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(x+y)^3 = ( x^3 + y^3)  + 3x^2y + 3xy^2

125      =  (x^3  + y^3)   + 3 (15 )                  (x+y)(xy) =   x^2 y + xy^2

80 = x³ + y³

AoPS copyright claim

Please show your logic.

Here is  Venn Diagram to help:

Question wording is a bit unclear    'if 45 people do not like both drinks'  could represent the  5 + 10 % in the circles

   so either 10% =45        or      15% = 45           Let me know what you find !     ~EP

AB = 2 x A x B

A =3   and   B =6

36 = 2 x 3 x 6

36 = 36

x =668

[100 x 668] = 1 mod 997.

60c  +  37,  where c=0, 1, 2, 3......etc.

The smallest  positive integer = 37

what is the modular inverse of 11 , modulo 1000?

The answer =91, because:

[91 x 11] mod 1000 = 1, which is the definition of "modular multiplicative inverse"

Solve like this:

1/45  + 1/105  = 1/t     solve for t = 31.5 min

or this 

1  /  ( 1/45  + 1/105)      = t

or this

(105 * 45)/(105+45) = t

My computer says there are 1,125 four-digit numbers with at least 3 even digits, including zero.

yeppers   

I wasn't quite sure whoops. so 18 and 191.1?

Time must be considered in 'average speed' questions

50 m / 40 m/hr = time  during 50 mile ride

20 m / 55 m/hr = time during  20 mile ride

total ride = 70 miles

total ride / total time = average

70 m /  ( 50/40 + 20/55) = 43.4 m/hr avg

x + 3y = -5.      is tha same as    y = -1/3 x - 5/3     slope, m = -1/3

parallel line is of the form   y = -1/3 x + b

   sub in the given point      2 = -1/3 (-7)  + b     shows b = -1/3

     so line eq becomes       y = - 1/3 x -1/3

       re-arranged                  y + 1/3 x = -1/3        multiply by 3

                                                x + 3y = - 1      now you can calc  B-A

Just use a calculator:

.075 * (.6 * 400) = .......

.20 * (.98 * 975)=.......

It was correct, thank you so much

The question doesn't make sense.

Where is the point E supposed to be ?

ACEG is a rectangle. If segment BE is 30, segment CG is 40, segment DF is 15 and \(\angle FDE=90^\circ\).
Find CE.

\(\text{Let $CE=x$} \\ \text{Let $GE=y$} \\ \text{Let $AB=z$} \\ \text{Let $BC=y-z$}\)

\(\begin{array}{|lrcll|} \hline 1): & x^2 + y^2 &=& 40^2 \\ & x &=& \sqrt{40^2-y^2} \\ \hline 2): & x^2+(y-z)^2 &=& 30^2 \quad | \quad x^2 = 40^2-y^2 \\ & 40^2-y^2+(y-z)^2 &=& 30^2 \\ & y^2 - (y-z)^2 &=& 40^2-30^2 \\ & \mathbf{y^2 - (y-z)^2} &=& \mathbf{700} \qquad (1) \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline 3): & \text{area $\triangle $ CDE} &=& \text{area $\triangle $ CFE} + \text{area $\triangle $ FEG} \\ & \dfrac{xy}{2} &=& \dfrac{x*15}{2} + \dfrac{y*DE}{2} \\ & xy &=& x*15 + y*DE \\ & && \boxed{\dfrac{DE}{15} =\dfrac{x}{y-z}\\ DE = \dfrac{15x}{y-z} } \\ & xy &=& x*15 + y*\left( \dfrac{15x}{y-z} \right) \quad | \quad : x \\ & y &=& 15 + \left( \dfrac{15y}{y-z} \right) \\ & (y-15)(y-z) &=& 15y \\ & y-z &=& \dfrac{15y}{y-15} \\ & \mathbf{(y-z)^2} &=& \mathbf{\dfrac{225y^2}{(y-15)^2}} \qquad (2) \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \mathbf{y^2 - (y-z)^2} &=& \mathbf{700} \quad | \quad \mathbf{(y-z)^2=\dfrac{225y^2}{(y-15)^2}} \\ y^2 - \dfrac{225y^2}{(y-15)^2} &=& 700 \quad | \quad *(y-15)^2 \\\\ y^2(y-15)^2 - 225y^2 &=& 700(y-15)^2 \\ y^2(y^2-30y+225) - 225y^2 &=& 700(y^2-30y+225) \\ y^4-30y^3+225y^2-225y^2 &=& 700y^2-700*30y+700*225 \\ y^4-30y^3 &=& 700y^2-700*30y+700*225 \\ \mathbf{ y^4-30y^3 -700y^2+700*30y-700*225} &=& \mathbf{0} \\ \Rightarrow \mathbf{y}&=&\mathbf{36.6659934977741} \qquad \text{WolframAlpha} \\ \hline x &=& \sqrt{40^2-y^2} \\ x &=& \sqrt{40^2-36.6659934977741^2} \\ \mathbf{x}&=&\mathbf{15.9876490087} \\ \hline \end{array}\)

