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Find the size of the smallest angle between the two hands of a clock displaying these times.

10:20, 6:55, 4:35, 2:27 and 11:09

1. angular speed = \(\omega\)

minutes hand: \(\omega_{\text{m}} = \frac{360^{\circ}}{1\ h}\)

hour hand: \(\omega_{\text{h}} = \frac{360^{\circ}}{12\ h}\)

2. angle between the two hands = \(\alpha\)

\(\begin{array}{rcll} \alpha &=& ( \omega_{\text{m}} - \omega_{\text{h}} ) \cdot t \\ \alpha &=& \Big( \frac{360^{\circ}}{1\ h} - \frac{360^{\circ}}{12\ h} \Big)\cdot t \\ \alpha &=& 360^{\circ}\cdot \Big( \frac{1}{1\ h} - \frac{1}{12\ h} \Big)\cdot t \\ \mathbf{\alpha} & \mathbf{=} & \mathbf{360^{\circ}\cdot \frac{11}{12\ h} \cdot t} \\ \end{array} \)

3. Solution for 10:20, 6:55, 4:35, 2:27 and 11:09

\(\begin{array}{|rcll|} \hline t=10:20=10.\bar{3}\ h & \alpha & = & 360^{\circ}\cdot \frac{11}{12\ h} \cdot 10.\bar{3}\ h \\ & \alpha & = & 3410^{\circ} \\ & \alpha & = & 3410^{\circ} \pmod {360^{\circ} } \\ & \mathbf{\alpha} & \mathbf{=} & \mathbf{170^{\circ}} \\\\ t=6:55=6.91\bar{6}\ h & \alpha & = & 360^{\circ}\cdot \frac{11}{12\ h} \cdot 6.91\bar{6}\ h \\ & \alpha & = & 2282.5^{\circ} \\ & \alpha & = & 2282.5^{\circ} \pmod {360^{\circ} } \\ & \mathbf{\alpha} & \mathbf{=} & \mathbf{122.5^{\circ}} \\\\ t=4:35=4.58\bar{3}\ h & \alpha & = & 360^{\circ}\cdot \frac{11}{12\ h} \cdot 4.58\bar{3}\ h \\ & \alpha & = & 1512.5^{\circ} \\ & \alpha & = & 1512.5^{\circ} \pmod {360^{\circ} } \\ & \mathbf{\alpha} & \mathbf{=} & \mathbf{72.5^{\circ}} \\\\ t=2:27=2.45\ h & \alpha & = & 360^{\circ}\cdot \frac{11}{12\ h} \cdot 2.45\ h \\ & \alpha & = & 808.5^{\circ} \\ & \alpha & = & 808.5^{\circ} \pmod {360^{\circ} } \\ & \mathbf{\alpha} & \mathbf{=} & \mathbf{88.5^{\circ}} \\\\ t=11:09=11.15\ h & \alpha & = & 360^{\circ}\cdot \frac{11}{12\ h} \cdot 11.15\ h \\ & \alpha & = & 3679.5^{\circ} \\ & \alpha & = & 3679.5^{\circ} \pmod {360^{\circ} } \\ & \mathbf{\alpha} & \mathbf{=} & \mathbf{79.5^{\circ}} \\\\ \hline \end{array}\)

Let's play mini-Sudoku!

We wish to place an "X" in four cells, such that there is exactly one "X" in each row, column, and 2x2 outlined box. For example:

In how many ways can we do this?
 

Each cell has a number:

\(\begin{array}{|r|r|r|r|} \hline I_1 & I_2 & II_1 & II_2 \\ \hline I_3 & I_4 & II_3 & II_4 \\ \hline III_1 & III_2 & IV_1 & IV_2 \\ \hline III_3 & III_4 & IV_3 & IV_4 \\ \hline \end{array}\)

Then we have 16 varying mini-Sudokus.

