \dfrac{\left(\dfrac{2017}{2} + \dfrac{2017}{3}+ \dfrac{2017}{4} + \dots + \dfrac{2017}{2016} +1\right) } {\left( \dfrac{2017}{2} + \dfrac{2017}{3} + \dfrac{2017}{4} + \dots + \dfrac{2017}{2016}+1 \right) } \times 2017 \\\\ &=& \mathbf{2017} \\ \hline \end{array}\)

physics help

a. mg

\(\begin{array}{|rcll|} \hline && 6.0\times 10^3\ kg \times \dfrac{1000\ g}{1\ kg} \times \dfrac{1000\ mg}{1\ g} \\ &=& 6.0\times 10^3\times10^3\times10^3\ mg \\ \mathbf{6.0\times 10^3\ kg} &=& \mathbf{6.0\times 10^9\ mg} \\ \hline \end{array} \)

b. Mg

\(\begin{array}{|rcll|} \hline && 6.0\times 10^3\ kg \times \dfrac{1\ Mg}{1000\ kg} \\ &=& 6.0\times 10^3\times \dfrac{1}{10^3}\ Mg\\ \mathbf{6.0\times 10^3\ kg} &=& \mathbf{6.0\ Mg} \\ \hline \end{array}\)

The expression \(a(b - c)^3 + b(c - a)^3 + c(a - b)^3\) can be factored into the form

\( (a - b)(b - c)(c - a) p(a,b,c)\), for some polynomial \(p(a,b,c)\). 

Find p(a,b,c)

\(\begin{array}{|rcll|} \hline a(b - c)^3 + b(c - a)^3 + c(a - b)^3 &=& -a^3b + a^3 c + a b^3 - a c^3 - b^3 c + b c^3 \\ (a - b)(b - c)(c - a) &=& -a^2b + a^2 c + a b^2 - a c^2 - b^2 c + b c^2 \\ \hline -a^3b + a^3 c + a b^3 - a c^3 - b^3 c + b c^3 &=& (-a^2b + a^2 c + a b^2 - a c^2 - b^2 c + b c^2)(a+b+c) \\\\ &=& -a^3b + a^3 c + a b^3 - a c^3 - b^3 c + b c^3 \\ && +a^2b^2-a^2b^2 \\ && +a^2c^2-a^2c^2 \\ && +b^2c^2-b^2c^2 \\ && +a^2bc-a^2bc \\ && +b^2ac-b^2ac \\ &&+c^2ab-c^2ab \\\\ \mathbf{a(b - c)^3 + b(c - a)^3 + c(a - b)^3} &=&\mathbf{ (a - b)(b - c)(c - a)(a+b+c) } \\ \hline \end{array}\)

\(\mathbf{p(a,b,c) = a+b+c} \)

Help please
\(\text{Find the value of } x = 1 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \ddots}}}}.\)

Thank you, CPhill !