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STEP1:Equation at the end of step 1

(((p2)-(9•(s2)))+6pq)+32q2

Equation at the end of step2:

(((p2) - 32s2) + 6pq) + 32q2

STEP3:

Final result :

p2 + 6pq - 9s2 + 9q2

Using  the  top  of  the clock  as   0°

The  minute  hand  moves  360°  in every hour......so in 45 min  = 3/4  hr  it  has  moved  360 * 3/4  = 270°  from the top of the clock

The  hour  hand  moves  30°    in one  hour

At  9,  the  hour  hand  is  at  270°....so in 3/4 of an hour  it  is  at   270 + (3/4)(30)   = 292.5°

So.....the  angle  between  them  is     292.5  -  270  =   22.5°

We can find  the slant   height of   the  side of the pyramid   as  sqrt  [ (base/ 2)^2  +  12^2 ]  =

sqrt  ( 5^2 + 12^2)   =   sqrt (169)  = 13

The  area of  the sides  = the  area of 4  congruent  triangles with  bases of  10 and slant  heights  = 13

So...the  area of  the sides =  4 (1/2) 10 * 13 =  260 cm^2

The  area of  the base is just  the area of  a square with a side of 10   = 10^2   =  100 cm^2

Total  surface area =   (260 + 100)  cm^2  =   360 cm^2

Triangle  AOB  is similar to triangle COD

Then   since  base  CD  =  6/5   of  base   AB,  then the  height  of  COD  = (6/5)  that of  triangle AOB

Then  there  11 equal parts  to  the height of the trapezoid  ...... and 11/2  = 5.5  equal parts on each side of the  midline

And  since the  height of AOB  contains 5  of  these......then  the height of triangle YOX  must = .5  of these

And triangles YOX  and COD  are similar

So

base  of  YOX  / base  of  COD     =   .5  / 6  =   1/12

So  XY  =(1/12) CD  =   30 / 12  =  5/2  in  =    2.5  in 

The probability is 2/15.

j  shifts  the   domain  of  f  to  the right  by  one unit...so.....the domain of j = [ 0, 2  ]

r^2  =    (2 - 0)^2  +  (5 - - 8)^2   =  4 + 13^2  =   4 + 169  =  173

(h, k)  =  the coordinates of  the  center  = ( 0 ,  -8)

So.....the equation  is

( x - 0)^2  +  ( y - - 8)^2  =  173       simplify

x^2  + (y + 8)^2   =  173

The  domain will include  all  x values  except    the ones  that   make  x^2  - 6x + 4   =   0

These two values, A and B,   will  be  the roots of this quadratic function.....by Vieta...... in the  form ax^2 +bx + c.....the  sum of the roots =    -b /a

So   we  have

-(-6) / 1 =    -6

So

A +  B  =   -6

The line  will only intersect  the parabola once  at  the  vertex  

To find  the  y coordinate  of  the  vertex  we  have    ( 3)  /(2 (-2))  =  -3/4

To  find  the x coordinate  we  have

-2(-3/4)^2 - 3(-3/4)  +  5      =  49/8

This  is   the "k" we  need

So....the line is    x = 49/8

See the graph here  : https://www.desmos.com/calculator/av4pagctlb

Rearrange  as

x^2 - 4x +  2x  +  y^2  - 6y  - 8y   ≤ 13      simplify

x^2  -  2x  +  y^2  - 14y  ≤  13       complete  the  square on x and y

x^2  - 2x + 1 + y^2  - 14y  + 49   ≤  13 + 1 + 49

(x - 1)^2   +  (y  - 7)^2  ≤  63

This is a circle centered at  ( 1, 7)    with a radius  of sqrt (63)

The inequality tells us that we  only  want to find  the area inside this circle   =  pi  (sqrt (63))^2  =   63 pi  units^2

A)

3x   +  y   =   13

-3x  - 18y  =  21

@cating

The problems from AoPS Mainly come from special classes, where the problems are not allowed to be shared. I don't think there is another way to access other problems either...

Probably the only thing we can actually do is to just watch out for problems during competition times.

f ( f(x) )   =

            1                                           1                                5x  - 2               5x  - 2

_________________  =            _____________  =   ___________   =   ________

5 (1 / (5x - 2)) - 2                          5  / (5x - 2) - 2          5 - 2 (5x - 2)          9  - 10x

The    x   not in the  domain   of this function    is  the  one  that makes   9  -10x  =  0

So....

