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Thanks to you too, heureka :D

Wow, interesting strategies!

1 / 175 =0.0057142857............................AND ENDS IN..........142857. 7 is 4000th decimal digit.

All  products  sold in  1993 = 

7032 + 5023 + 12024  + 4362  + 3540 + 5932  + 9620  =   47533

All products sold in  2003 =

14862 + 12564  + 25463  +  2753 + 4452  +  4976  + 4062  = 69132

%  increase  between    1993  and 2003 =     [ 69132 - 47533 ] / 47533   ≈  45.4% =  45%

All products sold in 2008  =

29086  + 25433 + 36963  + 8072  +  9833 + 7749 + 5183  =  122319

All products sold in  2019  =

94072  + 20745 + 21945 + 21308 + 19723  + 15257  + 7243  = 200293

% increase  between  2008 and 2019 =   [ 200293 - 122319 ] / 122319  ≈  63.7%  =   65%

So sales increase  by about  65 -45 =  20%  over the two periods

Call the rate of the  current, C

Call the rate of the  boat in still water, R

And with the  current, the boat travels at a rate of  R + C  mph

And against the  current the boat traravels at a rate of  R - C   mph

Rmember that  Distance / Rate  =   Time

And   time it took to travel  12 against the  current =  1 hour more than with the  current =  2 hours....so.....

12/ [ R + C ]  =  1        multiply both sides  by  R +  C

12 = R + C  →       12 - R  = C   (1)

And against  the current we have that

12/ ( R - C )  = 2     

12 = 2(R - C)                 divide through by 2  and sub (1)  in for  C

6 = R - (12 - R)          

6 = 2R- 12           

18 = 2R                    divide  both sides  by  2

9  (mph) =  R =   rate in still water

And the rate of the  current is   12 - 9 =  3 mph

THANK YOU TO!!!!

THANK YOU!

The corner of a unit cube is chopped off such that the cut runs through the three vertices adjacent to the vertex of the chosen corner. What is the height of the remaining cube when the freshly-cut face is placed on a table?

f(2) = 1/2.

Using the angle bisector theorem, the area of triangle XYZ works out to 108 + 144*sqrt(3).

That was for one question, but ok. Sorry

2) Evaluate the series

\(\dfrac{1}{3^1}+\dfrac{4}{3^2}+\dfrac{7}{3^3}+\dfrac{10}{3^4}+\cdots \).

\(\begin{array}{|rcll|} \hline (1+0*3)\left(1*\frac{1}{3}\right) +(1+1*3)\left(1*\frac{1}{3^2}\right) +(1+2*3)\left(1*\frac{1}{3^3}\right) +(1+3*3)\left(1*\frac{1}{3^4}\right) + \dotsb \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline \text{arithmetical sequence:} \\ {\color{red}1}+({\color{red}1}+1*{\color{orange}3})+({\color{red}1}+2*{\color{orange}3})+({\color{red}1}+3*{\color{orange}3})+\dotsb+\Big({\color{red}1}+(n-1)*{\color{orange}3}\Big)+\dotsb \\ \boxed{a={\color{red}1},\ d={\color{orange}3} \text{ is the common difference} } \\\\ \text{geometric series:} \\ {\color{blue}\frac{1}{3} } +{\color{blue}\frac{1}{3} }\left({\color{green}\frac{1}{3}}\right)^1 +{\color{blue}\frac{1}{3} }\left({\color{green}\frac{1}{3}}\right)^2 +{\color{blue}\frac{1}{3} }\left({\color{green}\frac{1}{3}}\right)^3 +\dotsb +{\color{blue}\frac{1}{3} }\left({\color{green}\frac{1}{3}}\right)^{n-1} +\dotsb\\ \boxed{ b={\color{blue}\frac{1}{3} },\ r={\color{green}\frac{1}{3}}\ \text{ is the common ratio} } \\ \hline \end{array}\)

