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help with geometry

\(\begin{array}{|rcll|} \hline \mathbf{ \overline{NM}*\overline{AM} } &=& \mathbf{ \overline{LM}*\overline{KM} } \\ (x-4)*(3+x-4) &=& (x-5)*(6+x-5) \\ (x-4)*(x-1) &=& (x-5)*(x+1) \\ x^2-x-4x+4 &=& x^2+x-5x-5 \\ -x-4x+4 &=& +x-5x-5 \\ -5x+4 &=& -4x-5 \\ -x &=& -9 \\ \mathbf{x} &=& \mathbf{9} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{\overline{KM}} &=& \mathbf{6+x-5} \\ \overline{KM} &=& x+1 \quad | \quad \mathbf{x=9} \\ \overline{KM} &=& 9+1 \\ \mathbf{\overline{KM}} &=& \mathbf{10} \\ \hline \end{array}\)

The address of the pic is: https://web2.0calc.com/img/upload/9de3171bf728a80bb304cee0/geometry-1.png

Help Please!

\(\text{Let $x$ and $y$ be integers. Show that $9x + 5y$ is divisible by 19 $\\$if and only if $x + 9y$ is divisible by 19.}\)

My attempt:

\(\begin{array}{|rcll|} \hline x + 9y &\equiv& 0 \pmod{19} \\ x + 9y &=& 19n,\ \qquad n\in \mathbb{Z} \\ \mathbf{x} &=& \mathbf{19n -9y} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline 9x + 5y &\overset{\color{red}?}{\equiv}& 0 \pmod{19} \quad | \quad \mathbf{x=19n -9y} \\ 9(19n -9y) + 5y &\overset{\color{red}?}{\equiv}& 0 \pmod{19} \\ 9*19n -81y + 5y &\overset{\color{red}?}{\equiv}& 0 \pmod{19} \\ 9*19n -76y &\overset{\color{red}?}{\equiv}& 0 \pmod{19} \\ 9*{\color{red}19}n -4*{\color{red}19}y & \equiv& 0 \pmod{19} \checkmark \\ \hline \end{array}\)

PLEASE HELP ON THIS!

My attempt:

\(\begin{array}{|rcl|l|} \hline 1 &:& 3 & \\ a &:& a+c & c \\ \hline 1 &:& 3 & 3-1 = 2 & \text{yes } \\ 2 &:& 6 & 6-2 = 4 & \text{yes } \\ 3 &:& 9 & 9-3 = 6 & \text{yes } \\ 4 &:& 12 & 12-4 = 8 & \text{yes } \\ \hline 5 &:& 15 & 15-5 = 10 & \text{no } \ c\ge10 \\ \vdots &:& \vdots & \vdots & \text{no } \ c\ge10 \\ \hline \end{array}\) \(\ldots \text{ so } \mathbf{a=\{1,2,3,4\} }\)

\(\begin{array}{|rcl|l|} \hline 3 &:& 5 & \\ a &:& a+b & b \\ \hline \mathbf{\color{red}3} &:& 5 & 5-3 = \mathbf{\color{red}2} & \text{yes } \\ \hline 6 &:& 10 & 10-6 = 4 & \text{no } \ a\ne \{1,2,3,4\} \\ \vdots &:& \vdots & \vdots & \text{no } \ a\ne \{1,2,3,4\} \\ \hline \end{array}\)\(\ldots \text{ so } \mathbf{a=3} \ \rightarrow \ \mathbf{b=2}\ \rightarrow \ \mathbf{c=6} \)

\(\begin{array}{|rcll|} \hline \mathbf{ 50\%*(b+d)} &=&\mathbf{ b } \\ 50\%* b+ 50\%*d &=& 100\% *b \\ 50\%*d &=& 100\% *b-50\% *b \\ 50\%*d &=& 50\%* b \\ d &=& b \quad | \quad \mathbf{b=2} \\ \mathbf{ d } &=& \mathbf{ 2 } \\ \hline \end{array}\)

check:

\(\dfrac{1}{3} * (3+6) = 3 \ (a=3\checkmark) \\~\\ \dfrac{1}{2} * (2+2) = 2 \ (d=2\checkmark)\\~\\ \dfrac{3}{5} * (3+2) = 3 \ (a=3\checkmark)\\~\\ \dfrac{1}{4} * (6+2) = 2 \ (d=2\checkmark)\)

Use the Euclidean algorithm to find integers x and y such that 164x + 37y = 1.
State your answer as a list with x first and y second, separated by a comma.

