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The correct answer is 72.

By the quadratic formula, the answer is 3 seconds.

I think it's 144 but I'm not 100% sure

\(\boxed{\color{red}{R}}\qquad \boxed{\color{white}{A}}\qquad \boxed{\color{red}{R}}\qquad \boxed{\color{white}{A}}\qquad \boxed{\color{red}{R}}\qquad \)

Here is a start, now work out how many combinations there are for the other 2 boxes.

And how many participants?

\(0=-15t^2+28t+144\\ t=\frac{-28\pm\sqrt{784+8640}}{-30}\\ t=\frac{-28\pm\sqrt{9424}}{-30}\\ t=\frac{+28+4\sqrt{589}}{30}\\ t\approx 4.17 sec\)

Hi Lizzie

Jessica bought 1.2 * 5 yards

and 

Derek bought  1.25* 3  yards

You can simplify those (with a calc if necessary)  and see which is bigger  

Hi Lina,

Welcome to Web2.0calc forum  

This is called partial fractions, it comes in useful when doing calculus.

\(\frac{3x+2}{x^2-x-12}=\frac{A}{(x-4)}+\frac{B}{(x+3)}\\ So\\ Ax+3A+Bx-4B=3x+2\\ (A+B)x+(3A-4B)=3x+2\\ A+B=3\qquad and \qquad 3A-4B=2\\ A=3-B\\ \text{sub to get B}\\ 3(3-B)-4B=2\\ -7B+9=2\\ B=1\\ A=3-1=2\\ \frac{3x+2}{x^2-x-12}=\frac{2}{(x-4)}+\frac{1}{(x+3)}\\ \)

You can finish :)

The answer is 7, because the number is 13579. 

The larger digits should have a lower place value to minimize the value of it

thanks

-6x = -8x-14\(-6x = -8x-14 | 2x=-14 | x=-7\)

2x=-14

x=-7

First I went online to find all the formulas that I need

You can make h=0  and k =0  so it is centred at the origin,  I'll use the blue one

a=94 and b=52

Plug those numbers in and you will have your formula.

Plug them into the c formula and you will have c

You should be able to finish it but you can ask questions if you want.

Were you asleep when you wrote the question? Read it carefully and see if there are any numbers missing!

What cards did you have in mind? Were you asleep when you wrote the question?

How many more times will this question be asked ?

And how many more times will  the answer be that a 3, 4, 8 triangle can't exist  ?

That is wrong.

Divide it up by cases and use stars and bars: 

If the twins get 0 candies: There are 10 stars and 2 bars for \({12 \choose 2} = 66\) cases

If they get 1 candy: There are 8 stars and 2 bars for \({10 \choose 2} = 45\) cases 

If they get 2 candies: there are 6 stars and 2 bars for \({8 \choose 2} = 28\) cases

If they get 3 candies: There are 4 stars and 2 bars for \({6 \choose 2} = 15\) cases

If they get 4 candies: There are 2 stars and 2 bars for \({4 \choose 2} = 6\) cases

If they get 5 candies: There are 0 stars and 2 bars for \({2 \choose 2} = 1\) case.

So, there are \(66 + 45 + 28 + 15 + 6 + 1 = \color{brown}\boxed{161}\) cases.

Now do the rest

a)

 Note that \({n - 1 \choose k - 1} = {(n-1)! \over (k-1)! (n-k)!}\) and \({n -1 \choose k} = {(n-1)! \over {k! (n-k-1)!}}\)

The LCD of both denominators is \(k!(n-k)!\). To get this, multiply the first denominator by k and the second denominator by n -k.

This gives us \({k(n-1)! \over k!(n-k)!} + {(n-k)(n-1)! \over k!(n-k)!} = {(n-k)(n-1)! + k(n-1)! \over k!(n-k)}\)

Now, factor the numerator into \((n-k+k)(n-1)! = n(n-1)! = n!\)

This leaves us with \({n! \over k!(n-k)!} = {n \choose k}\)

\((a_1, a_2) = (-1/3, 2 \sqrt{2}/3)\)

Sorry; I just tried that and its incorrect!

Plugging in different values of t, you can find that t = 5.5 works.