They are:

\(\begin{array}{|rrrrr|} \hline 1 & I_1 & II_3 & III_2 & IV_4 \\ 2 & I_1 & II_3 & III_4 & IV_2 \\ 3 & I_1 & II_4 & III_2 & IV_3 \\ 4 & I_1 & II_4 & III_4 & IV_1 \\ \hline 5 & I_2 & II_3 & III_1 & IV_4 \\ 6 & I_2 & II_3 & III_3 & IV_2 \\ 7 & I_2 & II_4 & III_1 & IV_3 \\ 8 & I_2 & II_4 & III_3 & IV_1 \\ \hline 9 & I_3 & II_1 & III_2 & IV_4 \\ 10 & I_3 & II_1 & III_4 & IV_2 \\ 11 & I_3 & II_2 & III_2 & IV_3 \\ 12 & I_3 & II_2 & III_4 & IV_1 \\ \hline 13 & I_4 & II_1 & III_1 & IV_4 \\ 14 & I_4 & II_1 & III_3 & IV_2 \\ 15 & I_4 & II_2 & III_1 & IV_3 \\ 16 & I_4 & II_2 & III_3 & IV_1 \\ \hline \end{array}\)

Simplify(-5+q)(-4p)-(-3q)(-3p-2)

\(\begin{array}{|rcll|} \hline && (-5+q)(-4p)-(-3q)(-3p-2) \\ &=& (5-q)(4p)+(3q)(-3p-2) \\ &=& 20p-4pq -9pq - 6q \\ &=& 20p-13pq - 6q \\ \hline \end{array}\)

What is the general form of the equation of a circle with its center at (-2, 1) and passing through (-4, 1)?

center at \((x_c=-2,\ y_c= 1 )\)

passing through point at \((x_p=-4,\ y_p= 1 )\)

\(\text{radius}^2 = r^2 = (x_c-x_p)^2+(y_c-y_p)^2\)

general form of the equation of a circle: \((x-x_c)^2 +(y-y_c)^2 = r^2\)

so we have:

\(\small{ \begin{array}{|rcll|} \hline (x-x_c)^2 +(y-y_c)^2 &=& r^2 \qquad | \quad r^2 = (x_c-x_p)^2+(y_c-y_p)^2 \\ (x-x_c)^2 +(y-y_c)^2 &=& (x_c-x_p)^2+(y_c-y_p)^2 \\ x^2-2x_c\cdot x + \not{x_c^2} + y^2-2y_c\cdot y + \not{y_c^2} &=& \not{x_c^2} -2x_cx_p + x_p^2 + \not{y_c^2}-2y_cy_p + y_p^2\\ x^2-2x_c\cdot x + y^2-2y_c\cdot y &=& -2x_cx_p + x_p^2 + -2y_cy_p + y_p^2 \\ x^2-2x_c\cdot x + y^2-2y_c\cdot y + 2x_cx_p - x_p^2 + 2y_cy_p - y_p^2 &=& 0 \\ x^2+ y^2 -2x_c\cdot x -2y_c\cdot y + 2x_cx_p + 2y_cy_p - x_p^2 - y_p^2 &=& 0 \\ x^2+ y^2 -2x_c\cdot x -2y_c\cdot y + x_p\cdot(2x_c-x_p) + y_p\cdot(2y_c-y_p) &=& 0 \\ \hline \end{array} } \)

The general form of the equation of a circle with its center \((x_c,y_c) \)and passing through point \( (x_p,y_p) \) is:

\(\mathbf{x^2+ y^2 -2x_c\cdot x -2y_c\cdot y + x_p\cdot(2x_c-x_p) + y_p\cdot(2y_c-y_p) = 0} \)

\(\small{ \begin{array}{|lrcll|} \hline x_c=-2,\ y_c= 1 \\ x_p=-4,\ y_p= 1 \\\\ & x^2+ y^2 -2x_c\cdot x -2y_c\cdot y + x_p\cdot(2x_c-x_p) + y_p\cdot(2y_c-y_p) &=& 0 \\ & x^2+ y^2 -2\cdot (-2)\cdot x -2\cdot 1\cdot y + (-4)\cdot[2\cdot(-2)-(-4)] + 1\cdot(2\cdot 1-1) &=& 0 \\ & x^2+ y^2 +4x -2y + (-4)\cdot(-4+4) + 1\cdot(2-1) &=& 0 \\ & x^2+ y^2 +4x -2y + 0 + 1\cdot 1 &=& 0 \\ & x^2+ y^2 +4x -2y + 1 &=& 0 \\ \hline \end{array} }\)

The equation of the circle is \(x^2+ y^2 +4x -2y + 1 = 0 \)

Find the diameter of a right cone with a slant height of  \(18\ cm\) and a surface area of \(208 \pi \ cm^2 \).

