All Answers 358335 Answers

Thanks for the help!

Find where the equation for Circle A = equation for Circle B

Uhhh....wait a minute.....   the two circles do NOT intersect:

https://www.desmos.com/calculator/fxvlkcnksv

For Richard to win in k games (k from 5 to 9; minimum of 5 games, maximum of 9, at 9 games at least 1 person has won 5 games) it is mandatory that Richard wins the last game and 4 of the first k-1 games. Vanessa wins k-5 of the first k-1 games. The probability of this is

(40/100)^5*(60/100)^(k-5).

There are k-1 choose 4 such configurations (Richard can win in any 4 of the first k-1 games), so there is a

(k-1 choose 4)*(40/100)^5*(60/100)^(k-5) probability that Richard wins in k games. We just sum this value for k from 5 to 9.

https://web2.0calc.com/questions/coordinates_85664

https://web2.0calc.com/questions/help-plz_42690#r1

This should help:

Hallo Guest!

\(2x(x - 10) - x(x - 5) = -50\ |\ +50\\ 2x^2-20x-x^2+5x+50=0\)

\(x^2-15x+50=0\)

         p           q

\(x=-\frac{p}{2}\pm\sqrt{(\frac{p}{2})^2-q}\)

\(x=7,5\pm\sqrt{56.25-50}\\ x=7.5\pm 2.5\\ x_1=5\\ x_2=10\)

\(\sum x=5+10=15\)

  !

2x+5y=-8

multiply by 3

6x+15y=-24      (1)

6x=16+a-15y

6x+15y = 16+a     (2)

These will be the same line and therefore have infinite simultaneous solutions if

16+a = -24

a= -40

Thanks EP

Tr1ck if you still have questions or concerns please keep asking .

We like interacting with people here   

Thanks catmg  

Hi Tr1

Think about this

3x + 5x  = 8x because you started with 3x and you added another 5x.  so you have (3+5)x

ctmeg had

z^4(z^3 + 8) - 2i(z^3 + 8) = 0

Here you have  z^4 lots of (x^3+8)  and from that you are taking away  2i lots of (z^3+8)

so you end up with

\((z^4-2i) \;\;lots\;\; of \;\;(z^3+8) = (z^4-2i) ( z^3+8) \)

Think about it.   

That's easy...

Try this one: what's the area of the largest possible circle that can be inscribed in the purple region?

( Inscribed circle is tangent to all 3 semicircles)

When reposting, you must include the address of the original post and ask people to answer there. 

(ask them to leave a post on the new thread saying they have done so.)

If it has been locked you can ask a moderator to unlock it for you. 

OK....had to look up how to work with this and how to rotate a hyperbola....so I learned something new today!)

Re-write into form    a x^2 + b xy + c y^2 + dx + ey + f = 0

x^2 +4xy + y^2  -1 = 0

a) Hyperbola

b) cot 2phi =  (a-c)/b  =  (1-1)/4 = 0

       phi = pi/4

c)  x = x' cos pi/4   -  y' sin pi/4    =   (x'-y')/sqrt2

      y = x' cos pi/4  + y' sin pi/4    =   (x'+y'/sqrt2

Sub these into the original equation   x^2                               + 4xy                          + y^2                      = 1

                                                         1/2 (x'^2 - 2x'y' + y'2)    + 4 ( x'^2 -y'^2)/2   + 1/2(x'^2+2x'y' + y'^2) =1

                                                             3x^2 -y'^2 =1

See graph:

https://www.desmos.com/calculator/w9jpq2jzpj

cos(∠ABM) = √(¹/₅) = 0.447213595

The first derivative will give you the slope at  x = 2     ( from the point 2,1)

10x (x^2-3)^4     when x = 2

10(2)(4-2)^4 = 20 = the  slope at x = 2

y = mx + b

y = 20x + b      for the point 2,1

1 = 20(2) + b      b = -39

y = 20x - 39

Here is a graph to confirm:

https://www.desmos.com/calculator/eqlln3qpv9

2xy+x^2+y^2=56 ->xy+1/2 x^2+1/2 y^2=28

[arctan(2) + arctan (8/3) ] / 2 = 66.43945

tan (66.43945)= bsector slope =2.2932  

Wait im kind of confused on how you went from this z^4(z^3 + 8) - 2i(z^3 + 8) = 0 to this (z^4 - 2i)(z^3 + 8) = 0. The answer is prob correct tho.

Hmm idk if its just me but the first equation cant be simplyfied and for the 2nd one i got it from x^2+y^2=56-2xy -> 2xy+x^2+y^2=56 ->xy+x^2+y^2=24. Now i dont think you can simplify it anymore so..... 

Answer: 7/3

Solution:

I'm not sure if this is correct but I think that it would just the average of the slopes. Which is 7/3.

This will be the area of the circle of the equation

(x-2)^2  + (y-3)^2  < =    73  + 4 + 9        The red numbers = r^2      Can you finish?

Hmmmmmm....

     x+y = 10    and    (x+y)² = 56     doesn't seem to have a solution.....am I missing something?

Hmmmm....not so sure about that !    What if the first card drawn is the ace of diamonds?

re-arrange....and divide through by 9 to get

x^2 - 2x      + y^2+4y    = - 88/9              complete the square for x and y

(x-1)^1       + (y+2)^2   = - 88/ 9  +1 + 4       the red total = r ^2      

                    You can finish? (probably not....there is something wrong with the equation in your Q!....Are you sure it is +88 and not -88?)