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1) Do you mean gravitational potential energy near the Earth?  If so the two factors are the objects mass and it's height above the Earth.

2) Increase an object's kinetic energy by increasing it's velocity or it's mass or both.

\(2(x-4)^2 + 8 = 2(x^2 - 8x+16)+8 = 2x^2-16x+32+8 = 2x^2-16x+40 \\ a=2,~b=-16,~c=40\\ 3ax^2+3bx+3c = 6x^2-48x+120\\ 6(x^2-8x+20) = 6((x-4)^2+4) = 6(x-4)^2 + 24 \\ h=4\\ \text{Having gone through all that it's pretty clear to see that this should be so }\\ \text{since multiplying all the constants by 3 just has the effect of scaling the parabola}\\ \text{it's location remains unchanged. The x coordinate of the vertex of the original parabola}\\ \text{was 4, and thus we expect that the scaled parabola's vertex also has an x coordinate of 4}\)

\(y^8-17y^4+16=0\\ (y^4-16)(y^4 -1) \\ (y^2-4)(y^2+4)(y^2-1)(y^2+1) = \\ (y-2)(y+2)(y+2i)(y-2i)(y+1)(y-1)(y+i)(y-i) \\ \text{and thus roots at }\\ y=\pm 2,~\pm 2i,~\pm 1,~\pm i\)

   4
{1, 2} | {2, 4} | | {3, 2} |(total: 3)
 {1, 2, 4} | {1, 3, 2} } | | {3, 1, 2} | | {3, 2, 4} |  {4, 1, 2} ... (total: 5)

{1, 3, 2, 4} | {1, 4, 3, 2} | | {3, 1, 2, 4} | | {3, 4, 1, 2} | | {4, 1, 3, 2} | ... (total: 5)

1 + 3 + 5 + 5=14 I get 14 integers divisible by 4

Flip the coin 10 times...let any 8 outcomes be heads  = C(10,8)  = 45  sequences

Flip the coin 10 times, let any 9 outcomes  be heads = C (10,9)  = 10  sequences

Let all 10 outcomes be heads  = 1  sequence

So.... 45 + 10 + 1  =  56 sequences

x  =  ounces of nuts

2x  ounces of raisins

x + 2x   = 24

3x  = 24    divid both sides by 3

x  = 8

2x  =  2(8)   =  16  ounces of raisins

Yep  !!!!

We can arrange the Spiderman comics in 6! ways. the Archie comic books in 5! ways and the Garfield comic books in 4! ways

And we can arrange these groupings in 3!  ways

So

( 6! * 5! * 4! ) * 3!   =   12,441,600 different orders   !!!

Here's the second one

A horizontal line  has "0" rise and an infinite run

So  the slope is

0 / run  =   0

Thanks EP...another way

[ "Old" area ] x scale factor^2  =

20 x  20.5 ^2   =

8405 in^2

Easy....both EP  and CU  are genuises   !!!!

\(\dfrac{vote_m}{votes_w}=\dfrac{125}{235}\\ \dfrac{vote_m}{3960+vote_m}=\dfrac{25}{47}\\ 47vote_m = 25vote_m + 99000\\ 22vote_m=99000 \\ vote_m = 4500\\ vote_w=8640\\ vote_t = 12960\)

Thanks! that makes sense but how do you reconize the type of problem so easiely? ugh I appulaud the smarts! 

82/4 = 20.5 scale factor, so the length is   5 x 20.5 = 102.5

Area = L x W  =   82 x 102.5  in^2

\(\text{Let's count the combination that have neither Penelope or Quentin first}\\ \text{there will be }\dbinom{20}{3}3! =6840 \text{ ways to assign the officers w/o P or Q}\\ \text{now if we have P we have Q so that just leaves }\dbinom{20}{1}=20 \text{ combinations}\\ \text{or }20\cdot 3!=120 \text{ different ways off assigning the officers using P and Q}\\ \text{this gets us a total of }6960 \text{ ways of assigning officers}\)

To solve this, we first have to find the biggest term in the sequence, this is obviously the square root of 400. Since we know that the sequence is increasing by 1 for each step, we can conclude that there are exactly 400 terms.

To get from A to C  we have this possible path

right up up

But we could select the  "ups" to be in any two of the three positions

So  C(3,2)  = 3 possible paths from A to C

To get from  C to B  we have this possible path

right  right right  right    up  up

But we could select the "ups" to occupy any two of the six positions

So C(6,2)  = 15 possible paths from C to B

So...the total possible paths from A to B  = 3 * 15  =  45

20^2 = 400         SO  there are 400 terms less than or equal to 20

thank you for verifying!

This can be solved using proportion then cross-multiplying. Let Jackson's shadow be X, and we are going to make a proportion with this. If you take the tree's length and divide it by the tree's shadow, it should be proportional to the quotient of Jackson's height and shadow. So we have the equation of 6/10.8=1.2/x, this we can easily solve by cross multiplying, and we get the answer of 2.16 meters.

i got 2.16, not sure if it is correct

I use AoPs, its a great way to prepare for math competitions

Yes, 60 is correct. We have 5 letters in total and the letter, "O" repeats twice, so 5!/2!=5*4*3=20*3=60 ways.

Total arrangements -  The number of arrangements where the ARE together

The can be together in  5  ways and for each of these they can be arranged in 2 ways

So

6!  -  2 * 5  =

6!  - 10

720 - 10  =

710  ways