Surface area of cone \( A = \pi r^2+\pi rl\)

\(\begin{array}{|rcll|} \hline \pi r^2+ \pi rl &=& A \quad & | \quad : \pi \\ r^2+rl &=& \frac{A}{\pi} \\ r^2+rl -\frac{A}{\pi} &=& 0 \\ r &=& \frac{-l\pm \sqrt{l^2-4\cdot (-\frac{A}{\pi}) } } {2} \\ r &=& \frac{-l\pm \sqrt{l^2+ \frac{4A}{\pi} } } {2} \\ 2r = d &=& -l\pm \sqrt{l^2+ \frac{4A}{\pi} } \quad & | \quad A = 280\pi \qquad l = 18 \\ d &=& -18\pm \sqrt{18^2+ 4\cdot 208 } \\ d &=& -18\pm \sqrt{1156 } \\ d &=& -18\pm 34 \\ d &=& -18 + 34 \\ d &=& 16 \\ \hline \end{array}\)

The diameter  is 16 cm

There is a row of Pascal's triangle that has three successive positive entries,"a" "b"  and "c"

such that "b" is double "c" 

and "a" is triple "c"

If this row begins "1,n,"  then find n.

Three successive positive entries:

\(\begin{array}{rcll} a&=&\binom{n}{k-1} \\ b&=&\binom{n}{k} \\ c&=&\binom{n}{k+1} \\ \end{array} \)

"b" is double "c" and "a" is triple "c"

\(\begin{array}{|rcll|} \hline a &= 3c &=& \binom{n}{k-1} \\ b &= 2c &=& \binom{n}{k} \\ c & &=& \binom{n}{k+1} \\ \hline \end{array} \)

\(\begin{array}{|lrcll|} \hline (1) & 2c &=& \binom{n}{k} \quad & | \quad c = \binom{n}{k+1} \\ & 2\cdot \binom{n}{k+1} &=& \binom{n}{k} \quad & | \quad \binom{n}{k+1}= ( \frac{n-k}{k+1} ) \binom{n}{k} \\ & 2\cdot ( \frac{n-k}{k+1} ) \binom{n}{k} &=& \binom{n}{k} \\ & 2\cdot ( \frac{n-k}{k+1} ) &=& 1 \\ & \mathbf{ n-k } & \mathbf{=} & \mathbf{ \frac{k+1}{2} } \\\\ (2) & 3c &=& \binom{n}{k-1} \quad & | \quad c = \binom{n}{k+1} \\ & 3\cdot \binom{n}{k+1} &=& \binom{n}{k-1} \quad & | \quad \binom{n}{k+1}= ( \frac{n-k}{k+1} ) \binom{n}{k} \\ & 3\cdot ( \frac{n-k}{k+1} ) \binom{n}{k} &=& \binom{n}{k-1} \quad & | \quad \binom{n}{k-1}= ( \frac{k}{n-k+1} ) \binom{n}{k} \\ & 3\cdot ( \frac{n-k}{k+1} ) \binom{n}{k} &=& ( \frac{k}{n-k+1} ) \binom{n}{k} \\ & 3\cdot ( \frac{n-k}{k+1} ) &=& \frac{k}{n-k+1} \\ & 3\cdot (n-k)\cdot (n-k+1) &=& k\cdot (k+1) \quad & | \quad \mathbf{ n-k } \mathbf{=} \mathbf{ \frac{k+1}{2} } \\ & 3\cdot ( \frac{k+1}{2} )\cdot ( \frac{k+1}{2} +1) &=& k\cdot (k+1) \\ & 3\cdot ( \frac{k+1}{2} )\cdot ( \frac{k+3}{2} ) &=& k\cdot (k+1) \\ & \frac34\cdot (k+1)\cdot (k+3) &=& k\cdot (k+1) \\ & \frac34 \cdot (k+3) &=& k \\ & \frac34 k + \frac94 &=& k \\ & k-\frac34 k &=& \frac94 \\ & \frac14 k &=& \frac94 \\ & \mathbf{ k } & \mathbf{=} & \mathbf{9} \\\\ & \mathbf{ n-k } & \mathbf{=} & \mathbf{ \frac{k+1}{2} } \\ & n-9 & = & \frac{9+1}{2} \\ & n-9 & = & 5 \\ & \mathbf{ n } & \mathbf{=} & \mathbf{ 14 } \\ \hline \end{array}\)

The three successive positive entries are:

\(a=3003 =\binom{14}{8} \\ b=2002 =\binom{14}{9} \\ c=1001 =\binom{14}{10} \\ \)

and n is 14.

A square wall is covered with square tiles.

There are 85 tiles altogether along the two diagonals.

How many tiles are there on the whole wall.

