Divineology

56 is divisible by 7, so this problem is very easy.

7 * 8 = 56

2 * 8 = [16] magnets Jason has

Squaring both sides of the original equation, we get $a^2 = 4+\sqrt{5+a}.$
Subtracting 4 from both sides and squaring again gives $(a^2-4)^2 = 5+a.$
Expanding this out and subtracting 5 + a from both sides, we have $a^4-8a^2-a+11 = 0.$
Similar manipulations on the other equations give $\begin{align*} b^4-8b^2-b+11 &= 0, \\ c^4-8c^2+c+11 &= 0, \\ d^4-8d^2+d+11 &= 0. \end{align*}$
(Note the signs carefully.)

Let $f(x)=x^4-8x^2-x+11$. Then a and b are roots of f(x), while c and d are roots of  $x^4-8x^2+x+11$, which is f(-x). It follows that -c and -d are roots of f(x). Since -c and -d are negative and hence distinct from a and b, we have four roots of f(x). Since f(x) has degree four, we know we have found all the roots.

By Vieta's formulas, the constant term of f(x) is the product of the roots. Thus, 

$\begin{align} abcd = (a)(b)(-c)(-d) = \boxed{11}. \end{align}$

We start with $2,000. He then uses $\frac14 + \frac18 = \frac38$ of his salary. This means he has $1 - \frac38 = \frac58$ of his salary left.

$$2000 \cdot \frac58 = $1,250$

Thus, he has $1,250 left.

This problem is fairly simple, you just need to assign values to $x$ and $y$, then solve the equation.
 

Using the equation $4x=3y$, we can assign $x=3$ and $y = 4.$

$\frac{2x+y}{3x+4y}$

$\frac{2(3)+4}{3(3)+4(4)}$

$\frac{6+4}{9+16}$

$\frac{10}{25}$

$\boxed{\frac{2}{5}}$

$11*101+110 = 1221$

That is incorrect.

The probability will always be % $50$ because each turn, there is a 50-50 chance that the coin will be heads or tails.

It didn't have any guts. Ha, ha.

Ok, I just wanted to make sure! 

Did you mean to add a slash here? 1/((1 - a)(1 - ab)) + 1/2*a/((1 - a)(1 - ab)) + 1/2*b$/$((1 - a)(1 - ab)) = (a + b + 2)/(2(1 - a)(1 - ab))

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