heureka

1.) Angle X and Angle Y are supplementary. Find the measure of each angle if Angle Measurement X= 6x -1 and Angle MEasurement Y=5x -17.

Two angles that sum to a straight angle \(180^{\circ} \)are called supplementary angles.

\(\begin{array}{rcl} X &=& 6x-1 \\ Y &=& 5x-17 \\ X+Y &=& 180^{\circ} = 6x-1 + 5x-17\\ 180^{\circ} &=& 11x-18\\ 180^{\circ}+18 &=& 11x\\ 198 &=& 11x\\ 11x &=& 198 \\ x &=& \frac{198}{11} \\ x &=& 18 \\\\ X &=& 6x -1\\ X &=& 6\cdot 18 -1 \\ X &=& 107^{\circ}\\\\ Y &=& 180^{\circ} -X \\ Y &=& 180^{\circ} -107^{\circ}\\ Y &=& 73^{\circ} \end{array}\)

2.) /P and /Q are complementary. The measure of /Q is 4 times the measure of /P. Find the measure of each angle.

Complementary angles are angle pairs whose measures sum to one right angle \( 90^{\circ}.\)

\(\begin{array}{rcl} Q &=& 4P\\ P + Q &=& 90^{\circ}\\ P + 4P &=& 90^{\circ}\\ 5P &=& 90^{\circ}\\ P &=& \frac{ 90^{\circ} }{5}\\ P &=& 18^{\circ}\\\\ Q &=& 90^{\circ} - P \\ Q &=& 90^{\circ} - 18^{\circ}\\ Q &=& 72^{\circ}\\ \end{array}\)

Very very Nice proof Berti !

Your answers are always wonderful

In a triangle ABC, AB = AC. D is the mid-point of BC, E is the foot of the perpendicular drawn from D to AC, and F is the mid-point of DE. Prove that AF is perpendicular to BE.

I. Definition:

\(\boxed{~ \begin{array}{lcl} \vec{a}= \vec{BC}\qquad |\vec{a}|= a \\ \vec{b}= \vec{AC}\qquad |\vec{b}|= b\qquad \vec{b}\cdot \vec{b} = b^2 \\ \quad \vec{a}\cdot \vec{b} = \frac{a}{2}\cdot a\\ \quad \vec{a}\cdot \vec{b} = \frac{a^2}{2}\\ \hline \vec{d}=\vec{AD}\\ \vec{d}=\vec{b}-\frac12 \vec{a} \\ \end{array} ~} \boxed{~ \begin{array}{lcl} \vec{f}=\vec{AE}\\ \vec{f}= \frac{ (\vec{b}\cdot\vec{d}) } {b^2} \vec{b}\\ \vec{f}= \frac{ (\vec{b}\cdot \left(\vec{b}-\frac12 \vec{a} \right) ) } {b^2} \vec{b}\\ \vec{f}= \frac{ (b^2 - \frac{ (\vec{a}\cdot \vec{b}) }{2} ) } {b^2} \vec{b}\\ \vec{f}= \frac{ (b^2 - \frac{ a^2 }{4} ) } {b^2} \vec{b}\\ \vec{f}= \left(1 - \frac{ a^2 }{4b^2} \right) \cdot \vec{b}\\ \end{array} ~}\\\)