CE is \(\approx \mathbf{16}\)

The measure of the altitude drawn from the vertex of the right angle of a right triangle to its hypotenuse is the geometric mean between the measures of the two segments of the hypotenuse.

Knowing that  x = sqrt( 10 * 6 ) = 7.745966692  

How many ways are there to split 14 people into seven pairs?

The first person of the 14 people can form a pair with \({\color{red}13}\) other people. We now have one pair less.
The first person of the remaining 12 people can form a pair with \({\color{red}11}\) other people. We now have two pairs less.
The first person of the remaining 10 people can form a pair with \({\color{red}9}\) other people. We now have three pairs less.
The first person of the remaining 8 people can form a pair with \({\color{red}7}\) other people. We now have four pairs less.
The first person of the remaining 6 people can form a pair with \({\color{red}5}\) other people. We now have five pairs less.
The first person of the remaining 4 people can form a pair with \({\color{red}3}\) other people. We now have six pairs less.
The first person of the remaining 2 people can form a pair with \({\color{red}1}\) other people. We now have seven pairs less.

There are \({\color{red}1}*{\color{red}3}*{\color{red}5}*{\color{red}7}*{\color{red}9}*{\color{red}11}*{\color{red}13} = \mathbf{135135}\) ways to split 14 people into seven pairs

triangle

\(\begin{array}{|rcll|} \hline \mathbf{\vec{BK}} &=& \mathbf{\dfrac{ \vec{BC} }{2} + \dfrac{ \vec{DA} }{2}} \\\\ && \boxed{ \vec{DA} = \vec{BA} - \dfrac{ \vec{BC} }{2} \\ \dfrac{\vec{DA}}{2} = \dfrac{\vec{BA}}{2} - \dfrac{ \vec{BC} }{4} } \\\\ \vec{BK} &=& \dfrac{ \vec{BC} }{2} + \dfrac{\vec{BA}}{2} - \dfrac{ \vec{BC} }{4} \\\\ \mathbf{\vec{BK}} &=& \mathbf{ \dfrac{ \vec{BC} }{4} + \dfrac{\vec{BA}}{2} } \\ \hline \end{array} \begin{array}{|rcll|} \hline \mathbf{\vec{BL}} &=& \mathbf{ \vec{BC} + \dfrac{ \vec{CE} }{2}} \\\\ && \boxed{ \vec{CE} = \dfrac{\vec{BA}}{2} - \vec{BC} \\ \dfrac{\vec{CE}}{2} = \dfrac{\vec{BA}}{4} - \dfrac{ \vec{BC} }{2} } \\\\ \vec{BL} &=& \vec{BC} + \dfrac{\vec{BA}}{4} - \dfrac{ \vec{BC} }{2} \\\\ \mathbf{\vec{BL}} &=& \mathbf{ \dfrac{ \vec{BC} }{2} + \dfrac{\vec{BA}}{4} } \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline 2A_{yellow} = 18 &=& \vec{BL} \times \vec{BK} \\ 18 &=& \left( \dfrac{ \vec{BC} }{2} + \dfrac{\vec{BA}}{4} \right) \times \left( \dfrac{ \vec{BC} }{4} + \dfrac{\vec{BA}}{2} \right) \\\\ 18 &=& \underbrace{ \left( \dfrac{ \vec{BC} }{2} \times \dfrac{ \vec{BC} }{4}\right) }_{=0} +\left( \dfrac{ \vec{BC} }{2} \times \dfrac{\vec{BA}}{2} \right) +\left( \dfrac{\vec{BA}}{4} \times \dfrac{ \vec{BC} }{4} \right) + \underbrace{\left( \dfrac{\vec{BA}}{4} \times \dfrac{\vec{BA}}{2} \right) }_{=0} \\\\ 18 &=& \left( \dfrac{ \vec{BC} }{2} \times \dfrac{\vec{BA}}{2} \right) +\left( \dfrac{\vec{BA}}{4} \times \dfrac{ \vec{BC} }{4} \right) \\\\ 18 &=& \left( \dfrac{ \vec{BC} }{2} \times \dfrac{\vec{BA}}{2} \right)-\left( \dfrac{\vec{BC}}{4} \times \dfrac{ \vec{BA} }{4} \right) \\\\ 18 &=& \dfrac{1}{4}*\left( \vec{BC} \times \vec{BA} \right)-\dfrac{1}{16}*\left( \vec{BC} \times \vec{BA} \right) \\\\ 18 &=& \dfrac{4}{16}*\left( \vec{BC} \times \vec{BA} \right)-\dfrac{1}{16}*\left( \vec{BC} \times \vec{BA} \right) \\\\ 18 &=& \dfrac{3}{16}*\left( \vec{BC} \times \vec{BA} \right) \\\\ \dfrac{16*18}{3} &=& \vec{BC} \times \vec{BA} \\\\ 16*6 &=& \vec{BC} \times \vec{BA} \\\\ 96&=& \vec{BC} \times \vec{BA} \\\\ \mathbf{ \vec{BC} \times \vec{BA} } &=& \mathbf{96} \quad | \quad 2\times \text{area } \triangle \text{ABC} = \vec{BC} \times \vec{BA} \\\\ 2\times \text{area } \triangle \text{ABC} &=& 96 \quad | \quad :2 \\\\ \mathbf{ \text{area } \triangle \text{ABC} } &=& \mathbf{48} \\ \hline \end{array}\)

