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Find all A and B such that (2x + 7)/(x^2 - 8x + 15) = A/(x - 3) + B/(x - 5) for all x besides 3 and 5.

Express your answer as an ordered pair in the form (A, B).

\( \frac{(2x + 7)}{(x^2 - 8x + 15) }\\ = \frac{(2x + 7)}{(x-3)(x-5) }\\~\\ \frac{2x + 7}{(x-3)(x-5) }=\frac{A}{x-3}+\frac{B}{x-5}\\ \frac{2x + 7}{(x-3)(x-5) }=\frac{A(x-5)}{(x-3)(x-5) }+\frac{B(x-3)}{(x-3)(x-5) }\\ \)

You can finish it.    

Well....of course......the combined area of the  six small circles is just  6 * pi (radius)^2  = 6 pi (3)^2  = 54pi

Note that  if   we  connect  the centers  of  the top two circles  and we  connect the center of the  large circle with both of these centers,, we  will  have an isosceles triangle  with one side of 6 , two sides of x  and an included angle  between these unknown sides of 60°

Using the Law of Cosines, we can find  x  thusly......

6^2  = 2x^2  -  2x^2  (cos 60)

36  = 2x^2 - 2x^2 (1/2)

36  = x^2

x  = 6

So.....the radius of the larger circle  = x + 3  =   6 + 3   =   9

So  the area  inside rhe larger circle  but  outside the smaller circles = 

pi  [ 9^2  - 54]  =

pi [ 81 - 54 ]  =

27 pi  units^2

Hint: You cannot divide by 0.    

Sorry for impatience

Call the point  where  BT  intersects the circle, A

And  we have  that

Measure of angle RTB  =  (1/2)  ( measure of minor arc RB - measure of minor arc SA)

But angle SOT  = measure of angle  SOA

And since SOA  is a  central angle, it has  the same measure as minor arc SA

And.....since ROB is also a central angle, it has the same measure as minor arc RB

Call the meaure of  minor arc SA =  angle SOT  = angle SOA  =  M

And call the measure of minor arc RB  = angle ROB =  4M

So  we have this equation

28  = (1/2) (4M - M)

56  = 3M

56/3  =  M

So  ,minor arc  SA = (56/3)°

And minor  arc RB   =  4 (56/3)  = (224/3)°

And  minor arc RS  =   180  - 56/3 - 224/3   = (260/3)° ≈  86.66°

I like your answer too guest.

It was good that you left it not quite finished too.

I like that Eternity bought my attention to your solution too.  

I do not think of similar figure solutions as often as I should.  It is good to get reminders of them.

\(\frac{2}{3}\div 4 = \frac{2}{3}\cdot \frac{1}{4} = \frac{2}{12} = \frac{1}{6}\)

You're right. It isn't correct. 

2/12

Call  the common difference between the angles,  D

And call the measure of the  smallest angle, A

And we  know  that the sum of the interior angles =  (5-2) (180)  = 3 *180  = 540°

So  we  have  that

A + (A + D )  + (A + 2D)  + (A + 3D) + (A + 4D)  = 540

5A + 10D  = 540    (1)

And we know that    (A + 4D)  - A  = 48

So, simplifying.......4D  = 48

D = 12    (2)

Sub  (2)  into (1)  and we have that

5A  + 10(12)  = 540

5A  + 120  = 540

5A  =  540 -120

5A  = 420

A = 420/5  =  84°

I just took a lengthy break.....I'm still hanging in here  ...LOL!!!!

Here's my  best  attempt :

Call  the vertices  (A,B,C,D,E,F,G,H)

The  number of possible  combos of choosing any two vertices  (without regard to order)  =  8C2  = 28

Note  that  the  vertices chosen - again disregarding order - where there  is only one  vertex bt\etween them  are :

ABC , BCD, CDE, DEF, EFG, FGH, GHA, HAB   =  

AC, BD, CE, DF, EG, FH , GA, HB

So....the probability =  8/28 =   2/7

that isn't correct

Let the number of 3-point questions = Q

Then, the number of 4-point questions =29 - Q

3Q  +  4[29 - Q] = 100, solve for Q

3Q  +  116 - 4Q   = 100

-Q  = 100 - 116

Q = 16 - Number of 3-point questions

29 - 16 = 13 - Number of 4-point questions.

Hey, GM......good to see you  !!!

Here's  the first one

Mean  age of the first  3 teachers   =  (30 + 40 + 50)/3 =  40

Mean age  of 4 teachers ( 30 + 40 + 50 + 39) / 4  = 39.75 

Variance of  three teachers

[(50  - 40)^2  + (40 -40)^2  + (30 -40)^2 ]  =  [10^2  + 0  + (-10)^2]    = 200

Variance  with 4th teacher

[ (50 - 39.75)^2  + (40 -39.75)^2  + (39 - 39.75)^2  + (30 - 39.75)^2 ]   =

[ 10.25^2  + .25^2  + (-.75)^2  + ( -9.75)^2  ]  = 200.75

So.....the mean  decreases and  variance  increases

@AboveGuest Very clever method... I didn't even think of using similar triangles!

Expanding on the solution, $\sqrt80$ can be simplified to $4\sqrt{5}$, so the number that goes in the green box is 4.

I believe the problem didn't take into account that the time zones are different, since if it did, then the problem would probably have told us which time zones to follow and what time zones the times it gave us were in.

And who downvoted the above post? This was an amazing attempt by a user, and if it was the wrong answer, at least congratulate them on trying!

Note that this could be a "trick" question.....

Look at the Canadian Time  zones, here : https://www.timeanddate.com/time/zone/canada

Note that the time  difference between Edmonton  and St. Johns  =  3 and 1/2   hrs

Thus when it is   8:45 PM  in St Johns, it is actually  just  5:15 PM in Edmonton

So....he  has  actually been traveling from   1:20PM to  5:15 PM  as viewed in  terms of  Edmonton time 

So....he   actually  only travels  5  minutes less than 4 hours =   3 hrs 55 min

St. John's Nfld is 3:30 hours ahead of Edmonton, AB.

When the plane leaves Edmonton at 1:20 pm, it is already 4:50 pm in St. John's.

8:45 pm - 4:50 pm =3:55 hours - duration of the flight

y / 20 = 4 / y

y² = 80

y = √80

Note that   √(xy)  =  x^(1/2) * y^(1/2)

So  we have

x^(1/2) * y^(1/2)  =  1  + x^2y

Using implicit differentiation

(1/2)x^(-1/2)* y^(1/2)  + x^(1/2) * (1/2) y^(-1/2) y'  =  0  + 2xy  + x^2 y'  =

y^(1/2) / [ 2x^(1/2) ] +  (x^(1/2) / [2 y ^(1/2) ]) y'  = 2xy + x^2 y'

Gather the y' terms on the left and everything else on the right we have that

y'  [ x^(1/2)/ [ 2y^(1/2)]  - x^2 ]   =  2xy  - y^(1/2) / [ 2x^(1/2) ]

y' =  dy/dx =   (  2xy  - y^(1/2) / [ 2x^(1/2) ] )  /  [ x^(1/2)/ [ 2y^(1/2)]  - x^2 ]

Hello Guest!

You're pretty lucky. I just happened to be on :D

So, since they're both PM, we don't have to do any conversions.

We know that 7 hours have passed, and also 25 minutes from 1:20 to 8:45.

Converting the 7hours to minutes and adding 25, we get 445 minutes. Javier's flight took 445 minutes to complete.

I hope this helped! 

Oh okay... Guess I'm learning something new every day!