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The probability is 5/2^10, so a + b = 15.

Sum divisible by 3         =[2 + 5 + 4 + 1] =12 / 36 =1/3

Sum < 7                         =[1 + 2 + 3 + 4 + 5]=15 / 36 =5 / 12

Divisible by 3 AND < 7    =[2 + 5] = 7 / 36

Let x be how much they have aged when the sum of their ages is 40.

12+x+8+x=40

20+2x=40

2x=20

x=10

Therefore Percy will be 8+10=18 years old when the sum of their ages is 40.

1. The area of triangle BXA is 18*sqrt(2).

2. The area of triangle PQR is 42*sqrt(3).

Let the inner circle's radius be x. We are solving for x in terms of r.

So connect the centers of two adjacent circles with radii r and the center of the small circle with radius x.

This is a isoceles right triangle with sides r+x, r+x, and 2r. 

We know that through the Pythagorean Theorem, the ratio of a side to the hypotenuse is \(1:\sqrt 2\).

So \(\sqrt 2(r+x)=2r\). Divide both sides by \(\sqrt 2\) and rationalize to get \(r+x=\sqrt 2 r\). Minus both sides by r and factor out r to get our answer: \(x = (\sqrt 2-1)r\)

If you connect the radii of each circle to its tangent points with the other circles, you have a square with side lenth 2r.

The diagonal of this square is \(2r\sqrt{2}\) by the pythagorean theorem. 

The diagonal is r+small circle diameter+r = \(2r\sqrt{2}\)

Subtracting both sides by 2r, the diameter of the samll circle should be \(2r\sqrt{2}-2r\) which means the radius should be \(\frac{2r\sqrt{2}-2r}{2}=\sqrt{2}r-r=r(\sqrt{2}-1)\).

Now I see it!  Arc EPF works out to 236 degrees.

Use the relationship distance = rate * time!  The distance works out to 55 miles.

No problem! Glad it helped :)

Sorry that's wrong

Using casework, there are 28 + 44 = 72 ways.

I count 20 triangles.

If Axex makes 60% of his shots, then he misses 40% of his shots.

Assuming that each of his shots is independent from each of his other shots, then it is correct to

multiply the probabilities together:  0.40 x 0.40  =  0.16

If they consist of single digit integers, they must be in the set:  { -9, -8, ..., -1, 0, 1, ..., 8, 9 }

If there is only one mode and that mode is 2, there must be no other digit used more than the 2 is used.

To get the largest possible median, try:  2, 2, 6, 7, 8, 9  which has a median of 6.5.

3sin(x) + 5cos(x) = 5

Using the identity:  sin²(x) + cos²(x)  =  1:

                    3sin(x) + 5cos(x) = 5

  3sin(x) + 5·sqrt( 1 - sin²(x) )  =  5

                 5·sqrt( 1 - sin²(x) )  =  5 - 3sin(x)

           [ 5·sqrt( 1 - sin²(x) ) ]²  =  [ 5 - 3sin(x) ]² 

                    25( 1 - sin²(x) ) )  =  25 - 30sin(x) + 9sin²(x)

                         25 - 25sin²(x)  =  25 - 30sin(x) + 9sin²(x)

                              - 25sin²(x)  =  - 30sin(x) + 9sin²(x)

                                            0  =  -30sin(x) + 34sin²(x)

               34sin²(x) - 30sin(x)  =  0

             sin(x)[34sin(x) - 30 ]  =  0

Either  six(x)  =  0     or     34sin(x) - 30  =  0

                  x  =  0                    34sin(x)  =  30

                                                    sin(x)  =  30/34

                                                           x  =  sin^-1(x)

For all the solutions, you will need to add  2·n·pi  to each of the above solutions.

Oh thanks, this explanation helps a lot more.

By stars and bars, the number of ways is C(51,5) = 2349060.

1. The area of the triangle is 23/10.

2. The coordinates of B are (13/5,62/5).

By similar triangles, the side length is 32/7.

N =  72.

hi! so umm I had already found those answers before and those were incorrect... also how did you get 210 and is there a way to solve problem 2 without sin? sorry for not mentioning that sooner..

~ Wolf

could you plese help with the venn diagrapm. I am so dumb

The definition of a horizontal asymptote is a horizontal line to which the graph becomes closer and closer as x becomes a very large or very small number, but never actually reaches it. So it's a y value that's undefined. A vertical asymptote is the same thing, but an undefined value for x. 

a) Since the vertical asymptote is x=3, we can see that c = -3, since placing x=3 into \(\frac{ax+b}{x-3}\) is undefined. (The denominator of a fraction is undefined for 0). 

Let f(x) equal y. Now we can find a by solving this equation for x.

\(x=\frac{3y-b}{a-y}\). Using the same thought process as above, we find that a = -4

Now we can use our values c = -3 and a = -4 and plug them in:

\(y=\frac{-4x+b}{x-3}\)

All we have to do is find b. We still have on piece of information we haven't used: the point (1, 0). Plug these x and y values in. 

\(0=\frac{-4+b}{1-3}\)

Now we can solve for b and find that b = 4. 

So putting everything together, we have our final answer: \(f(x)=\frac{-4x+4}{x-3}\)

You can do the same thing for part b.