CubeyThePenguin

$1 \frac{1}{2} + 2.51 + 1 \frac{17}{20}$

$= 1.50 + 2.51 + 1.85$

$= \boxed{\$5.86}$

considering only the time Nathan was actively working on the problem, he worked for 1 5/6 + 1 1/3 = 1 5/6 + 1 2/6 = 2 7/6 = 3 1/6 hours.

total possible sequences of flips: 2^10 = 1024 (there are only two choices for each flip, heads or tails, for ten flips)

8 heads: 10 choose 8 = 45 (the number of ways to arrange the sequence HHHHHHHHTT)

9 heads: 10 choose 9 = 10

10 heads: 10 choose 10 = 1

total probability: (45 + 10 + 1)/1024 = 56/1024 = 7/128

Look up information regarding the binomial theorem for further information.

$\binom{6}{4}(2x)^4 (2)^2 = \boxed{960x^4}$

The coefficient is 960.

Simply plug in the expression into the function.

a. f(1) = 3(1) - 2 = 1

b. f(t) = 3t - 2

c. f(x^2) = 3x^2 - 2

d. f(f(x)) = f(3x - 2) = 3(3x - 2) - 2 = 9x - 8

Let x be the number of apples Nicole has and y be the number of apples that Hannah has.

x/y = 1/3

(x + 4)/(y + 3) = 1/2

Rewrite the first equation to get y = 3x, then substitute into the second equation.

(x + 4)/(3x + 3) = 1/2

2(x + 4) = 3x + 3

2x + 8 = 3x + 3

5 = x

Nicole had 5 apples in the beginning, so Hannah had 15 apples in the beginning. In the end, Hannah had 15 + 3 = 18 apples.

1584 = 2^4 * 3^2 * 11

To be a perfect cube, a number's prime factors' exponents must all be multiples of 3. So, the smallest value of x is:

$\frac{2^6 \cdot 3^3 \cdot 11^3}{2^4 \cdot 3^2 \cdot 11} = \boxed{1452}$

In other terms, x = 1452 = 2^2 * 3 * 11^2. The product xy must be a multiple of 1584, so,

xy = 2^2 * 3 * 11^2 * y = 2^4 * 3^2 * 11 * n

y must have 2^2 * 3 as a factor, so the smallest value of y is 12.

Use the following algebraic identity:

(x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + xz + yz) = 9 + 2(18) = 45

$\frac{1}{x} + \frac{1}{y} = \frac{1}{17}$

$\frac{x + y}{xy} = \frac{1}{17}$

$17(x + y) = xy$

$0 = xy - 17x - 17y$

$289 = (x - 17)(y - 17)$

The possible values of x are 18, 34, and 306.