CPhill

u=-2i+j+5k

v=4i-3j-3k

The area   =    ll  u x v ll

u x v  =   det   [  i   j   k 

                         2  1  5

                         4  -3 -3 ]      write the first two columns again and take the det

[  i    j    k     i    j

  2   1    5    2  1

  4   -3  -3    4  -3 ]    

[  -3i + 20j  -6k ] - [ 4k -15i - 6j ] = 

[12i + 26j - 10k ]

And the area =  sqrt [ 12^2 + 26^2 + ( -10)^2]  = sqrt (920)  = sqrt (4 * 230) =  2sqrt (230)

f(x) = 6x^7 + 7x^6 

f'(x)  = 42x^6 + 42x^5      set to  0

42x^6 + 42x^5   = 0     factor

42x^5 ( x  + 1)  = 0 

Setting both factors to 0, we have that x = 0   and x = -1

When x = 0,  f(x) = 0

When x = -1, f(x) = 1

So.......the points are    (0, 0)  and ( -1, 1)   →     " b"

Let    (k, m, n), (r, s, t), and (u, v, w)   = (5,-1,2), (7,-4-2), (2,-6,3)

Then....the area can be calculated thusly :

(1/2)absolute value of    det  [ 1     1      1  

                                              r-k  s-m   t-n

                                              u-k  v-m  w-n ]  

(1/2) abs  det  [   1   1   1 

                           2  -3  -4

                          -3  -5  1 ]             write the first two columns again and calculate the determinant

(1/2)  1  1   1   1    1

          2 -3  -4  2   -3

         -3  -5  1 -3   -5

(1/2) abs (  [  1 * -3 *1  +  1* -4* -3 + 1*2*-5] - [ -3*-3*1 +  -5*-4*1 + 1*2*1]  )   = 

(1/2) abs  (  [ -3    + 12   -10 ]  - [ 9 + 20 + 2 ]  ) =

(1/2) abs(-32) =

(1/2 (32) = 

16 sq units

 x+y=0

 8x - 7y = 75

Rearranging the first equation produces  x = - y

Subbing this for x in the second equation, we have

8(-y) -  7y = 75

-8y- 7y  = 75

-15y  = 75    divide both sides by -15

y = -5

And x = -y   so x   = -(-5)  = 5

So  {x, y}   = {5, -5}

4. Given that xy = 3/2 and both x and y are nonnegative real numbers, find the minimum value of 10x+3y/5.

If xy = 3/2  then y = 3/[2x]

So we have

10x + (3/5)[3/(2x)]

10x + 9 / [10x]

10x + (9/10)x^-1      take the derivative  and set to 0 

10  - (9/10)x^-2   =  0       multiply through by x^2

10x^2 - (9/10)  = 0

10x^2  = 9/10     multiply throgh by 10

100x^2 = 9       divide though by 100

x^2  = 9/100     taking the positive root, x = 3/10

And taking the second derivative procuduces a positve result, so this is a minimum

And xy = 3/2.....so........3/10 * y   = 3/2  →   y /10 = 1/2  → y = 5

So {x, y} = { 3/10, 5}     produces a minimum ....where x,y > 0

V_sphere =  (4/3) * pi * (radius)^3   =  (4/3) *pi * (46)^3 ≈ 407,720.08 units^3

2. If t is a real number, what is the maximum possible value of the expression -t^2 + 8t -4?

This is similar to (1).....this parabola turns downward.......so the t value that produces the max is:

-8 / [2(-1)] =   -8 / -2  =    4

Putting this value back into the function we have the max value of

-(4)^2 + 8(4) - 4 =   -16 + 32 - 4 =    12

1. What real value of t produces the smallest value of the quadratic t^2 -9t - 36

Since this is  an upwards-turning parabola, the "t" that produces the smallest possible value  is given by :

-b/ [2a] =  -[9] / [2 (1) ] =   9/2 = 4.5

I'm not too sure about this, but I don't believe any such numbers exist......here's my reasoning :

Let the 4 digits  be a , b, c and d    

And assume that each digit, except a, can only have a possible range of 0 - 9, inclusive  [a only ranges from 1 - 9, inclusive]

So we have  that

a + b + c + d   = 9    

Rearrange this as   (a + c) + ( b + d)  = 9      (1)

And it can be shown that if  the alternating sums of a 4 digit number equal some multiple of 11 (or, 0)...then that number is divisible by 11

For instance   1331.....we have  1 - 3 + 3 - 1  =  -2 + 2 =  0     and 1331  =  11^3

Or, for instance,  9141    .....we have   9 -1 + 4 - 1  =  8 + 3  = 11   and 9141  = 831 * 11

So.... our 4 digit number is divisible by 11  if    ( a - b) + (c - d ) = 11n    where n is an integer

Rearranging this, we have     (a + c) - (b + d)  = 11n        (2)

Adding (1)  and (2)   we have that

2(a + c)  =  11n + 9 

Now......the left side is even, so n must be odd...now, let n = 1.....then

2(a + c)  = 11(1) + 9

2 (a + c)  = 20

a + c  = 10

But this means that, by  (1),  10 + ( b + d ) = 9  → ( b + d) = -1.....but, since b, d can only range from 0-9, this is impossible

And we can clearly see that if n is any odd integer > 1, then a + c > 10  which by (1), makes (b + d) negative. And this violates our assumption that b, d can only range for 0-9, inclusive.

And n cannot be any negative odd, either....for instance, if n = -1

2(a + c)  = 11(-1) + 9

2(a + c)  = -2

a + c = -1    and this is also impossible, by our assumption

And for any negative odd > 1, (a + c) < -1 which also violates our assumption

Thus, no "n" exists that makes (1) and (2) both true, so there is no 4 digit number that will sum to 9 and also be divisible by 11

P.S. - I'd like some other mathematician(s) to look at this.......!!!!

The sum is given by:

S =   2 * [1 - 2^8 ]  / [ 1 - 2]  =   510