9 -10x   = 0

9  = 10x

x =    9 /10

We notice the value under the square root must be positive to be in the domain. So we have $(x+3)^2-(x-8)^2\geq0$. We see that after simplifying, $22x-55\geq0$, where the minimum of $x$ is then $2.5$. Clearly, the smallest $x$ in the domain is $2.5$.

In triangle BCD, angle C = 90 degrees, CD = 3, and BD = sqrt(17) . What is tan B?

Hello Guest!

\(tan\ B=\dfrac{3}{\sqrt{17-3^2}}=\dfrac{3}{\sqrt{8}}=\dfrac{3\cdot \sqrt8}{8}=\color{blue}\dfrac{3\cdot \sqrt{2}}{4}=1.06066\)

  !

The function preserves 'parity' such that positive inputs give positive answers, and the converse for the negative. So therefore, such that f(f(x))=4 so a positive input gives 4. There are no such solutions possible. We let f(x)=5x+2 for positive inputs, so f(f(x))=25x+12, which we want to equal 4. However, then x has to be negative, which is not allowed, a contradiction.

Therefore, there are $\boxed{0}$ solutions to the equation.

It takes blue car 45 min to go 21 miles

21 m  / (3/4 hr) =   28 m/hr

When you state that adjacent cubes have different colors, then the layers must be like this:

First layer:

Second Layer

If we inverse the colours, then we can rotate the new cube 90 degrees to get the original cube we first drew.

So there is only $\boxed{1}$ possible cube.

1 m^2  = 10.7639   ft^2

 £12.99 * 1.28   =  $16.6272

Dollars / Sq Ft   =    16.6272  /  10.7639    =  $1.54  per sq ft

Cubic  yds =   (20)/3  *  (40/3) *  ( 5.5)  /3   =  162.9   cu yds  =  163  cu yds

So

163  *   12  =  $1956

Hi Guest!

It's disgusting how people cheat. 

I've seen way too many questions of competitions that I've done and are still running (there's usually a time frame for it to be completed). 

I try to avoid answering any questions when I know a competition I'm doing is happening since it'd be cheating if I had seen/solved the questions beforehand. 

However, I doubt there's any real method to prevent cheating other than reporting the person, and quite simply, the chance of someone recongnizing a question used for cheating before someone else answers it is quite low. 

Perhaps CPhill, Melody, Alan, Asinus, or any other moderator could contact websites like AOPS, AMC, or MathCounts to get a database of questions that are on tests that they can ban. But that'd be kinda hard to do. 

=^._.^=

I try not to answer AOPS  problems.....if I'm aware  that  they  ARE  AOPS  problems

Let  the speed  of  the two cars  be S   (mph)

So......the distance  the red car  goes in 4 hours  + 21 miles = the distance  the blue car goes in 4 + 3/4  hours  = 4.75 hours

So

4S  +  21  =  4.75 S        subtract  4S from each side

21  =  .75S       divide both sides  by  .75

S   =  28  (mph)

Another  way to see this is that the blue car  covers 21  miles in 3/4 hr

So   21/ (3/4)  =   84 / 3  = 28 mph

Let Z  be  the intersection point  of the diagonals

And  note  that triangle  XYZ  is similar to triangle RZS

So since  RS =  20  and XY  = 5, the  height of triangle RZS  is  4 times  that of  the height of  triangle XYZ

So....there are 5 equal parts  of the  distance  from SR to the midline

And one  of these parts is the  height of triangle XYZ

And 4 of the parts  = the height of triangle RZS

And  triangle  PZQ  is similar to triangle  RZS

And  there  are  also  5 of the same  equal parts from PQ  to the  midline

And so  there  are  6 equal  parts  to  the  height  of  triangle PZQ....the 5equal parts to  the midline and the one other equal part  of the height of triangle XYZ 

So.....the  height of  triangle  PZQ  to the height of triangle RZS =  6/4 = 3/2

So....the  base of  triangle PZQ = (3/2)  the  base of triangle RZS =  (3/2) 20  = 30  = PQ