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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

Your Question for all:

You can place weights on both side of weighing balance and you need
to measure all weights between 1 and 1000.
For example if you have weights 1 and 3,now you can measure 1,3 and 4
like earlier case, and also you can measure 2,by placing 3 on one side
and 1 on the side which contain the substance to be weighed.
So question again is how many minimum weights and of
what denominations you need to measure all weights from 1kg to 1000kg.

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

Anything divided by "0" is equal to "0"

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. 

2x=4

2x/2=4/2

x= 2 <=================Answer

8x-(3x+10)=15

8x-3x-10=15

8x-3x=15+10

5x=25

5x/5=25/5

x=5  <============Answer

What diameter should you make a copper wire so its resistance is 0.015 $$\Omega$$ per meter of length ? 

The resistance R of a copper wire with the length l can be calculated with the following formula:

 

where

R = resistance (ohms, Ω)

ρ = resistivity (ohm meter, Ω m)  Resistivity Copper: 1.7 x 10^-8 Ω m

l = length of conductor (m)

A = cross-sectional area of conductor (m²)

d  = the nominal diameter of the wire

 

We have:

$$\frac{R}{l}= 0.015\ \frac{\Omega}{m}$$

$$\boxed{\dfrac{R}{l}= \dfrac{4\cdot \rho}{\pi \cdot d^2}} = 0.015\ \frac{\Omega}{m}\\\\
\dfrac{4\cdot \rho}{\pi \cdot d^2}} = 0.015\ \frac{\Omega}{m}\\\\
d^2 = \dfrac{4\cdot \rho}{\pi \cdot0.015\ \frac{\Omega}{m}}} \quad | \quad \rho_{copper} = 1.7 \cdot 10^{-8}\ \Omega\ m\\\\\\
d^2 = \dfrac{4\cdot 1.7 \cdot 10^{-8}\ \Omega\ m}{\pi \cdot0.015\ \frac{\Omega}{m}}} \quad | \quad \sqrt{} \\\\\\
d = 2 \cdot 10^{-4} \sqrt{ \dfrac{1.7}{\pi\cdot 0.015} }\ m \\\\
d = 0.00120125135\ \mathrm{m} \\
d = 1.2\ \mathrm{mm}$$

 

 

I assume that 1 quarter is equal to 1/4? Is it not?

1/4 = .25

Solution:

$${\frac{{\mathtt{2\,076}}}{{\mathtt{0.25}}}} = {\mathtt{8\,304}}$$

Yes. If you add 1 to 2+2.

2+2 = 4

4+1 = 5 <======= Answer. Simple. Is it not?

some of this question appears to be missing. 

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

When you want to find an angle. 

Hey anonymous bro! I love you, thanks! It helped me a lot! My teacher said he had never seen such good math from me! Thank you, my mum said I could have my PS4 now! :D

hi bro! such simple question, we all know it's cos(log(10)*4 )^-1. 

Hope this helped, 

I am a Math Teacher at University of Getting Tickets.

C = (5/9)(F - 32)

Five people A,B,C,D,E are in a group that regularly play squash.

a) In how many ways can 2 people be selected from 5?    

This is an unordered selection. So the number of ways will be      $$\frac{5*4}{2}=\frac{20}{2}=10$$

This is because you can choose any one of the 5 first then any one of the remaining 4 second. Hence 5*4=20 BUT AB is the same as BA you cannot count it twice (this will happen for any pair you choose) so you have to divide by 2

[If you know about combinations it is 5C2]

b) List the possible selections of 2 people from the 5 above.

AB, AC, AD, AE

BC, BD, BE,

CD, CE,

DE

c) How many times would you expect A to have played in 35 games, if the 2 players are chosen at random each time?

Well of the 10 possibilities above, 4 of them include A So the probability that A is included in any individual pair is

$$\frac{4}{10} = \frac{2}{5}$$

So in 35 games A would be expected to play approx     

$$\frac{2}{5}\times 35 = 14\;\;times$$.

Area = base times height

36 = B*3

Base =12 inches

Thanks Alan but you can say that the other way around too

Imagine the set of all even integers.  There are an infinite number of them.  Take away the infinite number of all integers and you are left with minus an  infinite number of odd integers.  So here is an example of where infinity minus infinity equals minus infinity.

I am not saying that this is correct only that the logic of you example can be countered.

Haha no, cause if you had (3*3)+(5*2), you have to solve the paranteses before you do the addition, and you cannot switch places in twobdifferent paranteses :) BAM!

and 3*3+5*2=9+10=19 

Well...there are many problems that have remained unsolved......so any one of them might be the "hardest."  

But....one of the most difficult to solve was a problem known as "Fermat's Last Theorem"...it states that no known unique integers a,b,c exist such that a^n + b^n = c^n for n≥ 3.

This problem took more than 350 years to solve. {At least one hefty sum of money was offered for its solution). The solution was finally found by an English mathematician, Andrew Wiles, in the mid-1990s. Oddly enough, he found it so frustrating that he almost gave up.....!!!

BTW.....Wiles' proof is over 150 pages in length.....!!! Definitely not "light reading"