BuilderBoi

https://web2.0calc.com/questions/distributions_2

Take the first equation, and multiply it by 8. This gives us \(8x + 8y + 24x = 80\). 

Subtracting the second equation from this gives us \(12x + 19z = 73\). 

Now, take the third equation and multiply it by 19. This gives us \(19kx + 19z = 57\)

These two equations won't have any solutions when the coefficient of x in both equations is equal. This will mean that two of the same things sum to two different values, which is impossible. 

This means that \(12 = 19k\), meaning \(k = \color{brown}\boxed{12 \over 19}\)

There are 9 stars (chocolate) and 7 - 1 = 6 bars. 

This makes for \({15 \choose 6} = 5005\) cases. 

If Dhruve gets exactly 2, there are now 7 stars, and 6 - 1 = 5 bars, which makes for \({12 \choose 5} = 792\) cases. 

So, the probability is \({792 \over 5005} = \color{brown}\boxed{72 \over 455}\)

It says it's a quiz... 

Maybe you aren't supposed to get help.

By the distance formula, the distance from the origin to the graph is \(\sqrt{(y-0)^2 + (x-0)^2} = \sqrt{y^2 + x^2}\)

We know that \(y = {1 \over 2}x^2 - 4\), meaning \({1 \over 2}x^2 = y + 4\), so \(x^2 = 2y + 8\)

Substituting this gives us \(\sqrt{y^2 + 2y + 8}\). 

The minimum value occurs at \(y = -{b \over 2a} \), which is -1.

This means the minimum is \(\sqrt{(-1)^2 + 2(-1) + 8} = \color{brown}\boxed{\sqrt{7}}\)

\(99 \times 100 \div 2 = \color{brown}\boxed{4950}\)

c) If ab = 0, either a or b = 0. 

If a = 0, b = -1, so \((a, b)\) is \(\color{brown}\boxed{(0, -1)}\)

If b = 0, a = 1, so \(\color{brown}\boxed{(1, 0)}\)

These are the only solutions that work

b) We already proved that \(a^2 + b^2 + ab = 1\).

Now, we write this as \((a+b)^2 - ab = 1\)

Because \(ab = 0\), we know \((a+b)^2 = 1\), meaning \(a + b = \color{brown}\boxed{-1, 1}\) 

a) Note that \(a^3 - b^3 = (a-b)(a^2 + b^2 + ab)\)

Substituting what we have, we get \(1 = 1 (a^2 + b^2 + ab)\), which means that \(a^2 + b^2 + ab = 1\)

We can write \(a^2 + b^2 + ab \) as \((a - b)^2 + 3ab\). 

Using what we know, we have \(1^2 + 3ab = 1\), meaning \(ab = \color{brown}\boxed{0}\)

Because we are dealing with 4 coins, the easiest way (imo) is to list out all 16 cases and count the average

The 16 cases are: 

HHHH   HTHT   TTTT   THTT

HHHT   HTTH   TTTH   THHT

HHTH   HTTT   TTHT   THTH

HHTT   HTHH   TTHH   THTH

We see that 2 have a run of 4, 4 have a run of 3, 8 have a run of 2, and 2 have a run of 1.

So the expected value is \({{(2 \times 4) + (4 \times 3) + (8 \times 2) + (2 \times 1)} \over 16} = \color{brown}\boxed{2.375}\)