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Omg saaaaame y is school so boring 

US history...FUN

Dude chill and don't post things like this

http://www.rhymezone.com/r/rhyme.cgi?Word=Death&typeofrhyme=adv&org1=syl&org2=l&org3=y&loc=sel_change 

Has all you need. Trust me.

Not technically a holiday, but......Halloween.....!!!

http://wikirhymer.com/words/death

Unknown

More than perscribed by your Doctor.

BTW don't, it's not the way you want to go out, it's not worth it. Just don't even think about it.

Hi - what is the best way to solve simultaneous equations where one is quadratic and the other linear?

For example:

y=x+3

y=x²+3x

1. Formula line ( linear equation)

\(y_{\text{line}}=m\cdot x_{\text{line}}+b \)

2. Formula parabola (quadatic equation)

\(y_{\text{parabola}}=A\cdot x_{\text{parabola}}^2+B\cdot x_{\text{parabola}} + C \)

3. set equal  \(y_{\text{line}}=y_{\text{parabola}}=y_{\text{intersection}}\) :

\(\begin{array}{|rcll|} \hline m\cdot x_{\text{intersection}}+b &=& A\cdot x_{\text{intersection}}^2+B\cdot x_{\text{intersection}} + C \\\\ Ax_{\text{intersection}}^2+x_{\text{intersection}}(B-m)+C-b &=& 0 \\ x_{\text{intersection}_{1,2}} &=& \dfrac{m-B\pm \sqrt{(m-B)^2-4\cdot A \cdot(C-b)} }{2A} \\ y_{\text{intersection}_{1,2}} &=& m\cdot x_{\text{intersection}_{1,2}} + b \\ \hline \end{array} \)

4. Example:

\(\begin{array}{|rcll|} \hline y &=& x + 3 \quad & \quad m=1 \quad b = 3 \\ y &=& x^2+3x \quad & \quad A=1 \quad B = 3 \quad C = 0 \\ x_{\text{intersection}_{1,2}} &=& \dfrac{1-3\pm \sqrt{(1-3)^2-4\cdot 1 \cdot(0-3)} }{2\cdot 1} \\ x_{\text{intersection}_{1,2}} &=& \dfrac{-2\pm \sqrt{4+12} }{2} \\ x_{\text{intersection}_{1,2}} &=& \dfrac{-2\pm 4 }{2} \\ x_{\text{intersection}_{1}} &=& \dfrac{-2 + 4 }{2} \\ &=& 1 \\\\ x_{\text{intersection}_{2}} &=& \dfrac{-2 - 4 }{2} \\ &=& -3 \\\\ y_{\text{intersection}_{1}} &=& 1\cdot x_{\text{intersection}_{1}} + 3 \\ &=& 1\cdot 1 + 3 \\ &=& 4 \\\\ y_{\text{intersection}_{2}} &=& 1\cdot x_{\text{intersection}_{2}} + 3 \\ &=& 1\cdot (-3) + 3 \\ &=& 0 \\ \hline \end{array}\)

You've forgotten one of the minus signs Guest #1

f'(x) = 0 - 10*(-2)e^-2x  or  f'(x) = 20e^-2x 

.

d/dx (e^(-2x) )      = -2e^(-2x)      from the Chain Rule.                 

So f'(x) =  -20 e^(-2x)

jajco gowno platfusie

Coooooo chuju gnoju siusiaku ??!?!?

nie wiem stary 

How lucky!  We all got the same answer :DD

Probability Question

Daniel has a bag of marbles. he has twice as many black marbles as red marbles.

the rest are yellow.

he is going to take a marble at random from the bag.

yellow is 1/7

what is black and red probability?

\(\begin{array}{|rcll|} \hline b=2r \\ \hline \end{array}\)

\(\begin{array}{|lcrcll|} \hline \text{Probability yellow } = \frac17 & \Rightarrow & y+b+r &=& 7 & \text{ and } \quad y = 1 \\ & & y+b+r &=& 7 & \qquad b=2r \quad y = 1 \\ & & 1+2r+r &=& 7 \\ & & 3r &=& 6 \\ & & \mathbf{ r } & \mathbf{=} & \mathbf{2} \\ \text{Probability red:} & & \mathbf{ \frac{r}{y+b+r} } & \mathbf{=} & \mathbf{\frac27 } \\\\ & & y+b+r &=& 7 & \qquad r=2 \quad y = 1 \\ & & 1+b+2 &=& 7 & \\ & & b &=& 7-3 & \\ & & \mathbf{ b } & \mathbf{=} & \mathbf{4} \\ \text{Probability black:} & & \mathbf{ \frac{b}{y+b+r} } & \mathbf{=} & \mathbf{\frac47 } \\\\ \hline \end{array} \)

\(\begin{array}{|l|r|r|r|} \hline \text{Colour} & \text{yellow} & \text{black} & \text{red} \\ \hline \text{Probability} & \frac17 & \frac47 & \frac27 \\ \hline \end{array} \)

5.406 multiplied by 10 to the 12 power is equal to 5.406e + 12

Ten to the 12th power is equal to 10 being multiplied by itself 12 times. Ten to the power of 12 is 1,000,000,000,000, or one trillion.

health?

"Twenty-five percent of all automobiles undergoing an emissions inspection at a certain inspection station fail the inspection. 

(d) What is the probability that among 25 randomly selected cars, the number that pass is within 1 standard deviation of the mean value?"

I've assumed a binomial distribution.  i.e. each car either passes or fails.

.

You're right -- I became enamored of my simple model and overlooked something important about the set of integers {0 , 1 * 10^12-1}, which is that (unlike my model) nearly all of of them are 12 digits long.

Granted its limitations, a Python model shows perfect correspondence with the idea that the total number of '3' digits in the set {0,1 * 10^12-1} is equal to 1 * 10^12 (height) * 12 (width) / 10 (isolate a single digit) = 12 * 10^11

A general rule can be written for any chosen upper bound m:

t = (log(m)/log(10)) * m / 10

t = number of occurrences of a single digit in the set {1,9}

This rule doesn't work for zero, for a reason I haven't figured out.

n = b + r + y    n = total

b = 2r             twice as many black as red

y/n = 1/7       yellow probability

n = 3r + y

y/(3r + y) = 1/7

7y = 3r + y

6y = 3r

r = 2y

b = 4y

r/n = 2y/n = 2/7    probability of drawing a red

b/n = 4y/n = 4/7   probability of drawing a black

(I see Melody beat me to it.  Luckily, I got the same results as her!).

.

Colouryellowblackred

Probability1/7??

Daniel has a bag of marbles. he has twice as many black marbles as red marbles. the rest are yellow. he is going to take a marble at random from the bag. yellow is 1/7 what is black and red probability?

P(yellow) = 1/7

so

P(red or black) = 6/7

and there are twice as many black as red so that must be 

P(red)=2/7

P(black)=4/7

FV=PV[1 + R]^N

800=500[1 + R]^8 Divide both sides by 500

1.6 =[1 + R]^8 Take the 8th root of both sides

1 + R = 1.6^(1/8)

1 + R = 1.0605  Subtract 1 from both sides.

R = 1.0605 - 1 x 100 =6.05%