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2x - 5y = 10

x-intercept:  2*x - 5*0 = 10   so x = 5   x-intercept at point (5, 0)

y-intercept:  2*0 - 5*y = 10 so y = -2    y-intercept at point (0, -2)

Use Pythagoras to find distance between them:  r = √[(0-5)² + (-2-0)²]

Thinking about it further, I made the last few lines of the derivation of q more complicated than necessary.  Going back a few lines from the end we can do the following:

I cheated and used a calculator (well, Mathcad's calculation facilities).

Correct.  I just went for the simplest option!

H is the orthocenter of acute triangle ABC and the extensions of AH, BH, and CH intersect the circumcircle
of triangle ABC at A', B' and C'.
We know angle AHB:angle BHC:angle CHA=9:10:11.
Find angle A'B'C' in degrees.

\(\begin{array}{|rcll|} \hline \mathbf{\angle AHB:\angle BHC:\angle CHA} &=& \mathbf{9:10:11} \\\\ \frac{\angle AHB}{\angle CHA} &=& \frac{9}{11} \\ \angle AHB &=& \frac{9}{11} *( \angle CHA ) \\\\ \frac{\angle BHC}{\angle CHA} &=& \frac{10}{11} \\ \angle BHC &=& \frac{10}{11}*(\angle CHA ) \\\\ \angle AHB+\angle BHC+\angle CHA &=& 360^\circ \\ \frac {9}{11} *( \angle CHA )+\frac{10}{11}*(\angle CHA )+\angle CHA &=& 360^\circ \\ \frac {19}{11} *( \angle CHA )+\angle CHA &=& 360^\circ \\ \frac {30}{11} *( \angle CHA ) &=& 360^\circ \\ \angle CHA &=& \frac {11}{30} *360^\circ \\ \mathbf{\angle CHA} &=& \mathbf{132^\circ} \\ \hline \end{array}\)

\(\text{$\angle A'B'C'$ in degrees is $2*42^\circ = 84^\circ$}\)

What is the value of a+c?

Whenever you see a pecentage like this: 7 %, it means: 7 / 100 =0.07 . Now, you would simply multiply this by his total sales for the month:

0.07 x $58,965.00 = $4,127.55 -  This is his commission for that month.

thank u so much !! 

thank you so much !! very clear and was able to follow the steps :)

2/5 of a mile can be expressed as 0.4 of a mile.  Let's call it that, to make it easier for me. 

                                                                            We could do fractions, but decimals are easier. 

Achilles runs at 13 mph, and Turtle at 0.4 mph.  So Achilles is outdistancing Turtle by 12.6 mph. 

Do you see that?  In one hour, Achilles would have run 13.0 miles and Turtle would have run 0.4 mile. 

Well, they didn't run for an entire hour, or Achilles would have been 12.6 miles ahead of Turtle. 

How long did they run?  Distance = Speed x Time  ....  so let's start with that and see how far we get. 

Consider that 0.4 mile.  They ran long enough for that to be the Distance.  Let's set it up.

                                                                 D  =  S • T                                        (A "dot" means multiply.)

                                                                 0.4 mile  =  (12.6 miles/hour) • T 

Divide both sides by 12.6 miles/hour        (0.4 mile) • (12.6 miles/hour)  =  T 

Doing the division, we find that                 T  =  0.0317 hour                             (A little less than 2 minutes.) 

Why don't we go ahead and see if we can check that answer.... 

In 0.0317 hour, at 13 miles/hour, Achilles has run 0.412 mile 

In 0.0317 hour, at 0.4 mile/hour, Turtle has run     0.013 mile 

Subtract Turtle's distance from Achilles to confirm that Achilles is about 0.4 (aka 2/5) mile ahead. 

The distance doesn't come out to exactly 0.4 because we did a little rounding along the way. 

If you want a more precise answer, use a calculator and don't do any rounding. 

_.

