All Answers 358291 Answers

110=abs((30*6) - (5.5*M)), solve for M

M=140/11 =12.7272....minutes after 6 pm when he left home.

M =580/11 =52.7272...minutes after 6 pm when he came back.

52.7272 - 12.7272 =40 minutes that he was away.

https://web2.0calc.com/questions/mathcounts-amc8

That isn't correct.

Actually, 1210 does work as an automatic string, it's one 0, two 1's, one 2, and zero 3's. So 1210 is also an automatic string

The ratio is sqrt(2)/4.

20000/99=202 numbers are divisible by 99  

\( \sum_{1}^{202}\) 99n =2029797

Solve the system x/(x^2 + y^2) = 1 and y/(x^2 + y^2) = 2.

By the way, your first example it says if you insert 2352, then the machine prints out 0020. Which i'm pretty sure is incorrect. The machine will (I'm pretty sure) print out 0021, since there's a 3 as well.

There are 44 possible locations for the centroid.

Thanks so much!

Oh wow. Thanks dude!

You are mixing dimensions...

your rate is in mile per HOUR

your increase in time is in MINUTES.....convert 30 MINUTES to  1/2 hour and resolve for r

120/(r+8) + 1/2 = 120/r

This is an AoPS problem, isn't it? 

I got 9 as well. But it turns out the answer is 11:

 

This is the solution if you want to see their reasoning:

We can construct segments of length \( 1, \sqrt 2, \sqrt 5, \sqrt {10}, 2, 2\sqrt 2, \sqrt {13}, 3, 3\sqrt 2\)  (these can be obtained systematically by considering all lengths of segments from the top left dot to other dots). The following results can be obtained either by inspection or analysis of slope.

For each of \( 1, \sqrt 5, \sqrt {13}, 3\), there are zero isosceles triangles having it as its base length.
For \(\sqrt{10}\) there is one.
For each of \(2\sqrt 2, 3\sqrt 2\) there are two.
For each of \( 2, \sqrt 2\) there are three.
This gives a total of 11 non-congruent isosceles triangles.

Solve for real x: 

I once calculated the red area. I hope that was your question.

???

----------------------------

x = (x^2+y^2)

y = 2(x^2+y^2 )   from the first equation, this = 2x

Sub y = 2x in to the first equation

x = x^2 + 4x^2    re-arrange

5x^2-x = 0

x (5x-1) = 0      so x = 0 and  .2       sub these values into the equation (red ones) to find the corresponding 'y' values.....

126  135  144  153  162  216  225  234  243  252  261  315  324  333  342  351  414  423  432  441  513  522  531  612  621  Total =  25
 

25 out of a total of 6^3 =216

Therefore, the probability is: 25 / 216

This is a good one. I'll think about it later.

That's for the 2nd graders.  

cos A - sin A                      1 - sqrt (3)                

____________ =              ____________                       cross -  multiply    

cos A + sin A                      1 +  sqrt (3)

(cos A - sin A) ( 1 + sqrt (3))  =  (cos A + sinA) ( 1 - sqrt (3) )

cosA + sqrt (3)cos A  - sin A - sqrt (3)sin A    =   cos A - sqrt(3)cos A  + sin A  - sqrt (3) sin A

Simplify as

2sqrt(3) cos A  = 2sin A  

sqrt (3) cos A = sin A

sqrt (3) = sin A  / cos A

sqrt (3)  = tan A

arctan ( sqrt (3) ) =  A  =    60°

zvvvv

When we move this 5 units to the right, the function becomes

2y^2  +  (x - 5)^2   =  64

Subtracting the given function from this gives us

(x - 5)^2  = x^2      simplify

x^2 -10x + 25  =   x^2

-10x + 25

-10x = -25

x=  -25/-10   =   2.5   =  the x coordinate of the intersection 

Andusing either equation to find y, we have

2y^2 + (2.5)^2  =  64

2y^2  +   6.25  =  64     subtract    6.25 from both sides

2y^2  = 57.75     divide both sides by   2

y^2  = 28.875      take the positive root

y  = sqrt (28.875)

So   ( x , y)   =   (2.5, sqrt (28.875) )

See the graph here  :  https://www.desmos.com/calculator/3lkazxxzgq

tan  (BCE) = EB/ BC =  (1/2)S/ S  = (1/2)

So arctan (1/2)  = angle BCE  

EC =   sqrt  (S^2  + (1/2S)^2 ]  =  (S/2)sqrt (5)    =  [  sqrt (5) / 2 ] S

And angle EEF =  angle BCE  ≈ 26.565°   =   arctan (1/2)

So   

cos  (artan (1/2) )  =  (1/2S) / EF

2 / sqrt (5)  =  S/ (2EF)

2EF / S  =  sqrt (5) /2

EF = [ sqrt (5) / 4  ] S

So   tan ECF  =  EF/ EC  =     [ sqrt (5) / 4  ] S                 1

                                               _____________     =       ___  

                                                 [sqrt (5 )/ 2 ] S                  2

So  arctan (1/2) =   ECF

So  angle  BCF  =   angle BCE   +  angle  ECF  =    arctan (1/2)  +arctan (1/2)

And  angle  BCE  =  arctan (1/2)

So

Angle BCF                   [ arctan (1/2)  + arctan (1/2)  ]            2arctan (1/2)               2

__________  =           _________________________  =      ___________  =       ____

 Angle BCE                             arctan (1/2)                                arctan (1/2)               1