Use the Euclidean algorithm to find integers x and y such that 164x + 37y = 1. 

State your answer as a list with x first and y second, separated by a comma.

\(\begin{array}{|lrclcrcl|c|} \hline \text{Euclidean algorithm}& &&&& && \text{Remainder} &\text{Remainder} \\ 164x + 37y = 1 \\ \hline & 164 &:& 37 &=& 4*37 &+& 16 & \mathbf{16} = 164-4*37 \\ & 37 &:& 16 &=& 2*16 &+& 5 & \mathbf{5}=37-2*16 \\ & 16 &:& 5 &=& 3*5 &+& 1 & \mathbf{1=16-3*5} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{1} &=& \mathbf{16-3*5} \quad | \quad \mathbf{5}=37-2*16 \\ 1 &=& 16-3*(37-2*16) \\ 1 &=& 16-3*37+6*16 \\ 1 &=& 7*16-3*37 \quad | \quad \mathbf{16} = 164-4*37 \\ 1 &=& 7*(164-4*37) -3*37 \\ 1 &=& 7*164-28*37 -3*37 \\ \mathbf{1} &=& \mathbf{\mathbf{\color{red}7}*164\mathbf{\color{red}-31}*37} \\ \hline \end{array}\)

\(\mathbf{x=7,\ y=-31} \)

Using the Euclidian Algorithm, calculate \(\gcd(2^{60}-1, 2^{80}-1)\)

Formula: \( \boxed{ \gcd(a,b) = \gcd(b,a) \\~\\ \gcd(a,b) = \gcd(a-n*b,b) \\~\\ \gcd(a,a) = a }\)

\(\begin{array}{|rcll|} \hline && \mathbf{\gcd\left( 2^{60}-1,~ 2^{80}-1 \right)} \\ &=& \gcd\left( 2^{80}-1,~ 2^{60}-1 \right) \\ && \begin{array}{|rcll|} \hline 2^{80}-1 &=& 1+2+2^2+2^3+\ldots+2^{79} \\ 2^{60}-1 &=& 1+2+2^2+2^3+\ldots+2^{59} \\ \hline 2^{20}*(2^{60}-1) &=& 2^{20}+2^{21}+\ldots+2^{79} \\ \hline \mathbf{ (2^{80}-1)-2^{20}*(2^{60}-1)} &=& \mathbf{1+2+2^2+2^3+\ldots+2^{19}} \\ \mathbf{ (2^{80}-1)-2^{20}*(2^{60}-1)} &=& \mathbf{2^{20}-1} \\ \hline \end{array} \\ &=& \gcd\left( (2^{80}-1)-2^{20}*(2^{60}-1),~ 2^{60}-1 \right) \\ &=& \mathbf{ \gcd\left( 2^{20}-1,~ 2^{60}-1 \right) } \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline && \mathbf{ \gcd\left( 2^{20}-1,~ 2^{60}-1 \right) } \\ &=& \mathbf{ \gcd\left( 2^{60}-1,~ 2^{20}-1 \right) } \\ && \begin{array}{|rcll|} \hline 2^{60}-1 &=& 1+2+2^2+2^3+\ldots+2^{59} \\ 2^{20}-1 &=& 1+2+2^2+2^3+\ldots+2^{19} \\ \hline 2^{40}*(2^{20}-1) &=& 2^{40}+2^{41}+\ldots+2^{59} \\ \hline \mathbf{ (2^{60}-1)-2^{40}*(2^{20}-1)} &=& \mathbf{1+2+2^2+2^3+\ldots+2^{39}} \\ \mathbf{ (2^{60}-1)-2^{40}*(2^{20}-1)} &=& \mathbf{2^{40}-1} \\ \hline \end{array} \\ &=& \gcd\left( (2^{60}-1)-2^{40}*(2^{20}-1),~ 2^{20}-1 \right) \\ &=& \mathbf{ \gcd\left( 2^{40}-1,~ 2^{20}-1 \right) } \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline && \mathbf{ \gcd\left( 2^{40}-1,~ 2^{20}-1 \right) } \\ && \begin{array}{|rcll|} \hline 2^{40}-1 &=& 1+2+2^2+2^3+\ldots+2^{39} \\ 2^{20}-1 &=& 1+2+2^2+2^3+\ldots+2^{19} \\ \hline 2^{20}*(2^{20}-1) &=& 2^{20}+2^{21}+\ldots+2^{39} \\ \hline \mathbf{ (2^{40}-1)-2^{20}*(2^{20}-1)} &=& \mathbf{1+2+2^2+2^3+\ldots+2^{19}} \\ \mathbf{ (2^{40}-1)-2^{20}*(2^{20}-1)} &=& \mathbf{2^{20}-1} \\ \hline \end{array} \\ &=& \gcd\left( (2^{40}-1)-2^{20}*(2^{20}-1),~ 2^{20}-1 \right) \\ &=& \mathbf{ \gcd\left( 2^{20}-1,~ 2^{20}-1 \right) } \\ &=& 2^{20}-1 \\ &=& \mathbf{1048575} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline && \mathbf{\gcd\left( 2^{60}-1,~ 2^{80}-1 \right)} \\ &=& \mathbf{ \gcd\left( 2^{60}-1,~ 2^{20}-1 \right) } \\ &=& \mathbf{ \gcd\left( 2^{40}-1,~ 2^{20}-1 \right) } \\ &=& \mathbf{ \gcd\left( 2^{20}-1,~ 2^{20}-1 \right) } \\ &=& \mathbf{2^{20}-1} \\ &=& \mathbf{1048575} \\ \hline \end{array}\)

