my favourite film is fight club :D

and you :D

i have a girlfriend :)

it's good!

(no i havent... :( )

8 is the least

4, and what does sf mean?

If you can't solve that question, you probably couldn't find your brain even if you have one in which to put a bullet.

If I Only Had A Brain  . . .

I could wile away the hours
Conferrin' with the flowers
Consultin' with the rain
And my head I'd be scratchin'
While my thoughts were busy hatchin'
If I only had a brain

I'd unravel any riddle
For any individ'le
In trouble or in pain


Oh, I would tell you why
The ocean's near the shore
I could think of things I never thunk before
And then I'd sit and think some more

I would not be just a nuffin'
My head all full of stuffin'
My heart all full of pain
I would dance and be merry
Life would be a ding-a-derry
If I only had a brain

how do you do what? and dont say that word, the F one.

Here's another way of looking at it:

Let the distance betwen the cities be d miles

The speed, s, of the car is distance travelled divided by time taken or (d/6 miles)/(3/5 hours) = (d/6)*(5/3) miles per hour = 5d/18 miles per hour

At this speed the car will travel 5d/18 miles in one hour.

As a fraction of the distance between the cities this is (5d/18)/d or just 5/18

. 

As for how I found out 1/18,  I was finding 1/5

1/5 goes into 3/5 3 times,  so I divided 1/6 by 3

(1/6)/(3/1)

Flip and multiply

(1/6)*(1/3)=1/18

September price = S

October price = 1.2S = $1.5  so September price S = $1.5/1.2 = $1.25

Money earned in September = 0.4*900*$1.25 = $450

Money earned in October = 0.4*700*$1.5 = $420

.

3/5 is 2/5 less than one whole

We need to find how much they traveled in 2/5 of an hour and add that to 1/6

In 1/5 of an hour,  they traveled 1/18 of a mile

(1/18)*2=1/9

1/9+1/6=10/36

10/36=5/18

They traveled 5/18 in one hour

Can tackle it like this:

$$2^{x+4}-2^{x+1}=3$$

$$2^x2^4-2^x2^1=3$$

$$2^x(2^4-2^1)=3$$

$$2^x(16-2)=3$$

$$2^x\times14=3$$

$$2^x=\frac{3}{14}$$

$$\ln{2^x}=\ln{\frac{3}{14}}$$

$$x\times \ln2=\ln3-\ln14$$

$$x=\frac{\ln3-\ln14}{\ln2}$$

$${\mathtt{x}} = {\frac{\left({ln}{\left({\mathtt{3}}\right)}{\mathtt{\,-\,}}{ln}{\left({\mathtt{14}}\right)}\right)}{{ln}{\left({\mathtt{2}}\right)}}} \Rightarrow {\mathtt{x}} = -{\mathtt{2.222\: \!392\: \!421\: \!336\: \!447\: \!9}}$$

.

a.$${\mathtt{1.5}}{\mathtt{\,\times\,}}{\mathtt{0.2}} = {\frac{{\mathtt{3}}}{{\mathtt{10}}}} = {\mathtt{0.3}}$$

0.30+1.50=1.80

The price was $1.80

b.More in September because they sold more

$${\mathtt{900}}{\mathtt{\,\times\,}}{\mathtt{0.4}} = {\mathtt{360}}$$

They earned $360 in September

$${\mathtt{700}}{\mathtt{\,\times\,}}{\mathtt{0.4}} = {\mathtt{280}}$$

They earned $280 in October

360-280=80

They earned $80 more in September than October

Can anyone help me?

Since more than half of each row of cups must contain beans, each row will need at least five beans. This means that at least two of the empty cups in the first row and at least four of the empty cups in the second row must have beans added to them.

For the game operator to have the advantage over the player, the probability that the two cups selected will both contain beans must be less than 50%. The probability that the cups will both contain beans is the product of the probabilities of the cup from each row containing a bean. Adding beans to two of the empty cups in the first row and four of the empty cups in the second row will result in the probability that both cups selected contain beans being 58⋅58=2564≈39.06%, which is less than 50%.

