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Find all values of x such that 

\(\large{ \mathbf{\Big| ~ |x-17| - 3~ \Big| = \Big|~ |x-15| - 5 ~\Big|} }\)

\(\begin{array}{|rcll|} \hline \Big| ~ |x-17| - 3~ \Big| &=& \Big|~ |x-15| - 5 ~\Big| \quad \boxed{\text{Formula: } |a|^2 = (a)^2} \\ \left( ~ |x-17| - 3~ \right)^2 &=& \left(~ |x-15| - 5 ~\right)^2 \\ (x-17)^2-6|x-17|+9 &=& (x-15)^2-10|x-15|+25 \\ 10|x-15| &=& 6|x-17|+ (x-15)^2-(x-17)^2 + 16 \\ 10|x-15| &=& 6|x-17| +4x-48 \quad | \quad : 2 \\ \mathbf{5|x-15|} &\mathbf{=}& \mathbf{3|x-17| +2x-24} \\ \hline \end{array} \)

Here we have two different amounts.

To dissolve them, we must make a double case distinction.

Usually we did this one after the other.
For reasons of space, we start with the first case in which the content of the left amount
is greater than or equal to zero.

\(\begin{array}{|rcll|} \hline & \underline{x\ge 15:}& \\\\ & 5(x-15)\\ & =3|x-17|+2x-24 \\ \\ \underline{\text{for } x\ge 17:} && \underline{\text{for } x \lt 17: } \\\\ 5(x-15)=3(x-17)+2x-24 && 5(x-15)=-3(x-17)+2x-24 \\ 5x-75=3x-51+2x-24 && 5x-75=-3x+51+2x-24 \\ -75=-75~\text{true} && 6x=102 \\ x\ge 15 \wedge x\ge 17 && x=17 \quad (x\lt 17) ~\text{no solution} \\ \boxed{ x\ge 17 } && \\ \hline \hline & \underline{x\lt 15:}& \\\\ & -5(x-15)\\ & =3|x-17|+2x-24 \\ \\ \underline{\text{for } x\ge 17:} && \underline{\text{for } x \lt 17: } \\\\ -5(x-15)=3(x-17)+2x-24 && -5(x-15)=-3(x-17)+2x-24 \\ -5x+75=3x-51+2x-24 && -5x+75=-3x+51+2x-24 \\ -10x=-150 && -4x=-48 \\ x=15\quad (x\ge 17) ~\text{no solution} && x=12 \quad (x\lt 17) ~\text{solution}\\ && \boxed{ x=12 } \\ \hline \end{array}\)

\(\mathbf{\boxed{x=12 \lor x\ge 17}}\)

srsly?

The matrices

\(\large{ \mathbf{A} = \begin{pmatrix} 1 &x \\ y & 1 \end{pmatrix} \text{ and } \mathbf{B} = \begin{pmatrix} -1 &x \\ y & -1 \end{pmatrix} }\)

are inverses.

What is \(xy\)?

\(\begin{array}{|lrcll|} \hline & AB&=&I \\ & \begin{pmatrix} 1 &x \\ y & 1 \end{pmatrix} \begin{pmatrix} -1 &x \\ y & -1 \end{pmatrix} &=& \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \\\\ & \begin{pmatrix} -1+xy &x-x \\ -y+y & xy-1 \end{pmatrix} &=& \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \\\\ & \begin{pmatrix} xy-1 &0 \\ 0 & xy-1 \end{pmatrix} &=& \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline xy-1 &=& 1 \\ \mathbf{xy} & \mathbf{=} & \mathbf{2} \\ \hline \end{array}\)

Let \(\theta\) be the angle between a and b, where a and b are the vectors defined in

\(a = \begin{pmatrix} 2 \\ -6 \\ 5 \end{pmatrix} \quad \text{and} \quad b = \begin{pmatrix} -1 \\ -2 \\ 0 \end{pmatrix}\).
 

Find \(\cos^2 \theta\)

\(\begin{array}{|rcll|} \hline \mathbf{\cos(\theta)} &\mathbf{=}& \mathbf{\dfrac{\vec{a}\cdot \vec{b}}{|\vec{a}|\cdot |\vec{b}|} } \quad | \quad |\vec{a}| = \sqrt{2^2+(-6)^2+5^2}=\sqrt{65},\ |\vec{b}| = \sqrt{(-1)^2+(-2)^2+0^2}=\sqrt{5} \\\\ \cos(\theta) &=& \dfrac{\begin{pmatrix} 2 \\ -6 \\ 5 \end{pmatrix}\cdot \begin{pmatrix} -1 \\ -2 \\ 0 \end{pmatrix} } {\sqrt{65}\cdot \sqrt{5} } \\\\ \cos(\theta) &=& \dfrac{2\cdot (-1)+(-6)\cdot(-2)+5\cdot 0} {\sqrt{65}\cdot \sqrt{5} } \\\\ \cos(\theta) &=& \dfrac{10} {\sqrt{65}\cdot \sqrt{5} } \\\\ \cos^2(\theta) &=& \dfrac{100} {65\cdot 5 } \\\\ \cos^2(\theta) &=& \dfrac{100} {325 } \\\\ \mathbf{\cos^2(\theta)} &\mathbf{=}& \mathbf{\dfrac{4}{13}} \\ \hline \end{array}\)

V =   pi * (radius)^2 * height / 3 =    3.14 * ( 10)^2 * 6 / 3  =  628 in^3

THX to EP for catching my error....!!!!!

Find the limit:

\(\large{ \lim \limits_{x \to 0} ~\dfrac{\sin(5x)}{\sin(8x)} } \)

Apply L'Hopital's Rule and we have:

\(\begin{array}{|rcll|} \hline && \mathbf{\lim \limits_{x \to 0} ~\dfrac{\sin(5x)}{\sin(8x)}} \\\\ &=& \lim \limits_{x \to 0} ~\dfrac{5\cos(5x)}{8\cos(8x)} \quad | \quad \lim \limits_{x \to 0} ~ \cos(5x)= 1,\ \lim \limits_{x \to 0} ~ \cos(8x)= 1 \\\\ &=& \dfrac{ 5\cdot 1}{8\cdot 1} \\\\ &\mathbf{=}& \mathbf{\dfrac{ 5 }{8 }} \\ \hline \end{array}\)

Volume =   3.14 (1)^2 * 4.5  =   14.13 ft^3

The common ratio is 55/50 = 11/10 = 1.1  = r

a₁ =  500000

a_n = a₁ ( r)^(n - 1)   

a_n = 500000(1.1)^(n - 1)   =   (a)

2  Secant-Tangent Theorem

Angle ABS =  (1/2) ( arc AS - arc AP)

39 = (1/2) (arc AS - arc AP)

78 = arc AS - arc AP   (1) 

arc AS + arc AP =  360 - arc PS

arc AS + arc AP = 360 - 116

arc AS + arc AP = 244   (2)

Using (1) and (2)  we have that

arc  AS - arc AP  =  78

arc AS + arc AP  =  244       add these

2arcAS = 322

arc AS = 161°