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biggest common factor of   -8   12    -2      is   2

  biggest common factor of x^2   x^4   x^2   is x^2

   do similar for y and z

2x^2  y  z^3    ( -4y^3      +     6x^2   -     z^2 )       

radius 1  = 5         height = h

radius 2 =  9 m     height = 9/5 h

volume 1 = 12 tons/volume

1/3 pi 5^2 h = 12      remove the constants     25h = 12  tons     h = 12/25

1/3 pi 9^2 (9/5)h =                                          81(9/5)h = ?       81* 9/5 * 12/25 = 69.98=~~70 tons 

   (not positive of this answer!)

Since +2 shifts f(x) to the left by 2 units, shouldnt that be (a+2,b) instead of (a-2,b) which results in the solution of (a/3+2, 3b-2)? 

Edit: wait, it's supposed to be (a/3-2, 3b-2) because the -3 flips f(x) across the x axis right? 

Here s my take on it :

f( ? ) is defined only from  -8 thru 4     so the ? has to equate to this interval

f(2x-1)     must fit in the interval      2x-1 must not go outside of the interval

    so  2x-1 = - 8     means x = - 3.5

           2x-1 = 4     means   x = 2.5

so as far as I can say (I could be wrong)    the domain of g(x) = [-3.5,2.5]

By Hamilton's Theorem, the solutions are z = 4^{1/4}*e^(pi*i/4), 4^{1/4}*e^(pi*i/4 + pi/4), 4^{1/4}*e^(pi*i/4 + 2*pi/4), and 4^{1/4}*e^(pi*i/4 + 3*pi/4).  Since 4^{1/4} = sqrt(2) and e^(pi*i/4) = (1 + i)/sqrt(2), the first solution is 1 + i.  Then the other roots work out as

4^{1/4}*e^(pi*i/4 + pi/4) = 1 - i,

4^{1/4}*e^(pi*i/4 + 2*pi/4) = -1 - i, and

4^{1/4}*e^(pi*i/4 + 3*pi/4) = -1 + i.

If points (a,b) are on the graph of y = f(x), what points must be on the graph of y = -3f(3x+6)-2 = -3f(3(x+2))-2?  

The 2 in "x+2" horizontally shifts f(x) to the left by 2 units, so it becomes (a-2, b)

the 3 in  "3(x+2)" horizontally shrinks f(x) by a factor of 3, so it becomes ((a)/3-2, b)

the -3 in "-3f(3x+6)" vertically stretches f(x) by a factor of -3, so it becomes ((a)/3-2, -3b)

lastly, the -2 in "-3f(3x+6)-2" vertically shifts f(x) down by 2 units, so it becomes \(\displaystyle \boxed{\left(\frac{a}{3}-2, -3b-2\right)}\)

Area of ΔABD = AE [sqrt(AD² - AE²)]           (AD = AC)

I'll try...

Angle TQR = 180 - (160/8 * 4 +30)

   BUT four aces would be one card from each suit.....so I think you have to correct for that ...??

Geometric series   with r =2

   and 15 terms

11 22 44 88 16 16    32 32    64 64   .......

  2   4   8   16   32  

            a1 = 2        a15 = 2 * 2^{14 =} 32768 cents

Sum = a1 ( r^15 -1)/(r-1) = 65636 cents total     (656.36  dollars)

Try this.....   FIRST press and hold the Windows key   THEN press the   prt sc    key   .....the display should darken a bit and then re-brighten and the pic should be in your PICTURES / Screenshots folder       Pressing both keys simultaneously often does not work!!

Windows vista  ?     Windows 8  10  ?     Windows XP ?       The operating system.....

So 3/4 of the time, you win 2, and 1/4 of the time, you win -1. 

3/4*2 = 6/4

1/4 * -1 = -1/4

6/4 - 1/4 = 5/4

Hopefully that's the answer. :))

=^._.^=

wdym

U R Welcome !   

What operating system are you running on your dell laptop ?

Each draw has a 4 out of 52 chance of being an ace

4 / 52   *   4/52  *  4/52  *  4/52 = ____________

1 card from each suit: (1/4)^4

4 aces: (1/13)^4

(1/4)^4 + (1/13)^4

Note that angle  r  and angle  t   encompass the entire 360 degree circumference

  remember that inscribed angles inercept arcs 2 times the angle

so [  3x +5       +    5x +15 ] * 2 = 360      now you can solve for x    and can label three of the angles

Then ...all of the angles sum to 360 degrees

     can you solve for q  now ?

definitely, haha !

Prime factor of 105 -> 3*5*7

The number has 3, 5, 7, and 1

These are 4 distinct, so there are 4! ways

this is the only case b/c there is no multiplication of 3,5,7 that is 1 digit

The answer is 4!