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Yes that is correct.  Good Work!

Give yourself a point    

Thank you for you help! 

Basically what I did was I made all the inequalities.

I got

3n+2n+15>n+30 --------> n>15/2

3n+n+30> 2n+15 -------> n>15/4

2n+15+n+30>3n (no use)

I had to make a diagram for this next part. 

We know BC>AC because largest side is opposite largest angle, smallest side is opposite smallest angle. We also know AC>AB from similar reasoning. This means

n+30>2n+15 -------> n<15

2n+15>3n ------> n<15

There are 11 values of n that satisfy all of the inequalities

1) First draw it.

2) Then think about the size of the angles that are mentioned. They are all the same and they're easy to find.

3) Then use known ratios to work out WZ

4) then again to work out ZB

5) Then find the area.

Follow my instructions and if you are stuck then tell me what you have done I can help more.

Please no one answer over me!

A club with five men and six women wish to form a committee. The number of men must be between 1 and 3 (inclusive), and the number of women must be between 2 and 4 (inclusive). How many different committees can be formed?

Choose 1, 2 or 3 men from 5

How many ways to choose 1?

How many ways to choose 2?

How many ways to choose 3?

How many possible ways to choose the men altogether?

Choose 2,3,or 4 women from 6

Continue as I did for the men.

Multiply the two answers together.

Now you try :)     Ask questions if you want.

No one else answer over me thanks.

There are more than 4 values.

1) First draw up a triangle that looks possible.  

  All the angles are different so if you use a triangle that looks a bit like a 30,60, 90 degree triangle that will be good to work with.

2)  Label the 3 angles correctly according to their relative angle size

3)   Label the 3 sides 

4)   Now the smallest side is opposite the smallest angle etc so write 'smallest' and 'biggest' near the correct sides.

5) Now you know that smallest side < middle side      Solve the inequality

6) You also know that the middle side < longest side.   Solve the inequality.

7) So what do you know so far?

8) Now to be a triangle at all the sum of the 2 smallest sides must be longer than the long side.

   Form the inequality and solve it.

9) So what have you found.

If you have questions then ask.     You can take a photo of your pic on your phone and post it if need be. 

I am very happy to help more if necessary but i want to be convinced that you are trying to help yourself first.

Please no one else answer

A line segment is 8

Let the radius of a larger circle be   X

The radius of a smaller circle is:     Y = sqrt( X² - 4² ) 

The area of a larger circle    A₁ = X² * pi   

The area of a smaller circle    A₂ = Y² * pi      

The area of the green ring     A₃ = A₁ - A₂          

( I think that my dumb answer matches the question.) 

This question comes up regularly.

How about if all the people stand on the circumference of a circle.

Then you need to know how many lines must be drawn to join all the PAIRs  of people

The first person will have 39 people to shake hands with and then that person can be removed from the circle

The second person will have 38 people to shake hands with and then that person can be removed from the circle.

The third person has 37 people etc

So that is    39+38+37+ ........+1  = 39/2(2+38*1)= 780     (that is an AP)

OR

How many ways can I take pairs form the circle         40C2 = 780

I cannot be wrong twice .... OR can I ???

\(SA= 4 \pi r^2\)

Cros section through middle    \(A = \pi r^2\)

SA of a slice is  A + SA/8

SA of all 8 pieces = 8A+ SA

\(\text{Surface are of slices : surface are of whole orange}\\~\\ 8A+SA:SA\\~ 8\pi r^2+4\pi r^2:4\pi r^2\\ 12\pi r^2:4\pi r^2\\\)

And you can finish it.     

The number you want is = 219978 . Note: It was found by a computer code.

How many seats are there in the theatre?

What is the seating plan?

Are there any more restrictions?

Hi Loki,

To get from P to the midddle is a total of 5 steps, 3 up and 2 across

So that is     5!/(2!3!) ways 

  now do the same for the second bit and then multiply the two together.

