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Hi, some things are missing from this question.Check over carefully.

\(\frac{6}{7}=0.\overline{857142}\)

The six digits 857142 repeat. We divide 100 by 6 to get 16 r 4. The 6 digits are repeated only 16 times. The remaining four are not. The fourth number in this group is 1.

Edit: Oops somehow I got a two.

Area = h  x  b       and h = 3b-3

(3b-3) * b = 60

3b^2 - 3b - 60 = 0     solve for b      the h = 3b-3     (factor or use quadratic fromula to find 'b')

x   x+ 2    x+4   

4x-   ( x+4)  = 2 (x+2)+6

3x - 4 = 2x+10

x = 14                                14   16   18

x = original amount

2x-100 = after 1st temple

(2)(2x-100) - 100   after second temple = 4x-200-100 = 4x-300

2(4x-300) -100 = 8x-600-100 = 8x-700 after third temple

2 (8x-700) - 100 = 0    after last temple

16x-1500 = 0

16x = 1500

x = 93.75

More correctly:

Expected value

probability of a 3 is .8     .8 x3 = 2.4

                         9 is .2     .2 x9 = 1.8        expected value   2.4 + 1.8 = 4.2

add a nine    prob of a 3 is now   8/11     8/11 x 3 = 2.1818

                     prob of a 9 is now   3/11     3/11 x 9 = 2.4545       expected value   2.1818+2.4545 = 4.63

Is the second quadratic \(x^2+px+q+(x+a)(2x+p)=0\)? Or something else? please clarify.

10.

There are 10 numbers between 1 to 100 that have an odd number of factors: the perfect squares. That is what the problem translates to.

You are very welcome!

:P

Because in these problems it is usual to consider rotations as identical.

So one of the elements can go anywhere. Then that becomes the 'beginning' of the loop

I think expected would be   (3 + 3+ 3+ 3+ 3+3 +3 +3  +9 +9 )/10= 4.2

  add a nine :    (3 + 3+ 3+ 3+ 3+3 +3 +3  +9 +9 +9)/11  = 4.63

    add two nines     = 5

       etc ..... 

They each need to have at least one object. There are \({6 \choose 2}=15\) different ways for this to happen. 

Now, we see the different cases for the remaining 4 objects.

Sri   Godfrey

4          0        = \({4 \choose 4}=1\)

0          4        = (same as above) = 1

3          1        = \({4 \choose 3}=4\)

1          3        = (same as above) = 4

2          2        = \({4 \choose 2}=6\)

Total Cases: 16

Now we multiply 15*16 = 240 ways

Why don't you multiply 144*7 at the end of the problem to consider the $N$

From the right triangle with hypotenuse 25 and base 15, we know the height of the trapezoid is 20, by the Pythagorean Theorem.

Another right triangle is with hypotenuse 52, height 20, and base x+15.

So (x+15)^2+400=52^2

This simplifies to x^2+30x-2079=0

Factorizing, we get (x-33)(x+63).

Since x=-63 makes the area negative, x=33. 

Plug x into your area equation to get your answer.

If you think about it, this makes sense.... John runs a meter faster than David does....that is why he wins the 100 meter in the first place.

    adding ONE meter to each of their distances will result in John running THAT meter faster than David too....and John wins again !

\(\sum_{0}^{44}\) n =990

\(\sum_{0}^{45}\) n = 1035          HE MUST HAVE COUNTED '10'  TWICE !

1000 x 2 = 2000

Sqrt(2000) =~45

[45 x 46] / 2 =1035

1035 - 1000 =35 - The number that was counted twice

( x² + y² - z² )² - 4x²y²  =  ( x² + y² - z² )² - ( 2xy )²  

  difference of squares  =  [ ( x² + y² - z² ) + ( 2xy) ] · [ ( x² + y² - z² ) - ( 2xy) ]

  rewrite                         =  [ x² + y² - z² +  2xy ] · [ x² + y² - z² -  2xy ]

  rewrite                         =  [ x² + 2xy + y² - z² ] · [ x² - 2xy + y² - z² ]

  rewrite                         =  [ ( x² + 2xy + y² )  -  ( z )² ] · [ ( x² - 2xy + y² )  -  ( z )² ]

  rewrite                         =  [ ( x + y )²  - ( z )² ] · [ ( x - y )² -  ( z )² ]

  difference of squares  =  [ x + y + z ] · [ x + y - z ] · [ x - y + z ] · [ x - y - z ]  

Let  'x' be the time john runs 100                               'x' is also the time david runs 99

   rate = 100/x                                                                   rate = 99/x

     for 101 meters                                                               for 100 m

        time = 101m/ (100/x) = 101/100 = 1.01 x                      time =  100 m / (99/x) = 100/99  x = 1.010101x

So John will STILL win the race !    (by a hair !)      because david's time is greater......

Hmmmm ...   isnt it just   30 sec/120 sec = 1/4 ?

This is the same as distributing 7 identical balls into 3 distinct boxes.

The formula used for this kind of distribution is: (P - 1) nCr (K - 1) =(7 - 1) nCr (3 - 1) =6C2 =15 ways.

The distance is the same   d = rate * time     it is the TIME that is different

d = 30 k/h * t1

d = 6k/h *   t2        and they are equal 

d/30  = t1    and   d/6 =t2       

Total distance is 2 d

2d     =  Average *(t1 +t2)

2d = average * (d/30 + d/6)

2d = average ( (d+5d)/30)

2d = average * (6d/30)

2d * 30/6d = average

2*30/6 = average  = 10 k/hr

Let ABCDEFGH be a rectangular prism, where EF = 4, EH = 5, and EA = 6. Find the volume of the pyramid CFAH.

All the sides (including the base) are congruent.

The lengths of the sides are:    ( ≈ )    6.403,     7.211,     and      7.81

Area of the base                A ≈ 21.6564 u²

Height of the pyramid         h ≈ 5.5456

The volume of the pyramid CFAH             V = 1/3 * Ah ≈ 40.033 u³   

a)  V     (x·x)⁴  =  (x²)⁴  =  x⁸          x⁴·x⁴  =  x⁸

b)  F

c)  F

d)  V     (x + y)⁰  =  1

Opposite rays occur on a line. From point X, Y will be in one direction and Z will be in the opposite direction.