Melody

Derivative of (x^4)÷(y^8) with respect to x, given dy÷dx=v

$$\\\frac{d}{dx}\;\frac{x^4}{y^8}\\\\
=\frac{d}{dx}\;x^4*y^{-8}\\\\
=\;4x^3*y^{-8}+x^4*-8y^{-9}\;\;\frac{dy}{dx}\\\\
=\;4x^3*y^{-8}+x^4*-8y^{-9}*v\\\\
=\frac{4x^3}{y^{8}}+\frac{-8x^4v}{y^{9}}\\\\
=\frac{4x^3y-8x^4v}{y^{9}}\\\\
=\frac{4x^3(y-2xv)}{y^{9}}\\\\$$

Could a mathematician check  this please ?

It is undefined

Chris, this is one of your statements

"if we allowed for leading zeroes in both of the first two positions."

The question states that the digits must be unique.

I am standing by my answer.

At a party, everyone shook hands with everybody else. There were 66 handshakes. How many people were at the party?

Mmm

Let there be k people at the party.

The first person shook with k-1 people.

the second with a further k-2 people

the kth person did not shake with anyone new.

So the number of handshakes was

1+2+3+.....+(k-1)

this is the sum of an AP       S=n/2(a+L) =  $$\frac{k-1}{2}(1+(k-1))=\frac{k(k-1)}{2}$$

so

$$\\\frac{k(k-1)}{2}=66\\\\
k(k-1)=132\\\\
k^2-k-132=0\\\\$$

$${{\mathtt{k}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{k}}{\mathtt{\,-\,}}{\mathtt{132}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{k}} = {\mathtt{12}}\\
{\mathtt{k}} = -{\mathtt{11}}\\
\end{array} \right\}$$

Obviously there is not a neg number of people so there must be 12 people.

P(A)=0.43 given

Given: P(A) = 0.43      P(B) = 0.28       P(AB) = 0.17

Find the probability of:

a) P(A)  = 0.43                                  b) P(AB)=0.43+0.28-0.17

Look at the titles in the sticky notes on your right.

Try watching this.

I have only watched a part of it but it looks really good. 

This is just on the unit circle - if you understand the unit circle properly it makes many things in trig easier. 

some of this question appears to be missing. 

I suggest you ask a specific question but maybe this will help.