Math Is Fun Forum

I think the required probability is

(32 x 28 x 24)/(52 x 51 x 50) assuming the raminaing two cards can be either paired or unpaired, greater or lesser than nine.

Given two points, the equation of the line joining the points is
(y-y1)/(y2-y1) = (x-x1)/(x2-x1)
Substituting y1 = 5, y2 = -1, x1 = 3 and x2 = -3, the equation of the line is
x-y+2 = 0,
The slope of this line m=(y2-y1)/(x2-x1), i.e., m = 1.
The slope of a line perpendicular to this line would be -1
(Because, if two lines are perpendicular, the product of their slopes is -1).
The midpoint of U(3,5) and V(-3, -1) is given by
[(x1+x2)/2, (y1+y2)/2]
Therefore, the midpoint of the line UV is
(0,2).
The equation of a line of given slope passing through a given point is
(y-y1) = m(x-x1)
We know, m=-1 for the perpendicular line.
Therefore, the equation of the perpendicular bisector is
(y-2) = -1(x-0)
or
x+y-2 = 0
Substitute the x and y coordinates of point W(2,-1) in this equation.
We see that it does not satisfy the equation.
Therefore, the point does not lie on the perpendicular bisector.

Mathsy, all your solutions are correct! You had understood problem # k + 16 rightly!

Step 1 : Do Operations within the Parentheses

3 - (6 + 8) x 3 = 3 - 14 x 3 = 3 - 42 = -39
The first operation to be done is always the one within the Parantheses.

Step 2 : Do Operations with Powers and Roots

3 + 2^5 = 3 + 32 = 35 (Here ^ represents powers)
3 + √ 16 = 3 ± 4 = 7 or -1.

Step 3 : Do all multiplication and division operations from left to right
Step 4 : Do all addition and subtraction operations from left to right

50 ÷ 10 ÷ 2 = (50 ÷ 10 ) ÷ 2 = 5 ÷ 2 = 2.5
If it is done from the other direction, you get a different (and wrong) answer.

Problem # k + 23

A Pythagorean triple is a set of three numbers (a,b,c) such that
a² + b² = c². For example, (3,4,5).
Prove that three prime numbers cannot form a Pthagorean triple.

And solved it correctly! Well done, Kylekatarn!!!

And the solution to problem # k + 2 is a much smaller number!
It is

= {[k²(k+2)+k(k+2)]+k+1}mod6
= {[(k²+k)(k+2)] +k+1}mod6
={[(k(k+1)(k+2) + k+1}mod6

We know that k(k+1)(k+2) is divisble by 6 for any k>1, k∈N,
Hence the above is reduced to
(k+1)mod6

It is seen that it is true for k+1, hence, it is true for any value of k.

q.e.d

Great....
It would take some time to be familiar with the codes

by_Faizah=m-(1/3m-4)/4-3 
This should be written as   {[m-(1/3m-4)]/4}-3

= {[m - 1/3m +4]/4}-3

= {[2/3m+4]/4}-3

= {[(2m+12)/3]/4}/ - 3

= {[2m+12]/12} - 3

= {[2m+12]-36}/12

= {2m - 24}/12

= m/6 - 2

Hi Audrey, welcome to the forum.
We would be glad to help you. You can post your problems in the 'Help me' section.
To begin with, in Algebra, variables such as x,y,z or a,b,c etc. are used to solve problems.
A monomial is an algebraic expression containing one variable, like 3x+8 or 4x²-5x+9 etc.
A Binomial contains two variables like 3x+8y, 4x²-7y²etc.
A polynomial contains more than two variables.
The degree of a polynomial is the highest power of any variable.
There are certain expansions to be remembered.
Like (a+b)², (a-b)² etc. You can find them in your text books.
Believe me, Algebra is both fun and easy.

Put n=7.
7mod6=1
7^3mod6=343mod6=1
Put n=8
8mod6=2
8^3mod6=512mod6=2.
Assume this is true for k.
Therefore, kmod6=k^3mod6.
Try for k+1.
Lets say (k+1)mod6=m
(k+1)^3mod6 = (k^3 + 3k^2+3k+1)mod6.
= (k^3+3k^2+2k+k+1)mod6= (k^3+2k^2+k^2+2k+k+1)mod6
= [k(k^2+2)+k(k+2)+k+1]mod6...
Running out of time...gotta leave....

Very good work, very nice pictures.
Just a thought....R and r can be illustrated more clearly,initially I mistook them for the external and difference between external and internal radii (we are always more comfortable with 2 dimensions ) .
New forumlae to remember ......
Surface Area = 4 × π²  × R × r
Volume = 2 × π² × R × r² 
The best part was the metamorphosis of a torus into a sphere