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All EVEN numbers: 0,  2,  4,  6,  8

If you choose 2 numbers out of the first 6 primes {2, 3, 5, 7, 11, 13}, with repetitions allowed, you should have this many sums: [6 + 2 - 1] C 2 =7 C 2 =21 sums. You should get the following sums:

(4, 5, 6, 7, 8, 9, 10, 10, 12, 13, 14, 14, 15, 16, 16, 18, 18, 20, 22, 24, 26)>Total = 21 Sums

(6, 9, 12, 15, 18, 18, 24) = 7 sums that are divisible by 3.

Therefore, the probability is: 7 / 21 = 1 / 3

Paul and Jesse each choose a number at random from the first six primes. What is the probability that the sum of the numbers they choose is divisible by 3?  

The first six primes are 2, 3, 5, 7, 11, 13  

I see nothing in the problem that would prohibit Paul and Jesse  

from choosing the same number, so we have the following possible 

2+2, 2+3, 2+5, 2+7, 2+11, 2+13 

3+3, 3+5, 3+7, 3+11, 3+13  

5+5, 5+7, 5+11, 5+13  

7+7, 7+11, 7+13  

11+11, 11+13  

13+13 

The pairs that total a multiple of 3 (marked in purple above) are 

2+7,  2+13,  3+3,  5+7,  5+13,  7+11,  and  11+13  

  7 pairs that total a multiple of 3  

21 pairs possible 

So the probability is 7 out of 21   

               7               1  

             –––    =     –––  

              21              3  

_.

Yes thank you for your help!

I'm mortified.  It's bad enough that I made a mistake, but inexcusable that I gave mistaken advice. 

10/99 is 0.10101010 not 0.01010101.  What's that ailment where you see things backwards? 

Thanks for catching my mistake. 

_.

 London is 5 hours ahead of Philadelphia. When Julia leaves Philly at 11 am, it is already 4 pm, London time. Since her flight will take 7 hours, that means she will arrive at 11 pm, London time. So, by the time she gets to the swimming event, it will be 11:30 pm, London time, and very late for the 7 pm swimming event.
Therefore, Julia will not be able to see Prince William and Kate Middleton.

First you have to figure out the time difference between Philadelphia, New York and Tokyo.

Both New York and Philadelphia are in the same time zone. Tokyo is 14 hours ahead of both Philadelphia and New York.

1 - When Ben leaves Philadelphia at 2 pm Friday, it is already 4 am, on Saturday morning in Tokyo. Since Ben's direct flight will take 14 hours, that means he will arrive at 6 pm, Saturday evening, Tokyo time.

2 - When his brother, Matt, leaves N.Y. at 12 pm, Friday, it is already 2 am, Saturday morning, Tokyo time. If it were a direct flight, Matt would arrive 12 hours later, or 2 pm, Saturday afternoon, Tokyo time. But because of "layover" of 3 hours in Vancouver, then Matt should arrive at 5 pm, Saturday evening, Tokyo time.

3 - Therefore, Matt will arrive 1 full hour ahead of Ben, or 5 pm, Saturday evening, and Ben will arrive an hour later, or 6 pm, Saturday evening, Tokyo time.

Hi Melody!

I think:
In the term \(f(f^{-1}(2010)\) is 2010 the argument of the \(f^{-1}\).

Then applies:

\(f^{-1}(x)=f^{-1}(2100)=\sqrt[3]{x}+1=\sqrt[3]{2010}+1\\ \color{blue}f^{-1}(2100)\in\{-11.6202,13.6202\}\)

These values of the function \(f^{-1}\)are two arguments for the function f.

Please confirm whether this is the right or wrong idea.

Grüße

5X/7=15

Multiply by 7 on both sides

5X = 105

Divide by 5 on both sides

X=21

You are very welcome!

:P

I think they cancel each other out and the answer is just 2010

Looking at it a little more formally.

I think it means

\(2010=(x+1)^3\\ \sqrt[3]{2010}-1=x\\ f(\sqrt[3]{2010}-1)=(\sqrt[3]{2010}-1+1)^3=2010\)

Idk... But the answer I got was this

If Laura departs from 9 AM and her flight is 8 hours long... then
9AM + 8hrs = 5PM on Friday, at Rome
If it takes her a maximum of one hour to retrieve her bags and go through customs in Rome, it'll be 6 PM in advance of Siena...
If Laura proceeds to go to Siena, then it'll take her approximately 2 hours to take the train. It is more likely to tell Brittany to expect to see her by 8 PM on Friday.

Alright i iwll leave this one for now, not sure about it...

12 C 2 = 66 ways of subtracting one element from another. If you do that, this is what you will get:

(1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 8, 8, 8, 8, 9, 9, 9, 10, 10, 11)>Total = 66

Just count all the elements 3 and above to get your "sparse subsets".