\(\boxed{~ \begin{array}{lcl} \vec{e}=\vec{ED}\\ \vec{e}=\vec{d}-\vec{f}\\ \vec{e}=\left( \vec{b}-\frac12 \vec{a} \right) -\vec{f}\\ \frac{\vec{e}}{2}=\left( \vec{b}-\frac12 \vec{a} \right)\frac12 -\frac12 \vec{f}\\ \vec{AF} = \frac{\vec{e}}{2}+\vec{f} = \left( \vec{b}-\frac12 \vec{a} \right)\frac12 -\frac12 \vec{f}+\vec{f}\\ \vec{AF} = \left( \vec{b}-\frac12 \vec{a} \right)\frac12 +\frac12 \vec{f}\\ \vec{AF} = \left( \vec{b}-\frac12 \vec{a} \right)\frac12 +\frac12 \left(1 - \frac{ a^2 }{4b^2} \right) \cdot \vec{b}\\ \vec{AF} = \vec{b} \left[ \frac12 +\frac12 \cdot \left( 1 - \frac{ a^2 }{4b^2} \right) \right] -\frac14 \cdot \vec{a} \\ \vec{AF} = \vec{b} \left( \frac12 +\frac12 - \frac{ a^2 }{8b^2} \right) -\frac14 \cdot \vec{a} \\ \vec{AF} = \vec{b} \left( 1 - \frac{ a^2 }{8b^2} \right) -\frac14 \cdot \vec{a} \\ \end{array} ~} \boxed{~ \begin{array}{lcl} \vec{BE} = \vec{a}-(\vec{b}-\vec{f})\\ \vec{BE} = \vec{a}-\vec{b}+\vec{f}\\ \vec{BE} = \vec{a}-\vec{b}+\left(1 - \frac{ a^2 }{4b^2} \right) \cdot \vec{b}\\ \vec{BE} = \vec{a}-\vec{b}+ \vec{b} - \left(\frac{ a^2 }{4b^2} \right) \cdot \vec{b}\\ \vec{BE} = \vec{a} - \left(\frac{ a^2 }{4b^2} \right) \cdot \vec{b}\\ \end{array} ~}\)

II. Prove that AF is perpendicular to BE.

\(\boxed{~ \begin{array}{lcl} \vec{AF} = -\frac14 \cdot \vec{a} + \vec{b} \left( 1 - \frac{ a^2 }{8b^2} \right) \\ \vec{BE} = \vec{a} - \vec{b} \left(\frac{ a^2 }{4b^2} \right)\\ \text{Perpendicular, if } \ (~\vec{AF}\cdot \vec{BE}~) = 0 \\ \hline \\ (~\vec{AF}\cdot \vec{BE}~) = \left ( -\frac14 \cdot \vec{a} + \vec{b} \left( 1 - \frac{ a^2 }{8b^2} \right) \right) \cdot \left( \vec{a} - \vec{b} \left(\frac{ a^2 }{4b^2} \right) \right) \\ (~\vec{AF}\cdot \vec{BE}~) = -\frac14 \cdot \vec{a} \cdot \vec{a} + \frac14 \cdot \vec{a} \cdot \vec{b} \left(\frac{ a^2 }{4b^2} \right) + \vec{b} \cdot \vec{a} \left( 1 - \frac{ a^2 }{8b^2} \right) - \vec{b}\cdot \vec{b} \left( 1 - \frac{ a^2 }{8b^2} \right) \left(\frac{ a^2 }{4b^2} \right)\\ \qquad \vec{a} \cdot \vec{a} = a^2 \qquad \vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a} =\frac{a^2}{2} \qquad \vec{b}\cdot \vec{b}=b^2 \\ (~\vec{AF}\cdot \vec{BE}~) = -\frac{ a^2 }{4} + \frac14 \cdot \frac{a^2}{2} \left(\frac{ a^2 }{4b^2} \right) + \frac{a^2}{2} \left( 1 - \frac{ a^2 }{8b^2} \right) - b^2 \left( 1 - \frac{ a^2 }{8b^2} \right) \left(\frac{ a^2 }{4b^2} \right)\\ (~\vec{AF}\cdot \vec{BE}~) = -\frac{ a^2 }{4} + \frac{a^4}{32b^2} + \frac{a^2}{2} - \frac{ a^4 }{16b^2} - \left( 1 - \frac{ a^2 }{8b^2} \right) \left(\frac{ a^2 }{4} \right)\\ (~\vec{AF}\cdot \vec{BE}~) = -\frac{ a^2 }{4} + \frac{a^4}{32b^2} + \frac{a^2}{2} - \frac{ a^4 }{16b^2} -\frac{ a^2 }{4}+ \frac{ a^4 }{32b^2} \\ (~\vec{AF}\cdot \vec{BE}~) = -\frac{ a^2 }{4}-\frac{ a^2 }{4}+ \frac{a^2}{2} + \frac{a^4}{32b^2} + \frac{ a^4 }{32b^2} - \frac{ a^4 }{16b^2} \\ \color{red }(~\vec{AF}\cdot \vec{BE}~) \color{black }= -\frac{ a^2 }{2}+ \frac{a^2}{2} + \frac{a^4}{16b^2} - \frac{ a^4 }{16b^2} \color{red }= 0\\ \end{array} ~}\)

sin(x+15 degree) = 3 cos( x-15 degree)