\(\left(\dfrac{85+1}{2}\right)^2 = 43^2=1849\)

Solve for X given : 5.52 = -ln( -ln( -(1-1/X) ))

\(\begin{array}{|rcll|} \hline 5.52 &=& -\ln \Big( -\ln ( -(1-\frac{1}{x} ) ) \Big) \\ 5.52 &=& -\ln \Big( -\ln( \frac{1}{x}-1 ) \Big) \\ -5.52 &=& \ln \Big( -\ln( \frac{1}{x}-1 ) \Big) \quad & | \quad e^{()} \\ e^{-5.52} &=& -\ln( \frac{1}{x}-1 ) \\ -e^{-5.52} &=& \ln( \frac{1}{x}-1 ) \quad & | \quad e^{()} \\ e^{(-e^{-5.52})} &=& \frac{1}{x}-1 \\ 1+e^{(-e^{-5.52})} &=& \frac{1}{x} \\ \dfrac{1}{1+e^{(-e^{-5.52})} } &=& x \\ x &=& \dfrac{1}{1+e^{(-e^{-5.52})} } \\ x &=& \dfrac{1}{1+e^{(-0.00400584794)} } \\ x &=& \dfrac{1}{1+0.99600216476 } \\ x &=& \dfrac{1}{1.99600216476 } \\\\ \mathbf{x} & \mathbf{=} & \mathbf{0.50100146065} \\ \hline \end{array}\)

find angle c with trig

\(\begin{array}{|rcll|} \hline \tan(B) &=& \dfrac{c \cdot \sin(A)}{b-c\cdot \cos(A)} \quad & | \quad c = 75\ cm \quad b = 90\ cm \quad A = 20^{\circ} \\\\ \tan(B) &=& \dfrac{75 \cdot \sin(20^{\circ})}{90-75\cdot \cos(20^{\circ})} \\\\ \tan(B) &=& \dfrac{75 \cdot 0.34202014333}{90-75\cdot 0.93969262079} \\\\ \tan(B) &=& 1.31390875033 \\\\ B &=& \arctan(1.31390875033) \\\\ \mathbf{B} & \mathbf{=} & \mathbf{52.7256774464^{\circ}} \\ \hline \end{array}\)

The center of a circle is located at (−2, 7) . The radius of the circle is 2.

What is the equation of the circle in general form?

A circle can be defined as the locus of all points that satisfy the equation
\((x-h)^2 + (y-k)^2 = r^2 \)  ( Standard Form )
where r is the radius of the circle,
and h,k are the coordinates of its center.

The general Form is:
\(x^2+y^2 +ax+by+c = 0\)

Standard Form to general Form:

\(\begin{array}{|rcll|} \hline (x-h)^2 + (y-k)^2 &=& r^2 \\ x^2-2xh+h^2+y^2-2yk+k^2 &=& r^2 \\ x^2+y^2+x\cdot\underbrace{(-2h)}_{=a}+y\cdot\underbrace{(-2k)}_{=b}+\underbrace{h^2+k^2-r^2}_{=c} &=& 0 \\ \hline \end{array} \)

a,b and c ?

\(\begin{array}{|rcll|} \hline x^2+y^2+x\cdot\underbrace{(-2h)}_{=a}+y\cdot\underbrace{(-2k)}_{=b}+\underbrace{h^2+k^2-r^2}_{=c} &=& 0 \\\\ \color{red}a &\color{red}=& \color{red}-2h \\\\ \color{red}b &\color{red}=& \color{red}-2k \\\\ \color{red}c &\color{red}=&\color{red}h^2+k^2-r^2\\ \hline \end{array} \)

If we have h,k and r, we can calculate a,b and c:

\(\begin{array}{|lcll|} \hline \mathbf{x^2+y^2 +ax+by+c = 0} \\ a = -2h \\ b = -2k \\ c =h^2+k^2-r^2 \\ \hline \end{array}\)

\(h=-2\\ k=7\\ r=2\)

\(\begin{array}{|lcll|} \hline a = -2h \\ a = -2\cdot(-2)\\ \mathbf{a = 4} \\\\ b = -2k \\ b = -2(7) \\ \mathbf{a = -14} \\\\ c =h^2+k^2-r^2 \\ c =(-2)^2+7^2-2^2 \\ c =4+49-4 \\ \mathbf{c =49} \\\\ x^2+y^2 +ax+by+c = 0 \\ \mathbf{x^2+y^2 +4x-14y+49 =0} \\ \hline \end{array} \)

The equation of the circle in general form is: \(x^2+y^2 +4x-14y+49 =0\)