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The ratio of a circle's circumference to its diameter

c) calculate the volume of the solid obtained by rotating R about y-axis

The integration is made easier if we use the "shell" method

The volume is given  by :

      1

2pi ∫   x  *   (2x - x^2  - x) ] dx   =

     0

     1

2pi ∫ x  [x - x^2]  dx     =

     0

     1

2pi ∫  x^2  - x^3   dx   =

     0

                              1

2pi  [ x^3/3  - x^4/4]     =    2pi  [ 1/3  - 1/4]  =  2pi ( 1/12)  =  pi / 6    units ^3

                              0

90+32 = 122

If Mike ate 7 candy bars, of his 12 total. He'd still have 12 candy bars except 7 would be in his tummy.

Let us find the intersection points :

2x- x^2  = x

x = x^2    

x^2  - x  = 0

x( x - 1)  = 0      so   the intersection points are   x = 0   and x  = 1

a.  The area of the region is given by

1                                     1                                                   1

∫   2x - x ^2  - x  dx    =    ∫  x  - x^2   dx   =   [ x^2/2  - x^3/3]   =  1/2  - 1/3  =  1/6  units^2

0                                     0                                                   0

b) calculate the volume of the solid obtained by rotating R about X-axis

Using the "washer" method, this is given by  :

    1

pi ∫  [2x - x^2] ^2   -  [x]^2  dx    =

   0

   1

pi ∫  4x^2  - 4x^3 + x^4  - x^2  dx   =

   0

    1

pi  ∫  3x^2  - 4x^3 +  x^4   dx    =

    0

                                1

[ x^3   -  x^4 + x^5/5 ]      =   1 - 1  + 1/5   =     1/5  units ^3

                               0

Well you have all of the ingredients to substitute into the equation and solve for c

15.066 x 10^-11  =  1.674 x 10^-27    c^2

sqrt ( 15.066 x 10^-11/1.674x10^-27) = c = sqrt (9 x 10^16) = 3.00 x 10^8  m/s

2 1/8  x  1 1/6 x 3 =

17/8  x 7/6  x 3/1  =

(17x7x3)/(8x6x1) = 357/48= 119/16 = 7 7/16

I THINK you are asking what is  n^2 + n^2 + n^2    when n=3 ???       If that is what you are asking:

3^2 + 3^2 + 3^2 = 9 + 9 + 9 = 27

ANOTHER way    n^2 + n^2 + n^2 = 3 n^2 = 3 (3^2) = 3 (9) = 27

Just set the equation given to = 0 and solve for 't'   (when the equation = 0 the ball is on the ground)

0 = -16t^2 + 64t +80   (make things simpler...divide thru by 8)

0=-2t^2+8t+10              quadratic formula yields    5 sec     (or -1 which is invalid for a TIME)

Very nice, EP.....!!!!

Egads!  I did......   deleted it.   Thanx for keeping me honest, Chris !

EP.....I think you gave the calculation for the mean, not the median........

Just take the average of the two middle values.......for instance

1, 2, 3 ,4 ,5 ,6, 7 , 8

Median  =  [ 4 + 5] / 2  = 9/2  = 4.5

5a. For what positive integers n does $\(\left(x^2+\frac{1}{x}\right)^n\)$ have a nonzero constant term ??

This will occur  when n  = 3p   for p ≥ 1

5b. For the values of n that you found in part (a), what is that constant term?

The term that will procuce a constant  is given by :   C(3p,2p)(x^2)^p*(1/x)^(2p) 

Another option is to graph (plot) the x =2t^2-16t+35    and you will se that the graph never reaches zero..(i.e. the car never reaches the tree)....the low point (nadir) is at t=4 with a value of 3 (as found in the previous solution)..

SInce you need to use the quadratic formula    ( -b +- sqrt(b^2-4ac))  / 2a

For the first one a =1 b=-3 c=-10         (3 +- sqrt(9- (-40)  )/2(1)  = ( 3 +- sqrt49 )/2    = (3+-7) / 2 = x =  5 or -2

You can now do the other one!

(7x-7y) ^2 =

49x^2  -49xy -49xy +49y^2 =

49x^2 - 98xy +49y^2

In both problems, factor each one ...

x² - 3x - 10  =  0

(x - 5)(x + 2)  =  0

x - 5  =  0     or     x + 2  =  0

x  =  5          or     x  =  -2

2s² - 3x - 2  =  0

(2s + 1)(s - 2)  =  0

2s + 1  =  0     or     s - 2  =  0

2s  =  -1          or     s  =  2

s  =  -1/2         or     s  =  2

I hardly ever survive math class. It's hard to stay awake.

4a.Expand $\left(x^2+\frac{1}{x}\right)^3$. (Write the terms with higher degree first, so for example an x^2 term would come before x or 1/x.)

[ x^2  +  1/x] ^3   =

(x^2)^3  +  (3)(x^2)^2 (1/x)  + (3)(x^2)(1/x)^2   + (1/x)^3   =

x^6   +  3x^3   + 3    + (1/x)^3  =

x^6  + 3x^3  + (1/x)^3  + 3

4b. Find the constant term in the expansion of \($\Big(x^2+\frac{1}{x}\Big)^4$\)

There is no constant term  in this expansion.......

3.Find the constant term in the expansion of \[\Big(z - \frac{2}{\sqrt{z}}\Big)^9.\]

[ z  - 2/sqrt(z)]^ 9

The constant term  occurs   at    C(9,6) (z)^3 [2/ sqrt(z)]^6   =  84 z^3 [ 64/ z^3]  =

84 * 64   =

5376

2. Use the Binomial Theorem to write $(3+sqrt{5})^4$ in the form $a + bsqrt{5}$ for some positive integers a and b.

(3 + sqrt(5) ) ^4   =

3^4   +  (4)(3)^3 * sqrt(5)  + (6)(3)^2* [sqrt(5)]^2  +  (4)(3)(sqrt(5))^3  +  (sqrt(5))^4   = 

81  + 108sqrt (5)    +  54 *5   +  60sqrt(5)   +  25  =

81  +  54 *5  + 25 + 108sqrt(5)  + 60sqrt(5)  =

81  +  270  + 25   + 168sqrt(5)  =

376  + 168sqrt(5)