CPhill

10.   If &x  is defined as &x = x + 5    and  #x  is defined as #x = x^2       what  is # (&4)

&4   = &x  =  x + 5  =   4 + 5  =  9

And

# (&4)  =  # ( 9) =  # x =  x^2  =  9^2    =  81

Thanks guest....I was assuming that the integers were unique....but....if not.....your answer is correct

4. The red parabola shown is the graph of the equation  x = ay^2  + by + c   find c

Since the point    ( -2,0)  is on the graph, we have that

So

-2  = a(0)^2 + b(0)  + c

-2  =  0   +  0   + c

-2  = c 

3. The product of three even consecutive positive integers is twenty times their sum. What is the sum of the three integers?

Let  the integers be     (n - 2), n (n + 2)

So

n (n - 2) (n + 2)  =  20 [ (n - 2) + n + (n + 2) ]

n (n^2 - 4)  =  20 [ 3n ]

n^3 - 4n  = 60n

n^3   - 60n - 4n  = 0

n^3  - 64n  =  0

n ( n^2 - 64)  = 0

Either n  =  0  (reject)   or   n^2 - 64  = 0  ⇒    n  = 8

So   the three integers  are    6, 8 10

The product of these = 480

The sum is 24

And 20 times this sum  = 480

2.  The largest product obtainable  is    (1998/2 + 1) (1998/2 - 1)  =  (1000)(998) = 998000

Since there are no sign changes to  p(x).....there are no positive roots

Since  p(-x)  has 3 sign changes, there are either 3  negative roots  or 1 negative root

Since the bounds  are   -6  and 1, he only  needs to test

- 1/4, -1/2 , -1, -5/4, -2, -5/2, -4, -5

Here's a few

1.  2x^2 + 6x - 14  =  0   divide through by  2

x^2  + 3x  - 7  = 0

The sum of the roots  =  a + b  = -3/1  = -3

The product of the roots =  ab  = (-7)/1  = -7

And we have

(2a - 3) (4b - 6) =  12ab -12b - 12a + 24  =   12ab - 12 (a + b) + 24   =  12(-7) - 12(-3) + 24 =

-84 + 36 + 24  =

-84 + 60  =

-24

Even degree polynomials can cross the x axis 

See the following :  https://www.desmos.com/calculator/fkd1kpcoui

The function that you have   ( x^2 + 3 - 1  =  x^2 + 2)   won't cross the x axis because it is the basic graph of y = x^2  shifted up 2 units.....so.....this will never cross the  x axis

First one

2* 2 * 2 * 3 * 5    =   8 * 3 * 5   =  24 * 5   = 120

Sum

2,2,2,3,5

2 + 2 + 2 + 3 + 5   =   14

Second one

2^7 * 3^2 * 5^3  =  144000

Sum 

2,2,2,2,2,2,2,3,3,5,5,5

2 + 2 + 2 + 2 + 2 + 2 + 2 + 3 + 3 + 5 + 5 + 5  =  35

Even though the answers don't differ much....heureka's is correct

The first answer  would be the angular wirth of a chord of 0.6 units if this chord were in a circle with a radius of 31.9 units......however.....this chord is not quite 31.9 units from the center of the circle....so the angular width is just a little larger than the true measure

We can verify this by using the Law of Cosines

arccos [ (.6^2  - 2(31.9^2))/ ( -2*31.9^2) ] = θ  ≈ 1.077679450357°

Heureka's  answer moves this chord outside the circle, but still tangent to the circle.....and it will be exactly 31.9 units from the  circle's center