\(\small{ \boxed{~ \text{Formula: }\quad \begin{array}{lcl} \sin{ (x+y) } &=& \sin{(x)}\cdot \cos{(y)} + \sin{(y)} \cdot \cos{(x)} \\ \cos{ (x-y) } &=& \cos{(x)}\cdot \cos{(y)} + \sin{(x)} \cdot \sin{(y)} \\ \end{array} ~}\\ \begin{array}{rcl} \sin{ ( x + 15^{\circ} ) } &=& 3 \cos{ ( x - 15^{\circ} ) } \\ \sin{(x)}\cdot \cos{(15^{\circ})} + \sin{(15^{\circ})} \cdot \cos{(x)} &=& 3 \cdot [~ \cos{(x)}\cdot \cos{( 15^{\circ})} + \sin{(x)} \cdot \sin{( 15^{\circ})} ~] \\ \sin{(x)}\cdot \cos{(15^{\circ})} + \sin{(15^{\circ})} \cdot \cos{(x)} &=& 3 \cdot \cos{(x)}\cdot \cos{( 15^{\circ})} + 3 \cdot \sin{(x)} \cdot \sin{( 15^{\circ})} \quad | \quad : \cos{(x)} \quad x\ne 90^{\circ}\\ \tan{(x)}\cdot \cos{(15^{\circ})} + \sin{(15^{\circ})} &=& 3 \cdot \cos{( 15^{\circ})} + 3 \cdot \tan{(x)} \cdot \sin{( 15^{\circ})} \\ \tan{(x)}\cdot \cos{(15^{\circ})} - 3 \cdot \tan{(x)} \cdot \sin{( 15^{\circ})} &=& 3 \cdot \cos{( 15^{\circ})} - \sin{(15^{\circ})} \qquad | \qquad : \cos{(15^{\circ})} \\ \tan{(x)} - 3 \cdot \tan{(x)} \cdot \tan{( 15^{\circ})} &=& 3 - \tan{(15^{\circ})} \\ \tan{(x)}\cdot \left[~ 1 - 3 \cdot \tan{( 15^{\circ})} ~ \right] &=& 3 - \tan{(15^{\circ})} \\ \tan{(x)} &=& \frac{ 3 - \tan{(15^{\circ})} } { 1 - 3 \cdot \tan{( 15^{\circ})} } \\ \tan{(x)} &=& \frac{ 2.7320508076 } { 0.1961524227 } \\ \tan{(x)} &=& 13.9282032303 \\ \mathbf{ x } & \mathbf{=} & \mathbf{ 85.8933946491^{\circ} \pm k\cdot 180^{\circ} \qquad k \in Z } \end{array} }\)

6(в степени 8-х)=216. Найдите корень уравнения.

I assume the question is:

  \(\begin{array}{rcl} 6^{8-x} &=& 216 \\ 6^{8-x} &=& 6^3 \\ 8-x &=& 3 \\ x &=& 8-3 \\ x &=& 5 \end{array}\)

How do I solve this in hand: diff(f(x), x) = 4*x^3-4/x^2

I assume the question is:  \(\dfrac{d\ \left( 4\cdot x^3 - \frac{4}{x^2 } \right) } {dx} = \ ?\)

\(\begin{array}{rcl} y &=& 4\cdot x^3 - \frac{4}{x^2 } \\ y &=& 4 \cdot \left( x^3 - \frac{1}{x^2} \right)\\ y' &=& 4 \cdot [ \left( x^3 \right)' - \left( \frac{1}{x^2} \right)' ] \\ y' &=& 4 \cdot [ \left( x^3 \right)' - \left( x^{-2} \right)' ] \\ \boxed{~ \begin{array}{lcl} \text{rule:} \qquad \left(x^n\right)' &=& n x^{n-1} \end{array} ~}\\ y' &=& 4 \cdot [ \left( 3\cdot x^{3-1} \right) - \left( (-2)\cdot x^{-2-1} \right) ] \\ y' &=& 4 \cdot [ \left( 3\cdot x^{3-1} \right) + \left( 2\cdot x^{-2-1} \right) ] \\ y' &=& 4 \cdot [ 3x^{2} + 2\cdot x^{-3} ] \\ y' &=& 4 \cdot [ 3x^{2} + \frac{2}{x^3} ] \\ y' &=& 12x^{2} + \frac{8}{x^3} \\ \end{array}\)