 Add beans to two of the empty cups in the first row and four of the empty cups in the second row.

The lower the probability of both revealed squares being red, the fewer winners the contest will have. The probability of revealing a red square in a section is the number of red squares in the section divided by the total number of squares in the section. The probability that the both revealed squares will be red is the product of the probabilities of revealing a red square in each section.

Changing three of the non-red squares in the first section to red and leaving the second section as is will result in the probability that both revealed squares will be red being 55⋅49=2045=49≈44.4%. Changing two of the non-red squares in the first section to red and changing one of the non-red squares in the second section to red will result in the probability that both revealed squares will be red being 45⋅59=2045=49≈44.4%. Changing one of the non-red squares in the first section to red and changing two of the non-red squares in the second section to red will result in the probability that both revealed squares will be red being 35⋅69=1845=25=40%. Leaving the first section as is and changing three of the non-red squares in the second section to red will result in the probability that both revealed squares will be red being 25⋅79=1445≈31.1%. The lowest probability of both revealed squares being red occurs when the first section is left as is and three of the non-red squares in the second section are changed to red, so this is the best option to add three red squares and still minimize the number of winners.

 Leave the first section as is and change three of the non-red squares in the second section to red.

$${{\mathtt{2}}}^{\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{4}}\right)}{\mathtt{\,-\,}}{{\mathtt{2}}}^{\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)} = {\mathtt{3}} \Rightarrow {\mathtt{x}} = {\mathtt{\,-\,}}{\frac{\left({ln}{\left({\mathtt{2}}\right)}{\mathtt{\,-\,}}{ln}{\left({\frac{{\mathtt{3}}}{{\mathtt{7}}}}\right)}\right)}{{ln}{\left({\mathtt{2}}\right)}}} \Rightarrow {\mathtt{x}} = -{\mathtt{2.222\: \!392\: \!421\: \!336\: \!448\: \!1}}$$

Besides......I got a cake......yum.....probably rich though

And apparently he likes marbles......

And we're friends!

MELODY!  STOP TEASING!  I NEED TO HAVE THAT....THAT....THAT WHATEVER IT IS!

With what now?

.....................................x= -2

$${\frac{{\mathtt{14}}}{{{\mathtt{x}}}^{{\mathtt{2}}}}}{\mathtt{\,-\,}}{\mathtt{9}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{6}}}{{\mathtt{x}}}}{\mathtt{\,-\,}}{\mathtt{3}} = {\frac{{\mathtt{8}}}{{\mathtt{x}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{211}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}{{\mathtt{15}}}}\\
{\mathtt{x}} = {\frac{\left({\sqrt{{\mathtt{211}}}}{\mathtt{\,-\,}}{\mathtt{1}}\right)}{{\mathtt{15}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = -{\mathtt{1.035\: \!055\: \!936\: \!422\: \!263\: \!3}}\\
{\mathtt{x}} = {\mathtt{0.901\: \!722\: \!603\: \!088\: \!93}}\\
\end{array} \right\}$$

(2X^2+63y^2)(2x^2-63y^2)

Assuming x = X:

set: a = 2X^2 and b =  63y^2

(a + b)(a - b) = a^2 - b^2=

(2X^2)^2 - (63y^2)^2 =

4X^4 - 3969y^4

Assuming x =/= X:

(2X^2+63y^2)(2x^2-63y^2) 

Set: a = 2X^2, b= 63y^2 and c = 2x^2

(a + b)(c - b) = ac - ab + bc - b^2 = a(c - b) + b(c - b) =

(2X^2)(2x^2 - 63y^2) + (63y^2)(2x^2 - 63y^2) =

4(X^2)(x^2) - 126(X^2)(y^2) + 126(x^2)(y^2) - 3969 y^4 

CPHILL!  WHAT A TRAITOR!

You already know that you are apart of our pack,  and you cross enemy grounds?  How dare you?

Happy B-Day Old Man!

I know you love those!

And a flaming football! LOL