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Ok, thanks

Here's a hint: Complete the square.

Here's a hint: Complete the square.

1. Use trig to find height of ramp

2. Find the gravitational potential energy when the sled is at top of ramp

All of the potential energy will be converted to kinetic energy when it reaches the bottom of ramp.

3. Set potential energy equal to kinetic energy and solve for velocity

You now know how fast the sled is going when it reaches the rough horizontal surface

shortcut: acceleration on rough surface = g * coeficient of friction (you should figure out why if you are not familiar)

4. Divide initial speed bby acceleration on rough surface to find time on rough surface before the sled stops

5. Plug time, acceleration and initial velocity into kinetic formula and solve for d:    d = vt + (a/2)t^2

There is a trig identity    tan (a+b) =   (tan a + tanb )/(1- tan a tan b)

so this equals     tan (20 + 10) = tan30 = sin 30 / cos 30 =  1/2 / sqrt3/2 = 1/sqrt3 = sqrt3/3

To make the reading simpler, let  x  =  ( sqrt(3) + sqrt(2) )   --->   a  =  x^-3

and                                         let  y  =  ( sqrt(3) - sqrt(2) )   --->   b  =  y^-3

This means that  x · y  =  ( sqrt(3) + sqrt(2) ) · ( sqrt(3) - sqrt(2) )  =  3 - 2  =  1

      and  x³y³  =  ( xy )³  =  ( 1 ) =  1

1 / (a + 1)  =  1 / ( x^-3 + 1 )  =  x³ / ( 1 + x³ )                (multiplying both the numerator and denominator by x³)

1 / (b + 1)  =  1 / ( y^-3 + 1 )  =  y³ / ( 1 + y³ )                (multiplying both the numerator and denominator by y³)

1 / (a + 1)  +  1 / (b + 1)  =  x³ / ( 1 + x³ )   +   y³ / ( 1 + y³ ) 

using the common denominator of  ( 1 + x³ ) · ( 1 + y³ ) :

                                       =  { [ x³( 1 + y³ ) ] + [ y³( 1 + x³ ) ] }  /  [ ( 1 + x³ )( 1 + y³ ) ]

multiplying out both the numberator and denominator:

                                             [ x³ + x³y³ + y³ + x³y³ ] / [ 1 + x³ + y³ + x³y³ ]

since x³y³  =  1:                   [ x³ + y³ + 2 ] / [ x³ + y³ + 2 ]  =  1.

since   y = 2x+4    substitute that into the second equation for 'y'

3 ( 2x+4) -4x= 6

6x+12-4x = 6

2x +12 = 6

2x = -6

x = -3          then use this value in one of th equations to find 'y''

Volume of the cone =  1/3  pi r^2 h     r = 9   h = 14

Volume of a sphere = 4/3 pi r^3       you only need 1/2 of this  so   1/2  4/3 pi r^3 = 4/6 pi r^3      r = 9

add them together   

Rectangle QRTU has dimensions |-4| + 2 = 6 , times |-3| + 4 = 7, so area 42.

The triangle is 3 *7, so 21/2 = 10.5

42+10.5 = 52.5

it is a rectangle of area   6 x 7

  and a right triangle area = 1/2( 3 x 7)       add these two together   

Where was it posted? i cant find it

Posted before.  Look it up.

From right triangle on the left   height = 12

From right triangle on left the base 2 =   5 + 15  + 16

area = 1/2 (base1 + base 2) x height

        = 1/2 (15 +(15+5+16) x 12 = 306 units²

6 people must shake hands with 5 other people   = 30 handshakes

   BUT each hand shake involves TWO people    30/2 = 15 handshakes

Distance is 2d

rate x time = distance

36t1 = 48(3.5-t1)

t1 = 2 hrs         2 hrs at 36 km/hr = 72 km

Let thhe distance between A and B be \(d\).

Since distance/speed=time, \(\frac{d}{36}+\frac{d}{48}=\frac{7}{2}\).

Multiplying both sides by 144, we get \(4d+3d=504\)

\(7d=504\)

\(d=72\)

SAMBHRANT

The "word" must begin and end with the letter "A"

So, it could be just AA -- 1 "word"

But, it could be three letters long: 

     AxA where x represents any of the other 7 letters  -- 7 "words"

Or, it could be four letters long:

     AxxA with 7 choices for the first x and 6 choices for the second x  -- 7·6 = 42 "words"

Or, it could be five letters long:

     AxxxA with 7 choices for the first x and 6 choices for the second x and 5 choices for the third x

              --  7·6·5 =  210 "words"

and, you can take it from here.

You'll need to add up all the individual possibilites to get the total number of possibilities.

A very interesting question ...

The integers:   ..., -3, -2, -1, 0, 1, 2, 3, ...

If you arranged them as:  0, 1, -1, 2, -2, 3, -3, 4, -4, ...

and added them this way:  (0) + (1 + -1) + (2 + -2) + (3 + -3) + ...

       you would get:             0 + 0 + 0 + 0 + ...  =  0

but, if you added them this way:  (0 + 1) + (-1 + 2) + (-2 + 3) + (-3 + 4) + ...

       you would get:                       1 + 1 + 1 + 1 + 1 + ...  which tends to positive infinity

and, if you added them this way:  (0 + -1) + (1 + -2) + (2 + -3) + (3 + -4) + ...

       you would get:                        -1 + -1 + -1 + -1 + ...  which tends to negative infinity

So, there is no particular sum  --  no sum ...

You might want to try to rearrange the integers to get other "answers" ...

1. By power of a point, CQ = 14.

2. By power of a point, AB = 8.

3. By power of a point, ST*TU = SV*VW ==> 3*15 = SV*3  ==> SV = 15, so SW = 18.

For k = 4, all x = y is a solution, so this is the answer.

The volume of pyramid CFAH works out to 1/4*4*5*6 = 30.

By just looking at it:

2    and   8         in one year they will be     3   and   9      trying to find a mathematical solution.....

x = y^3    and

x+1 = (y+1)^2

 x+1 = y^2+2y+1

x = y^2+2y

y^3 = y^2 + 2y

y^3 - y^2-2y=0

y(y^2-y-2)=0

y(y-2)(y+1) = 0     y = 2   0 (throw out)  -1 (throw out)   

so y = 2   x = 8 

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