S = second person

1 /70  -  1 / 120 = 1/S, solve for S

S = 168 minutes to clean the jellyfish tank.

It will go from Quadrant 2      to  Quadrant 3   and will be   -5, -7

"I don't do math"   better learn 'to do'  math or  " I don't do math"  will never get far in this world. 

     and I won't get much further if I make MISTAKES at math !!    Corrected my typo

1st person rate   =    1 tank / 120 m

2nd person rate   =   1 tank / x m

1 tank /  ( 1 tank/120m + 1tank/xm)   = 70 m

1 / (1/120 + 1/x)  = 70

1 = 70 ( 1/120 + 1/x)  

1 - 70/120   = 70/x

x= 168 m

From 11 to 35   is   24       1/3 of this is   8        11 + 8 = 19

from -13 i    to   - i   is   12 i     1/3 of this is 4 i     -13 i + 4 i = -  9 i

Consider the series. 

\(\sum \limits_{n=1}^{\infty} \dfrac{1+4n} {4^{n+1}}\)

\(\begin{array}{|rcll|} \hline \mathbf{\sum \limits_{n=1}^{\infty} \dfrac{1+4n} {4^{n+1}}} &=& \sum \limits_{n=1}^{\infty} \dfrac{1+4n} {4^{n}*4} \\\\ &=& \dfrac{1}{4} \left( \sum \limits_{n=1}^{\infty} \dfrac{1+4n} {4^{n}} \right) \\\\ &=& \dfrac{1}{4} \left[ \sum \limits_{n=1}^{\infty} (1+4n) \left( \dfrac{1}{4}\right)^{n} \right] \\\\ &=& \dfrac{1}{4} \left[ -1 + \sum \limits_{n=1}^{\infty} \Big(1+4(n-1)\Big) \left( \dfrac{1}{4}\right)^{n-1} \right] \\\\ \mathbf{\sum \limits_{n=1}^{\infty} \dfrac{1+4n} {4^{n+1}}} &=& \mathbf{\dfrac{1}{4}( -1 + s )} \qquad \boxed{s=\sum \limits_{n=1}^{\infty} \Big(1+4(n-1)\Big) \left( \dfrac{1}{4}\right)^{n-1} } \\ \hline \end{array}\)

Infinite arithmetico-geometric sequence:

\(\begin{array}{|rcll|} \hline s &=& \sum \limits_{n=1}^{\infty} \Big(1+4(n-1)\Big) \left( \dfrac{1}{4}\right)^{n-1} \\ \hline \end{array}\)

\({\color{red}a}{\color{blue}b }+(a+d){\color{blue}br }+{\color{red}(a+2d)}{\color{blue}br^2 }+{\color{red}(a+3d)}{\color{blue}br^3 }+\cdots + {\color{red}\Big(a+(n-1)d \Big)}{\color{blue}br^{n-1} } + \cdots\)

\(\begin{array}{|lrcll|} \hline \text{arithmetical sequence:}\quad 1+5+9+13+17+\ldots \\ {\color{red}1}+({\color{red}1}+1*{\color{orange}4})+({\color{red}1}+2*{\color{orange}4})+({\color{red}1}+3*{\color{orange}4})+\dotsb+\Big({\color{red}1}+(n-1)*{\color{orange}4}\Big)+\dotsb \\ \boxed{a={\color{red}1},\ d={\color{orange}4} \text{ is the common difference} } \\\\ \text{geometric series:}\quad 1+1*\frac{1}{4^1}+1*\frac{1}{4^2}+ 1*\frac{1}{4^3}+ 1*\frac{1}{4^4}+\cdots \\ {\color{blue}1} +{\color{blue}1}*\left({\color{green}\frac{1}{4}}\right)^1 +{\color{blue}1}*\left({\color{green}\frac{1}{4}}\right)^2 +{\color{blue}1}*\left({\color{green}\frac{1}{4}}\right)^3 +\dotsb +{\color{blue}1}*\left({\color{green}\frac{1}{4}}\right)^{n-1} +\dotsb\\ \boxed{ b={\color{blue}1},\ r={\color{green}\frac{1}{4}}\ \text{ is the common ratio} } \\ \hline \end{array} \)