The area of the triangle ABC is \(\mathbf{48\ \text{cm}^2}\)

1/A+1/B =1/3,  
1/B+1/C =1/4,  
1/A+1/C  =1/6, solve for A, B, C 
A =8 hours,  B =24/5 hours,  C = 24 hours
1/8 + 5/24 +1/24 =1/C, solve for C
C = 8/3 =2 2/3 hours - for 3 taps to fill the container.

Can someone help find \(\begin{bmatrix}\sqrt3/2 & -1/2 \\1/2 & \sqrt3/2\end{bmatrix}^{2018}\)

I assume the matrix is \(\begin{bmatrix} \dfrac{\sqrt3}{2} & -\dfrac{1}{2} \\\\ \dfrac{1}{2} & \dfrac{\sqrt3}{2} \end{bmatrix}^{2018}\), then this matrix is a rotation matrix counterclockwise through an angle of \(30^\circ\)

\(\begin{array}{|rcll|} \hline \begin{bmatrix} \dfrac{\sqrt3}{2} & -\dfrac{1}{2} \\\\ \dfrac{1}{2} & \dfrac{\sqrt3}{2} \end{bmatrix}^{2018} &=& \begin{bmatrix} \cos(30^\circ) & -\sin(30^\circ) \\\\ \sin(30^\circ) & \cos(30^\circ) \end{bmatrix}^{2018} \\ \hline \end{array} \) Rotation through an angle of \(30^\circ \times 2018\)

\(\begin{array}{|rcll|} \hline 30^\circ \times 2018 &=& 60540^\circ \\ &=& 60540^\circ-168\times 360^\circ \\ &=& \color{red}\mathbf{60^\circ} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \begin{bmatrix} \cos(30^\circ) & -\sin(30^\circ) \\\\ \sin(30^\circ) & \cos(30^\circ) \end{bmatrix}^{2018} &=& \begin{bmatrix} \cos( {\color{red}\mathbf{60^\circ}} ) & -\sin( {\color{red}\mathbf{60^\circ}} ) \\\\ \sin( {\color{red}\mathbf{60^\circ}} ) & \cos( {\color{red}\mathbf{60^\circ}} ) \end{bmatrix} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \begin{bmatrix} \cos( {\color{red}\mathbf{60^\circ}} ) & -\sin( {\color{red}\mathbf{60^\circ}} ) \\\\ \sin( {\color{red}\mathbf{60^\circ}} ) & \cos( {\color{red}\mathbf{60^\circ}} ) \end{bmatrix} &=& \begin{bmatrix} \dfrac{1}{2} & -\dfrac{\sqrt3}{2} \\\\ \dfrac{\sqrt3}{2} & \dfrac{1}{2} \end{bmatrix} \\\\ \begin{bmatrix} \dfrac{\sqrt3}{2} & -\dfrac{1}{2} \\\\ \dfrac{1}{2} & \dfrac{\sqrt3}{2} \end{bmatrix}^{2018} &=& \begin{bmatrix} \dfrac{1}{2} & -\dfrac{\sqrt3}{2} \\\\ \dfrac{\sqrt3}{2} & \dfrac{1}{2} \end{bmatrix} \\ \hline \end{array} \)