[(+2 3/4)*(-1 1/11)]/(+1 1/5)

\(\begin{array}{rcl} \dfrac{ \left( +2 \frac34 \right) \cdot \left( -1 \frac{1}{11} \right) } { +1 \frac15 } &=& \dfrac{ \left( \frac{ 2\cdot 4+3 }{4} \right) \cdot \left( - \frac{ 1\cdot 11+1 }{11} \right) } { \frac{ 1\cdot 5+1 }{5} } \\ &=& \dfrac{ \left( \frac{11}{4} \right) \cdot \left( - \frac{12}{11} \right) } { \frac{6}{5} } \\ &=& -\dfrac{ \left( \frac{11}{4} \right) \cdot \left(\frac{12}{11} \right) } { \frac{6}{5} } \\ &=& -\dfrac{ \left( \frac{12}{4} \right) \cdot \left(\frac{11}{11} \right) } { \frac{6}{5} } \\ &=& -\dfrac{ 3 }{ \frac{6}{5} } \\ &=& - \frac{3 \cdot 5}{6} \\ &=& - \frac52 \\ &=& - 2\frac12 \\ &=& - 2,5 \\ \end{array}\)

Will can paint a house in 3 hours. Sam can paint a house in 5 hours. How long will it take for them to paint it together?

\(\begin{array}{lll} \text{Will painting } & \text{per hour} & \frac13 \text{ house }\\ \text{Sam painting } & \text{per hour} & \frac15 \text{ house }\\\\ \end{array}\\ \boxed{~ \begin{array}{lll} \frac13 \frac{\text{house}}{\text{hour}} \cdot t + \frac15 \frac{\text{house}}{\text{hour}} \cdot t & =& 1\ \text{ house } \end{array} ~}\\ \begin{array}{rcl} \\ t \cdot \left( \frac13 \frac{\text{house}}{\text{hour}} + \frac15 \frac{\text{house}}{\text{hour}} \right) & =& 1\ \text{ house }\\ t \cdot \left( \frac13 + \frac15 \right) \frac{\text{house}}{\text{hour}} & =& 1\ \text{ house }\\ t \cdot \left( \frac13\cdot \frac55 + \frac15\cdot \frac33 \right) \frac{\text{house}}{\text{hour}} & =& 1\ \text{ house }\\ t \cdot \left( \frac{5}{15} + \frac{5}{15} \right) \frac{\text{house}}{\text{hour}} & =& 1\ \text{ house }\\ t \cdot \left( \frac{8}{15} \right) \frac{\text{house}}{\text{hour}} & =& 1\ \text{ house }\\ t & =& \frac{15}{8}\ \text{ house }\frac{\text{hour}}{\text{house}}\\ t & =& \frac{15}{8}\ \text{ hour }\\ t & =& 1\ \text{ hour } 52.5\ \text{ minutes }\\ \end{array}\)

The sum is 1+11+111+1111+1111... find the sum of n

Sorry

New edit, without mistake:

\(\small{ \begin{array}{rcl} a_1 &=& 1 \\ a_2 &=& 11\\ a_3 &=& 111\\ a_4 &=& 1111\\ \dots \\ \end{array} \qquad \begin{array}{rcl} a_1 &=& 1 \\ a_2 &=& 1 + 10^1\\ a_3 &=& 1 + 10^1+ 10^2\\ a_4 &=& 1 + 10^1+ 10^2+10^3\\ \cdots \\ a_n &=& 1 + 10^1+ 10^2+10^3+10^4+\cdots + 10^{n-2}+ 10^{n-1}\\ \end{array}\\\\ \begin{array}{rcl} \\ S_n &=& ( n - 0 )\cdot 10^0 +( n - 1 )\cdot 10^1 +( n - 2 )\cdot 10^2 +( n - 3 )\cdot 10^3 +( n - 4 )\cdot 10^4\\ & + & \cdots +[ n - (n-2) ]\cdot 10^{n-2} +[ n - (n-1) ]\cdot 10^{n-1}\\ S_n &=& n\cdot 1 + n\cdot 10^1 + n \cdot 10^2 + n\cdot 10^4 + \cdots + n \cdot 10^{n-2} + n \cdot 10^{n-1} \\ & -& 0\cdot 10^0 - 1 \cdot 10^1 - 2\cdot 10^2 - 3\cdot 10^3 - 4\cdot 10^4 -\cdots -(n-2)\cdot 10^{n-2} -(n-1)\cdot 10^{n-1}\\ S_n &=& n\cdot \underbrace{( 1 + 10^1 + 10^2 + 10^4 + \cdots + 10^{n-2} + 10^{n-1} )}_{\text{geometric series}} - \underbrace{\sum \limits_{r=1}^{n} { (r-1)\cdot 10^{r-1} }}_{\text{'Arithmetic-geometric' series}}\\ S_n &=& n\cdot \left( \dfrac{10^{n}-1}{10-1} \right) - \sum \limits_{r=1}^{n} { (r-1)\cdot 10^{r-1} }\\\\ S_n &=& n\cdot \left( \dfrac{10^{n}-1}{10-1} \right) - s_n\\\\ \end{array}\\ \begin{array}{lclcl} \hline s_n = \sum \limits_{r=1}^{n} { (r-1)\cdot 10^{r-1} } &=& 0\cdot 10^0 + &1& \cdot 10^1 + 2\cdot 10^2 + 3\cdot 10^3 +4\cdot 10^4 +\cdots + (n-1)\cdot 10^{n-1} \\ 10\cdot s_n &=& &0&\cdot 10^1+ 1\cdot 10^2 + 2\cdot 10^3 + 3\cdot 10^4 +\cdots + (n-2)\cdot 10^{n-1} + (n-1)\cdot 10^{n} \\ \hline \end{array}\\ \begin{array}{rcl} s_n- 10 s_n &=& \underbrace{1\cdot 10^1 + 1\cdot 10^2+1\cdot 10^3 +\cdots + 1\cdot 10^{n-2} + 1\cdot 10^{n-1}}_{\text{geometric series}} - (n-1)\cdot 10^n \\\\ -9 s_n &=& 10\cdot \left( \dfrac{10^{n-1}-1}{10-1} \right)- (n-1)\cdot 10^n \\\\ -9 s_n &=& 10\cdot \left( \dfrac{10^{n-1}-1}{9} \right)- (n-1)\cdot 10^n \\\\ - s_n &=& 10\cdot \left( \dfrac{10^{n-1}-1}{9\cdot 9} \right)- \dfrac { (n-1)\cdot 10^n}{9} \\ \hline \end{array}\\ \begin{array}{rcl} S_n &=& n\cdot \left( \dfrac{10^{n}-1}{9} \right) - s_n\\\\ \end{array}\\\\ \begin{array}{rcl} \boxed{~ S_n = n\cdot \left( \dfrac{10^{n}-1}{9} \right) + 10\cdot \left( \dfrac{10^{n-1}-1}{9\cdot 9} \right)- \dfrac { (n-1)\cdot 10^n}{9} \\\\ \text{or }\quad S_n = \dfrac{1}{81}\cdot [~ 10\cdot (10^n - 1) - 9n ~]\\\\ \text{Example: } \\ \qquad S_4 = \dfrac{1}{81}\cdot [~ 10\cdot (10^4 - 1) - 9\cdot 4 ~]\\ \qquad S_4 = \dfrac{1}{81}\cdot [~ 10\cdot (10000 - 1) - 36 ~]\\ \qquad S_4 = \dfrac{1}{81}\cdot [~ 10\cdot (9999) - 36 ~]\\ \qquad S_4 = \dfrac{1}{81}\cdot [~ 99990 - 36 ~]\\ \qquad S_4 = \dfrac{1}{81}\cdot [~ 99954 ~]\\ \qquad S_4 = 1234 ~} \end{array} }\)