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

\(\begin{array}{|rcll|} \hline \mathbf{s} &=& \mathbf{\left(\dfrac{b}{1-r}\right) \Bigg( a+d\left( \dfrac{r}{1-r} \right) \Bigg)} \\\\ && \boxed{a={\color{red}1},\ d={\color{orange}4} \text{ is the common difference} } \\ &&\boxed{ b={\color{blue}1 },\ r={\color{green}\frac{1}{4}}\ \text{ is the common ratio} } \\\\ s &=& \left(\dfrac{{\color{blue}1} }{1-{ \color{green}\frac{1}{4}} }\right) \Bigg( {\color{red}1} + { \color{orange}4} \left( \dfrac{ {\color{green}\frac{1}{4} }}{1-{\color{green}\frac{1}{4}} } \right) \Bigg) \\\\ s &=& \dfrac{4}{3}\left( {\color{red}1} + \dfrac{4}{3} \right) \\\\ s &=& \dfrac{4}{3} *\dfrac{7}{3} \\\\ \mathbf{s} &=& \mathbf{\dfrac{28}{9}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{\sum \limits_{n=1}^{\infty} \dfrac{1+4n} {4^{n+1}}} &=& \mathbf{\dfrac{1}{4}( -1 + s )} \quad | \quad s=\dfrac{28}{9} \\\\ &=& \dfrac{1}{4}\left( -1 + \dfrac{28}{9} \right) \\\\ &=& \dfrac{1}{4}\left( \dfrac{19}{9} \right) \\\\ \mathbf{\sum \limits_{n=1}^{\infty} \dfrac{1+4n} {4^{n+1}}} &=& \mathbf{\dfrac{19}{36}} \\ \hline \end{array}\)

For squares. see:

https://www.purplemath.com/modules/specfact.htm

For cubes see:

https://www.cliffsnotes.com/study-guides/algebra/algebra-ii/factoring-polynomials/sum-or-difference-of-cubes

The graph should help:

omega doesnt have to be e^i*(2pi/7) it can be e^i*(4pi/7)

Hi Alan

Could you tell me how you get from the last but one line to the last one, (without the use of a calculator). ?

Determine the maximum volume of a box with surface area 3000cm2. The box has a square base and no top. 

Let the time Gilbert spent driving =T
The time Brian spent driving          =11.6 - T

45T + 65(11.6 - T) =604, solve for T
T = 7.5 hours that Gilbert spent driving.
11.6 - 7.5 =4.1 hours that Brian spent driving.

How to solve these algebraically? 

a)

b)

How to solve these algebraically? 

Hello Guest!

b)

\(\color{BrickRed}2,1x^{0,25}+5=7\\ x^{\frac{1}{4}}=\frac{2}{2,1}\\ x=(\frac{2}{2,1})^4 \)

\(x=(\frac{20}{21})^4=(\frac{2^2\cdot 5}{3\cdot 7})^4\)

\(x=\frac{2^8\cdot 5^4}{3^4\cdot7^4}=0,8227...\)

  !

Look at the dashed lines.  They intersect at the point where g(x) = -10.  The corresponding x-value is shown.

Solving for g(x)>f(x) means looking to find the range of x values for which the blue line is higher than the red line.

How to solve these algebraically? 

Hello Guest!

a)

\(\color{BrickRed}-2+5z^{-\frac{3}{2}}=133\\ 5z^{-\frac{3}{2}}=135\\ z^{-\frac{3}{2}}=27\\ z^{-3}=27^2\\ z^3=\frac{1}{27^2}\\ z=\frac{1}{\sqrt[3]{(3^3)^2}}\)

\(z=\frac{1}{9}\)

  !