MathProblemSolver101

I guess that works, too!

Anytime!

The values of $x$ and $y$ are always positive, and $x^2$ and $y$ vary inversely. If $y$ is 10 when $x$ is 2, then find $x$ when $y$ is 4000.

$y = \frac{k}{x^2}, k \in \mathbb{R}$ 

$k = x^2 \cdot y = 2^2 \cdot 10 = 40$

$y = \frac{40}{x^2}$

$4000 = \frac{40}{x^2}$

$100 = \frac{1}{x^2}$

$100x^2 = 1$

$x^2 = \frac{1}{100}$

$x = \frac{1}{10}$

Hint: Evaluate each of $\binom{n}{k-1}, \binom{n}{k}, \binom{n}{k+1}$ (which can be found in heureka's answer if you're stuck.)

Hint 2: You know $\frac{\binom{n}{k}}{\binom{n}{k-1}} = \frac{n - k + 1}{k},$ you have $\binom{n}{k} = \binom{n}{k-1} \left(\frac{n - k + 1}{k}\right)$ and $\binom{n}{k-1} = \frac{k \binom{n}{k}}{n-k + 1}$. Evaluate $\binom{n}{k+1}$ in terms of $\binom{n}{k}$

\(\begin{align*} \frac{100\%}{120\%} \cdot 2.52 &= \frac{5}{6} \cdot 2.52 \\ &= 5 \cdot 0.42 \\ &= \boxed{2.1} \end{align*}\)

Use binomial thoerem

4+1=5th row of pascal triangle

1,4,6,4,1

1 x^4

4x^3y

6x^2y^2

4xy^3

1 y^3

use this expansion of (x+y)^4 

Thank you, melody !

Cos 80 = h / 12

solve for h

Yup, seems like this guy is solving more advanced questions than he asks and he is plagiarizing other people's answers to build rep.

1/7 is around 14% I think