The sum is 1+11+111+1111+1111... find the sum of n

\(\small{ \begin{array}{rcl} a_1 &=& 1 \\ a_2 &=& 11\\ a_3 &=& 111\\ a_4 &=& 1111\\ \dots \\ \end{array} \qquad \begin{array}{rcl} a_1 &=& 1 \\ a_2 &=& 1 + 10^1\\ a_3 &=& 1 + 10^1+ 10^2\\ a_4 &=& 1 + 10^1+ 10^2+10^3\\ \cdots \\ a_n &=& 1 + 10^1+ 10^2+10^3+10^4+\cdots + 10^{n-2}+ 10^{n-1}\\ \end{array}\\\\ \begin{array}{rcl} \\ S_n &=& ( n - 0 )\cdot 10^0 +( n - 1 )\cdot 10^1 +( n - 2 )\cdot 10^2 +( n - 3 )\cdot 10^3 +( n - 4 )\cdot 10^4\\ & + & \cdots +[ n - (n-2) ]\cdot 10^{n-2} +[ n - (n-1) ]\cdot 10^{n-1}\\ S_n &=& n\cdot 1 + n\cdot 10^1 + n \cdot 10^2 + n\cdot 10^4 + \cdots + n \cdot 10^{n-2} + n \cdot 10^{n-1} \\ & -& 0\cdot 10^0 - 1 \cdot 10^1 - 2\cdot 10^2 - 3\cdot 10^3 - 4\cdot 10^4 -\cdots -(n-2)\cdot 10^{n-2} -(n-1)\cdot 10^{n-1}\\ S_n &=& n\cdot \underbrace{( 1 + 10^1 + 10^2 + 10^4 + \cdots + 10^{n-2} + 10^{n-1} )}_{\text{geometric series}} - \underbrace{\sum \limits_{r=1}^{n} { (r-1)\cdot 10^{r-1} }}_{\text{'Arithmetic-geometric' series}}\\ S_n &=& n\cdot \left( \dfrac{10^{n-1}-1}{10-1} \right) - \sum \limits_{r=1}^{n} { (r-1)\cdot 10^{r-1} }\\\\ S_n &=& n\cdot \left( \dfrac{10^{n-1}-1}{10-1} \right) - s_n\\\\ \end{array}\\ \begin{array}{lclcl} \hline s_n = \sum \limits_{r=1}^{n} { (r-1)\cdot 10^{r-1} } &=& 0\cdot 10^0 + &1& \cdot 10^1 + 2\cdot 10^2 + 3\cdot 10^3 +4\cdot 10^4 +\cdots + (n-1)\cdot 10^{n-1} \\ 10\cdot s_n &=& &0&\cdot 10^1+ 1\cdot 10^2 + 2\cdot 10^3 + 3\cdot 10^4 +\cdots + (n-2)\cdot 10^{n-1} + (n-1)\cdot 10^{n} \\ \hline \end{array}\\ \begin{array}{rcl} s_n- 10 s_n &=& \underbrace{1\cdot 10^1 + 1\cdot 10^2+1\cdot 10^3 +\cdots + 1\cdot 10^{n-2} + 1\cdot 10^{n-1}}_{\text{geometric series}} - (n-1)\cdot 10^n \\\\ -9 s_n &=& 10\cdot \left( \dfrac{10^{n-1}-1}{10-1} \right)- (n-1)\cdot 10^n \\\\ -9 s_n &=& 10\cdot \left( \dfrac{10^{n-1}-1}{9} \right)- (n-1)\cdot 10^n \\\\ - s_n &=& 10\cdot \left( \dfrac{10^{n-1}-1}{9\cdot 9} \right)- \dfrac { (n-1)\cdot 10^n}{9} \\ \hline \end{array}\\ \begin{array}{rcl} S_n &=& n\cdot \left( \dfrac{10^{n-1}-1}{9} \right) - s_n\\\\ \end{array}\\\\ \begin{array}{rcl} \boxed{~ S_n = n\cdot \left( \dfrac{10^{n-1}-1}{9} \right) + 10\cdot \left( \dfrac{10^{n-1}-1}{9\cdot 9} \right)- \dfrac { (n-1)\cdot 10^n}{9} \\\\ ~} \end{array} }\)