Formula:
sum of a infinite arithmetico-geometric sequence
\(\begin{array}{|rcll|} \hline \mathbf{s} &=& \mathbf{\left(\dfrac{b}{1-r}\right) \Bigg( a+d\left( \dfrac{r}{1-r} \right) \Bigg)} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{s} &=& \mathbf{\left(\dfrac{b}{1-r}\right) \Bigg( a+d\left( \dfrac{r}{1-r} \right) \Bigg)} \\\\ && \boxed{a={\color{red}1},\ d={\color{orange}3} \text{ is the common difference} } \\ &&\boxed{ b={\color{blue}\frac{1}{3} },\ r={\color{green}\frac{1}{3}}\ \text{ is the common ratio} } \\\\ s &=& \left(\dfrac{{\color{blue}\frac{1}{3}} }{1-{ \color{green}\frac{1}{3}} }\right) \Bigg( {\color{red}1} + { \color{orange}3} \left( \dfrac{ {\color{green}\frac{1}{3} }}{1-{\color{green}\frac{1}{3}} } \right) \Bigg) \\\\ s &=& \frac{1}{3}*\frac{3}{2} \Bigg( {\color{red}1} + \dfrac{1}{1-{\color{green}\frac{1}{3}} } \Bigg) \\\\ s &=& \frac{1}{2} *\frac{5}{2} \\\\ \mathbf{s} &=& \mathbf{\frac{5}{4}} \\ \hline \end{array}\)

\(\mathbf{\dfrac{1}{3^1}+\dfrac{4}{3^2}+\dfrac{7}{3^3}+\dfrac{10}{3^4}+\cdots} = \mathbf{\dfrac{5}{4}}\)

False because supplementary angles are two angles whose sum is 180 degrees while complementary angles are two angles whose sum is 90 degrees.

Choice D

Because Segment MB is an angle bisect angle CMB and angle AMB are congruent. So you do this:

x + 25 = 3x + 5

   -   5          - 5

x + 20 = 3x

- x          - x

20 = 2x

/ 2    /2

10 = x

1) Find
\(\dfrac{1}{4} + \dfrac{1}{4^2} + \dfrac{1}{4^3} + \dotsb\) .

Geometric progression: \( a,\ ar,\ ar^2,\ ar^3,\ ar^4,\ \ldots\)

\(\begin{array}{|rcll|} \hline \mathbf{\text{Geometric progression: } a,\ ar,\ ar^2,\ ar^3,\ ar^4,\ \ldots,\qquad a=\dfrac{1}{4},\ r=\dfrac{1}{4} } \\ \hline \dfrac{1}{4},\ \dfrac{1}{4}*\left(\dfrac{1}{4}\right)^1,\ \dfrac{1}{4}\left(\dfrac{1}{4}\right)^2,\ \dfrac{1}{4}\left(\dfrac{1}{4}\right)^3,\ \dfrac{1}{4}\left(\dfrac{1}{4}\right)^4,\ \ldots \\ \hline \end{array}\)

Infinite geometric series, the sum formula: \(\dfrac{a}{1-r}\)

\(\begin{array}{|rcll|} \hline \mathbf{\dfrac{a}{1-r}} &=& \dfrac{\frac{1}{4}}{1-\frac{1}{4}} \quad |\quad a=\dfrac{1}{4},\ r=\dfrac{1}{4} \\\\ &=& \dfrac{4}{4*3} \\\\ &=& \mathbf{\dfrac{1}{3}} \\ \hline \end{array}\)

\(\mathbf{\dfrac{1}{4} + \dfrac{1}{4^2} + \dfrac{1}{4^3} + \dotsb} = \mathbf{\dfrac{1}{3}}\)

Let the freighter's speed =S

The steamboat's speed    =S + 8

116/[S +8] = 100 / S, solve for S

S =50 km/h - the freighter's speed

50 + 8 = 58 km/h - the steamboat's speed

If you work out the exact number of months that took him to pay the $5000  loan @150 per month, you will see that it took 38.50 months to pay off the loan. Each month he paid $150. So, just multiply the two together:

$150 x 38.50 months =$5,775 which includes both principal + interest.

