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I just didn't realize it... but heres the gist of my way thats basically the same as CPHILL I didn't get it at first because instead of 3^2 I thought it was 32...

The prime factorization of 144000 is2^7*3^2*5^3 , so the minimum sum in this case is 35 .

Thanks Melody and Cphill and mathhemath and waffles!

Widgets produced on Monday  =  widgets per hr   * hrs worked  = wt   =  (2t) * t  = 2t^2 

Widgets produced on Tuesday  =  (w + 4) *  (t -2)  =  wt + 4t - 2w - 8  = 2t^2  + 4t - 2(2t) - 8  =  2t^2  - 8

So  he produces 8 more widgets on Monday than Tuesday

Oh thank you.

 A price for a good or service that is equal to the cost of production.

really?

63 + 6 x 27 = ?

thanks for the help

A laissez-faire policy - basically "leave alone" - is one that does not interfere with business. 

Therefore the only answer that fits is B.  Anything that contains government control does not.

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c  = √ [a^2 + b^2]     square both sides

c^2  =  a^2 + b^2     subtract b^2 from both sides

c^2 - b^2  =  a^2       take the positive root of both sides

√[c^2 - b^2 ] = √a^2

√[c^2 - b^2 ]  = a

Note  that   at x = -1,  y = 0   so  x = -1  is a root

And note that when x = 0, y  =3    and when x = 1, y = -1....so....the graph must cross the x axis between x = 0 and x = 1.....so...we have another root  on this interval

Finally.....at x = 1, y =-1  and at x = 2, y =4   ...so.....the graph must again cross the  x axis ......so...we have another root

Lastly  at x = x, y= 4  and at x =3, y = -1...so....the graph crosses the x axis one last time...so....we have one more root

So.... since the function crosses the x axis 4 times....we have 4 roots...so....the degree is 4

We have  one negative zero   at x  = -1

And the other three roots are positive

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 

2)

1998 / 2 = 999

999^2 =998,001

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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

Thx for the halp!! 

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

ohhh thank you !

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

Your account has not been deleted because I still have all your old messages :)

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

Hi DS it is really good to see you here again!

There might be a problem but I think it is more likely that you have forgotten your password.

Maybe it is different from what you remember???