Answer is 64

If \(f(x)=x^3+3x^2+3x+1 \), find \(f(f^{-1}(2010)) \)

Hello Guest!

\(f(x)=x^3+3x^2+3x+1 \)

\(f(x)=y=x^3+3x^2+3x+1\\ f(x)=y=(x-1)^3\\ f^{-1}(x)=x=(y-1)^3\\ f^{-1}(x)=y=\sqrt[3]{x}+1\)

\(f^{-1}(2010)=\sqrt[3]{2010}+1=\pm 12.6202+1\\ f^{-1}(2010)\in \{13.6302,-11.6202\}\)

\(f(f^{-1}(2010))=f(13.6302, -11.6202) \)

\((x-1)^3=(13.6302-1)^3=2010\)

\((x-1)^3=(-11.6302-1)^3=-2010\)

\(f(f^{-1}(2010))\in \{-2010,2010\}\)

  !

This is impossible.....the point  (3,2)  is  on this line.......

0.1111111111111 = 1 / 9

0.0101010101  = 1 / 99

[1 /9  +  1 /99] = 12 / 99 = 4 / 33

This is an interesting problem......

Let R be the radius of the larger circle  and r  the radius of the smaller circle

Note  that  F = (0, R)   and   H = (R,0)

FH =   sqrt ( R^2 + R^2 )  =  R*sqrt (2)

Likewise BC = r*sqrt (2)

Midpoint of  FH  =  (R/2  R/2)

Distance  from O  to the midpoint =  R/sqrt (2)

Midpoint of BC =   (r/2, r,2) 

Distance  from O to this midpoint  = r/sqrt (2)

The  quadrilateral is a  trapezoid  with  height =  (R + r) / sqrt (2)

And  the sum of its bases =  sqrt (2) (R + r)

Area of  quadrilateral =   (1/2) (R + r)  / sqrt (2)  *  ( sqrt (2) (R + r) )  =  (1/2) ( R + r)^2

Area  between the circles  =  pi (R^2  - r^2)

Area of quadrilateral / area between circles =   2/3

So

(1/2) (R + r)^2             2

___________  =       ____

pi (R^2 - r^2)               3

(1/2) (R + r) (R + r)           2

_______________  =      ___

pi ( R + r) ( R - r)               3

(1/2) (R + r)                2

_________  =         ________      cross-multiply

pi (R - r)                      3

(3/2) ( R + r)  = 2pi ( R - r)

(3/2)R  + (3/2)r  =  2pi R  - 2pi r

( 2pi + 3/2  ) r =  (2pi - 3/2) R

r / R  =  (2pi - 3/2)  / (  2pi + 3/2 ) ≈  .6145

We don't have  a right triangle here, so the Pythagorean Theorem is no good to us

OA =   sqrt ( 5^2 + 2^2)  = sqrt  (29)

OB  = sqrt ( 6^2  + 8^2)  = sqrt (100)  = 10

AB = sqrt [ ( 6 - 5)^2  + (-8-2)^2  ]  =  sqrt [ 1^2  + 10^2 ]  = sqrt (101)

Using the Law of Cosines , we have

[ sqrt (101)]^2  - 10^2  - [sqrt (29)]^2

______________________________   = cos (AOB)

  -2 (10)  sqrt (29)

101 - 100 - 29

____________  = cos (AOB)

-20sqrt (29)

-28

_________  =  cos (AOB)

-20sqrt (29)

7 / ( 5 sqrt (29) )  = cos (AOB) 

arccos  [  (7/(5sqrt (29)  ] ≈  74.9°

[ AOB ]  =  (1/2) (10) (sqrt (29)_  sin ( 74.9°)      =    26 units^2

I don't think you need to know where the plane started from, given that you know that it's an 8 hour flight.

No....Laura is flying to Siena.....

She lands in Rome and takes a train  to Siena

We need  to  know where she boards the plane  that flies to Rome....

The line will have  the  form  y = 3x  + b

In standard form we have   3x - y  + b = 0

Using the  formula for the distance  from a point to a line  we can solve for b thusly  :

l  3(0) - 0  + b l

______________    =   4

sqrt ( 3^2 + 1^2)

   b

________   =   4

sqrt (10)

b = 4 sqrt (10) =    the height of the triangle

So  the eqation of the  line is

y = 3x  +  4sqrt (10)

Letting   y = 0, we can find  the base of the triangle  as

3x = -4sqrt (10)

x =    -(4/3)sqrt (10)

The absolute value of this is the base length

So....area of the triangle  is

(1/2)  (base) (height)  =

(1/2) (4/3)sqrt (10) (4 sqrt (10)  = 

(1/2) ( 4/3) ( 4) ( 10)  =

80/3  units^2