I did not do it algebraically I just drew a pic with GeoGebra and got GeoGebra to estimate the area.

Your answer looks spot on to me  

Square MATH has sides of length 4, and N is the midpoint of TH.
A circle with radius 2 and center N intersects a circle with radius 4 and center M at points P and H.
What is the distance from P to MH.

\(\begin{array}{|rcll|} \hline \mathbf{\text{Circle with }r=4 \\ \text{ and center } M(0,4):} \\ \hline x^2+(y-4)^2 &=& 4^2 \\ \hline \end{array} \begin{array}{|rcll|} \hline \mathbf{\text{Circle with }r=2 \\ \text{ and center } N(2,0):} \\ \hline (x-2)^2+y^2 &=& 2^2 \\ \hline \end{array} \)

\(\begin{array}{|lcll|} \hline \mathbf{\text{Intersects at }P(x_P,\ y_P):} \\ \begin{array}{|rcll|} \hline x_P^2+(y_P-4)^2 &=& 16 \\ x_P^2+y_P^2-8yP + 16 &=& 16 \\ x_P^2+y_P^2-8yP &=& 0 \\ \mathbf{x_P^2+y_P^2} &=& \mathbf{8y_P} \\ \hline \end{array} \begin{array}{|rcll|} \hline (x_P-2)^2+y_P^2 &=& 4 \\ x_P^2-4x_P + 4 +y_P^2 &=& 4 \\ x_P^2 +y_P^2-4x_P &=& 0 \\ \mathbf{x_P^2 +y_P^2}&=& \mathbf{4x_P} \\ y_P^2 &=& 4x_P -x_P^2 \\ \mathbf{ y_P } &=& \mathbf{\sqrt{ 4x_P -x_P^2}} \\ \hline \end{array} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \mathbf{x_P^2+y_P^2} &=& \mathbf{8y_P} \quad | \quad \mathbf{x_P^2 +y_P^2=4x_P},\ \mathbf{ y_P =\sqrt{ 4x_P -x_P^2}} \\ 4x_P &=& 8\sqrt{ 4x_P -x_P^2} \quad | \quad :4 \\ x_P &=& 2\sqrt{ 4x_P -x_P^2} \quad | \quad \text{square both sides} \\ x_P^2 &=& 4(4x_P -x_P^2) \\ x_P^2 &=& 16x_P -4x_P^2 \\ 5x_P^2 &=& 16x_P \\ 5x_P^2-16x_P &=& 0 \\ x_P(5x_P-16) &=& 0 \quad | \quad x_P \neq 0~! \\\\ 5x_P-16 &=& 0 \\ 5x_P &=& 16 \\ x_P &=& \dfrac{16}{5} \\ \mathbf{x_P} &=& \mathbf{3.2} \\ \hline \end{array}\)

The distance from P to MH is \(\mathbf{3.2}\)

The corner of a unit cube is chopped off such that the cut runs through the three vertices adjacent to the vertex of the chosen corner. What is the height of the remaining cube when the freshly-cut face is placed on a table?

Let the area of a single side of the cube be A = 25 cm².

The length of the edge is  E = 5 cm

Let D be diagonal of a side of the cube.    D = 5 / sin(45°) = 7.071067812              

Angle after cutting the corner is: tan(β) = E / (D/2)       β = 54.7356°

The height of the remaining cube is   H = D * sin(β)  = 5.7735 cm  

Find the standard form of the equation of a circle with center ​(1​,2​) and touching the​ x-axis. Graph the equation.  

Square MATH has sides of length 4, and N is the midpoint of TH. A circle with radius 2 and center N intersects a circle with radius 4 and center M at points P and H. What is the distance from P to MH.