Solve \(\large \sqrt{x+14-8\sqrt{x-2}} + \sqrt{x+23-10\sqrt{x-2}} = 3\)

Substitute: \(\begin{array}{|rcll|} \hline \mathbf{ y } &=& \mathbf{\sqrt{x-2}} \qquad \text{or} \qquad y^2 &=& x-2 \\ \mathbf{ x } &=& \mathbf{y^2+2} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \sqrt{x+14-8\sqrt{x-2}} + \sqrt{x+23-10\sqrt{x-2}} &=& 3 \\\\ \sqrt{y^2+2+14-8y} + \sqrt{y^2+2+23-10y} &=& 3 \\\\ \sqrt{y^2-8y+16} + \sqrt{y^2-10y+25} &=& 3 \\\\ && \boxed{ y^2-8y+16 = (y-4)^2=(4-y)^2} \\ && \boxed{ y^2-10y+25 = (y-5)^2=(5-y)^2} \\ \hline \end{array} \)

\(\begin{array}{|lrcll|} \hline \text{Case }1: \\ & y^2-8y+16 = (y-4)^2 \\ & y^2-10y+25 = (y-5)^2 \\ \hline & \sqrt{(y-4)^2} + \sqrt{(y-5)^2 } &=& 3 \\\\ & y-4 + y-5 &=& 3 \\\\ & 2y-9 &=& 3 \\\\ & 2y &=& 9+3 \\\\ & 2y &=& 12 \\\\ & \mathbf{y} &=& \mathbf{6} \\ \hline \end{array} \begin{array}{|lrcll|} \hline \text{Case }2: \\ & y^2-8y+16 = (y-4)^2 \\ & y^2-10y+25 = (5-y)^2 \\ \hline & \sqrt{(y-4)^2} + \sqrt{(5-y)^2 } &=& 3 \\\\ & y-4 + 5-y &=& 3 \\\\ & \mathbf{1} &\ne& \mathbf{3} \qquad \text{no solution!} \\ \hline \end{array}\\ \begin{array}{|lrcll|} \hline \text{Case }3: \\ & y^2-8y+16 = (4-y)^2 \\ & y^2-10y+25 = (5-y)^2 \\ \hline & \sqrt{(4-y)^2} + \sqrt{(5-y)^2 } &=& 3 \\\\ & 4-y + 5-y &=& 3 \\\\ & -2y+9 &=& 3 \\\\ & 2y &=& 9-3 \\\\ & 2y &=& 6 \\\\ & \mathbf{y} &=& \mathbf{3} \\ \hline \end{array} \begin{array}{|lrcll|} \hline \text{Case }4: \\ & y^2-8y+16 = (4-y)^2 \\ & y^2-10y+25 = (y-5)^2 \\ \hline & \sqrt{(4-y)^2} + \sqrt{(y-5)^2 } &=& 3 \\\\ & 4-y + y-5 &=& 3 \\\\ & \mathbf{-1} &\ne& \mathbf{3} \qquad \text{no solution!} \\ \hline \end{array}\\\)

