Melody

\(2^{2^{2^{-892^{2^{2^{\frac{2^2}{\frac{8}{\frac{98}{\frac{99999}{9999999}}}}}}}}}} {nl} {nl} \)

mmm   interesting :///

Chloe and Jack play 3 games. The probaiblity that Chloe wins any game is 3/5. What is the probaility that Chloe wins for the first time in the third game?

P(lose, lose, win) = 2/5 *2/5 * 3/5 = 12/125

Hi Juriemagic,

It is good to see you again  

Equation 1: \(x^2-y^2=0 \)

Equation 2: \(x^2-4x-9=4y\)

Mmm  This is an unusual one.

It always helps if you can picture what the graphs look like

\(x^2-y^2=0\\ x^2=y^2\\ y=\pm x\\ \mbox{This is a big X on the number plane centred on (0,0)}\\ \mbox{I can also see that the other one is a concave up parabola.}\\ \mbox{anyway that means}\\ x^2-4x-9=4x\qquad or \quad x^2-4x-9=-4x\\ x^2-8x-9=0 \qquad or \quad x^2-9=0\\ (x-9)(x+1)=0 \qquad or \quad (x-3)(x+3)=0\\ x=9\;\;\;or\;\;\;x=-1 \qquad or\qquad x=3\;\;\;or\;\;\;x=-3\\ \mbox{Now I am going to check if y should be pos or neg by subbing into equ2}\\ x=9 \qquad 81-36-9>0\;\;\;so\;\;\;y>0\;\;\;y=9\;\;\;(9,9)\\ x=-1 \qquad 1+4-9<0\;\;\;so\;\;\;y<0\;\;\;y=-1\;\;\;(-1,-1)\\ x=3 \qquad 9-12-9<0\;\;\;so\;\;\;y<0\;\;\;y=-3\;\;\;(3,-3)\\ x=-3 \qquad 9+12-9>0\;\;\;so\;\;\;y>0\;\;\;y=3\;\;\;(-3,3)\\ \)

So the 4 points of intersection (simuiltaneous solutions)  are  (9,9), (-1,-1),  (3,-3),  and   (-3,3)

Here is the graph to show you what is happening

https://www.desmos.com/calculator/lvjgdscib6

what is cavemans favorite sandwich?

I expect that would be a club sandwich ://      

There are 4 squares that are 1 by 1

and there is 1 square that is 2 by 2

That is 5 squares.

Draw the dots on a peice of paper.  It is super easy. :)

cos (2x) times cos (4x) 

\(cos(2x)=cos^2x-sin^2x\\ cos(2x)=cos^2x-(1-cos^2x)\\ cos(2x)=2cos^2x-1\\ ...\\ cos(4x)=cos^2(2x)-sin^2(2x)\\ cos(4x)=[cos^2x-sin^2x]^2 -[2sinxcosx]^2\\ cos(4x) =[cos^2x-(1-cos^2x)]^2 -[4sin^2xcos^2x]\\ cos(4x) =[2cos^2x-1]^2 -[4(1-cos^2x)cos^2x]\\ cos(4x) =[4cos^4x-4cos^2x+1] -[4cos^2x-4cos^4x]\\ cos(4x) =4cos^4x-4cos^2x+1-4cos^2x+4cos^4x\\ cos(4x) =8cos^4x-8cos^2x+1\\ ...\\ cos(2x)(cos4x)=(2cos^2x-1)(8cos^4x-8cos^2x+1)\\ cos(2x)(cos4x)=2cos^2x(8cos^4x-8cos^2x+1)-1(8cos^4x-8cos^2x+1)\\ cos(2x)(cos4x)=6cos^6x-16cos^4x+2cos^2x-8cos^4x+8cos^2x-1\\ cos(2x)(cos4x)=6cos^6x-24cos^4x+10cos^2x-1\\ \)

That is if I did not make any careless mistakes.    

(-64)^2\3

\(((-64)^{1/3})^2 = (-4)^2=16 \)

Are E and R both just variables or is E really the number e ???

If there are 2 variables then you can't solve it with just one equation.

However ,if your question to find R given this then..

log(e) = 1.2 + 1.81 ln(R)

1= 1.2 + 1.81 ln(R)

-0.2=  1.81 ln(R)

(-0.2/1.81)=  ln(R)

R=e^(-0.2/1.81)

e^(-0.2/1.81) = 0.8953888036358034

18ab2 – 4ac2 – 10ac  If a= 3 and b=1

\(18ab^2 – 4ac^2 – 10ac\qquad If \qquad a= 3 \;\;and \;\;b=1\\ 18*3*1^2 – 4*3c^2 – 10*3c\\ 54 – 12c^2 – 30c\\ – 12c^2 – 30c+54\\\)

That is all you can do unless you know the value of c      

(14.08-2.003) x 1.2 / 6.3-2.01

I'm using the forum calc to check the answer :)

(14.08-2.003) x 1.2 / 6.3-2.01 = 0.2903809523809523809523809523809523809523809523809524