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Thanks very much Alan,    

I just differentiated inverse tan incorrectly.

It was a careless mistake that I kept overlooking    

\((x^2 + y^2)^2 = x^4 + 2(xy)^2 + y^2\\\\ 1 = \frac{5}{8}+2(xy)^2\)

Can you take it from here?

Have a look at: https://www.youtube.com/watch?v=0KPeDLSLKro 

how much power does a water pump expend when it lifts 40.0 liters of water per minute to a height of 10.0 meters?

Hello Guest!

\(N=\dfrac{F\cdot s}{t}=\dfrac{40kg\cdot9.81\frac{m}{s^2}\cdot 10m}{60s}\cdot \dfrac{N\cdot s^2}{kg\cdot m}=65.4\cdot \dfrac{Nm}{s}=\color{blue}65.4\ W\)

  !

Since it is on the x axis, the point will be of the from  (x,0 )   

  Use the distance formula :for point 1 and point 2 and equate them

(-2 - x )^2 + 0^2  = x^2 + (8)^2

x^2 + 4x + 4 = x^2 + 64

4x -60 = 0

x = 15

Maybe I misunderstood your delema?  It doesn't matter though.  Time to move on.  :)

oh yes there is a logical reason, . 

Any angle in a semicircle will be 90 degrees.  

This is an extension of the  theorem that: 

  Any angle on circumference subtended from an arc will always be half that as will be subtended to the centre.

(my wording may leave a little to be desired)

I knew AB was a diameter but I did not think to use the centre in my logic.    

YZ = 10.5

or...

1 / tan(20º) 

AB = 3 * 16 = 48 

I solved this problem before!  $x_0 + y_0 = 27$

I think the main thing that causes this is the line segment CM being an angle bisector of the right angle ACB.

Angle AMB must be 90 degrees! 

No matter how much you increase/decrease AC and BC, AM = BM !!!

To me, this is the phenomenon.

There must be some logical explanation.

$AB = 42$.

2 distinct vertices of an octagon are chosen at random. What is the probability that they are opposite each other?     

One vertex can be any of them.  The one opposite can be only 1 of the remaining 7.  So, 1 out of 7.  

If 2 cards from a standard deck are selected randomly, what is the probability that either two kings or at least 1 space occurs? (There are 4 Aces, 4 kings and 52 total cards in a standard deck.)     

or at least 1 space occurs  –  Should this be 1 Ace, and autocorrect bit you?  

Let c = Cheko's toy cars and k = Keddy's toy cars. 

c + k = 370
0.25c = 0.2k + 25

20c = -20k + 7400 
25c = 20k + 2500

------------------------

45c = 9900

c = 220

c + k = 370
(220) + k = 370
k = 150 

220 - 150 = 70

Cheko collected 70 more cars than Keddy. 

Possibles   are     12  13   14   15   16    24  26   and 36

  out of 6c2 = 15 choices            8/15

Need more info....   exponent of what ?

Here is the value of the exponent for 10    (it is just  log₁₀ 65 )

log(65) = 1.8129133566428555

so 10^{1.8129133566428555}    = 65     

4x-15 = 6x+31    when they are equal, it is a point of intersection of these two lines ....subtract 4x from both sides to get

-15 = 2x + 31      now subtract 31 from both sides to get

-46 = 2x              now divide both sides by two

x = - 23                      this ould be the x value where   f(x) = g(x)

If what you put is true wouldn't 4x-15=6x+31 be your answer?

Is that your way of saying thanks to apsig.. ?

111 numbers ARE divisible by 9 out of the thousand  ( 9   18  27 ....... 990  999)

means     1000-111 = 889 numbers ARE NOT  divisible by '9'

If AM = BM then these 2 arcs have the same length.

Yes that is obvious, but you have not said why  AM=BM    That is what I keep asking you.

Thanks for including the address of Chris's explanation. Now i understand.

The explanation is that since AB is a  diameter, the radius and the centre of the circle are easy to determine.

From that information, all else can be determined.

Ok yea, that is the correct answer.