\(\begin{array}{|rcll|} \hline \text{First solution}: \\ y= 6 \\ \hline x &=& y^2+2 \\ x &=& 6^2+2 \\ \mathbf{x} &=& \mathbf{38} \\ \hline \end{array} \begin{array}{|rcll|} \hline \text{Second solution}: \\ y= 3 \\ \hline x &=& y^2+2 \\ x &=& 3^2+2 \\ \mathbf{x} &=& \mathbf{11} \\ \hline \end{array}\)

A and B are angles in the interval \(0 < A, B < 45^\circ\).
If \(\cos(A + B) = \dfrac{4}{5}\) and \(\sin(A - B) = \dfrac{5}{13}\), find \(\tan(2A)\).

\(\begin{array}{|rcll|} \hline \cos(A + B) &=& \dfrac{4}{5} \\ A + B &=& \arccos\left( \dfrac{4}{5} \right) \qquad (1) \\ \hline \end{array} \begin{array}{|rcll|} \hline \sin(A - B) &=& \dfrac{5}{13} \\ A - B &=& \arcsin\left( \dfrac{5}{13} \right) \qquad (2) \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline (A + B) +(A - B) &=& \arccos\left( \dfrac{4}{5} \right)+\arcsin\left( \dfrac{5}{13} \right) \\ 2A &=& \arccos\left( \dfrac{4}{5} \right)+\arcsin\left( \dfrac{5}{13} \right) \\ \tan(2A) &=& \tan\Bigg( \arccos\left( \dfrac{4}{5} \right)+\arcsin\left( \dfrac{5}{13} \right) \Bigg) \\ \tan(2A) &=& \tan\Bigg( 36.8698976458^\circ+22.6198649480^\circ \Bigg) \\ \tan(2A) &=& \tan\Bigg( 59.4897625939^\circ \Bigg) \\ \tan(2A) &=& 1.69696969697 \\\\ \mathbf{\tan(2A)} &=& \mathbf{\dfrac{56}{33}} \\ \hline \tan(2A) &=& \tan\Bigg( 59.4897625939^\circ \Bigg) \\ 2A &=& 59.4897625939^\circ \\ \mathbf{A} &=& \mathbf{29.7448812969^\circ} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline (A + B) -(A - B) &=& \arccos\left( \dfrac{4}{5} \right)-\arcsin\left( \dfrac{5}{13} \right) \\ 2B &=& \arccos\left( \dfrac{4}{5} \right)-\arcsin\left( \dfrac{5}{13} \right) \\ \tan(2B) &=& \tan\Bigg( \arccos\left( \dfrac{4}{5} \right)-\arcsin\left( \dfrac{5}{13} \right) \Bigg) \\ \tan(2B) &=& \tan\Bigg( 36.8698976458^\circ-22.6198649480^\circ \Bigg) \\ \tan(2B) &=& \tan\Bigg( 14.2500326978^\circ \Bigg) \\ \tan(2B) &=& 0.25396825397 \\\\ \mathbf{\tan(2B)} &=& \mathbf{\dfrac{16}{63}} \\ \hline \tan(2B) &=& \tan\Bigg( 14.2500326978^\circ \Bigg) \\ 2B &=& 14.2500326978^\circ \\ \mathbf{B} &=& \mathbf{7.12501634890^\circ} \\ \hline \end{array}\)

Question 1.

A mathematician works for \(t\) hours per day and solves \(p\) problems per hour, where \(t\) and \(p\) are positive integers and \(1< p < 20\).
One day, the mathematician drinks some coffee and discovers that he can now solve \(3p+7\) problems per hour.
In fact, he only works for \(t-4\) hours that day, but he still solves twice as many problems as he would in a normal day.
How many problems does he solve the day he drinks coffee?