Please let locki finish it him/herself

10000003 , 10000012 , 10000021 , 10000030 , 10000102 , 10000111 , 10000120 , 10000201 , 10000210 , 10000300 , 10001002 , 10001011 , 10001020 , 10001101 , 10001110 , 10001200 , 10002001 , 10002010 , 10002100 , 10003000 , 10010002 , 10010011 , 10010020 , 10010101 , 10010110 , 10010200 , 10011001 , 10011010 , 10011100 , 10012000 , 10020001 , 10020010 , 10020100 , 10021000 , 10030000 , 10100002 , 10100011 , 10100020 , 10100101 , 10100110 , 10100200 , 10101001 , 10101010 , 10101100 , 10102000 , 10110001 , 10110010 , 10110100 , 10111000 , 10120000 , 10200001 , 10200010 , 10200100 , 10201000 , 10210000 , 10300000 , 11000002 , 11000011 , 11000020 , 11000101 , 11000110 , 11000200 , 11001001 , 11001010 , 11001100 , 11002000 , 11010001 , 11010010 , 11010100 , 11011000 , 11020000 , 11100001 , 11100010 , 11100100 , 11101000 , 11110000 , 11200000 , 12000001 , 12000010 , 12000100 , 12001000 , 12010000 , 12100000 , 13000000 , 20000002 , 20000011 , 20000020 , 20000101 , 20000110 , 20000200 , 20001001 , 20001010 , 20001100 , 20002000 , 20010001 , 20010010 , 20010100 , 20011000 , 20020000 , 20100001 , 20100010 , 20100100 , 20101000 , 20110000 , 20200000 , 21000001 , 21000010 , 21000100 , 21001000 , 21010000 , 21100000 , 22000000 , 30000001 , 30000010 , 30000100 , 30001000 , 30010000 , 30100000 , 31000000 , 40000000 , Total =  120 such numbers.

sorry.. that's not right :(

hi guest!

the formula for the sum is \(\dfrac{a_1+a_n}{2}\cdot n\)

where n is the number of terms

there are 200-101=99. 99+1=100 terms.

so, the sum is \(\dfrac{101+200}{2}\cdot 100=\boxed{15050}\)

i hope this helped you!

:)

I worked out the question :). Thanks for our warning though I never asked for the question to be solved I just wanted to be guided in the right direction but its all good.

||a - b|| = 14/3.

The probability is 5/18.

Between 1 and 1000, number 7 appears 300 times if you count the repeated digits, such as 77, 777.....etc.

Between 1 and 1000, number 7 appears 271 times if you do not count the repeated digits.

The point that must be on the graph of y = h(x) is (9,12), so the answer is 9 + 12 = 21.

There are 6*24 = 144 paths from P to Q.

Two of the numbers {1, 2, 3, ..., 100} are chosen at random.  What is the probability that their sum is odd? 

To obtain an odd sum, requires one even number and one odd number. 

If you have two even numbers, or two odd numbers, the sum will be even. 

So pull one number.  It can be anything, because the second number will determine whether the sum is odd or even. 

Once you have your first number, if it's even you need to pull an odd number.  All fifty of them are left. 

                                                      if it's odd you need to pull an even number.  All fifty of them are left. 

So, there are 99 numbers left – remember, you've already pulled one out – and 50 of them are the one you want. 

Your chance of getting one of the numbers to make an odd sum is 50/99 

_.

Simplify \(\dfrac { \sqrt { 6 } -2\sqrt { 3 } }{ \sqrt { 6 } +2\sqrt { 3 } } \)

Hello Guest!

\(\dfrac { \sqrt { 6 } -2\sqrt { 3 } }{ \sqrt { 6 } +2\sqrt { 3 } } \times \dfrac { \sqrt { 6 } -2\sqrt { 3 } }{ \sqrt { 6 } -2\sqrt { 3 } }=\dfrac{6-4\cdot \sqrt{6\cdot 3}+4\cdot 3}{6-4\cdot 3}=\dfrac{18-12\cdot \sqrt{2}}{-6} \)

\(=-3+2\cdot \sqrt{2}\)

  !