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Thank you so much!!!

78 cents      3 q  + 3 p      or    7 d   8 p

28 cents      1  q    3 p       or    2d    8 p

Need more info re: trajectory of the ball....

Let A = (4,-1), B = (6,2), and C = (-1,2). There exists a point X and a constant k such that for any point P:

PA^2 + PB^2 + PC^2 = 3PX^2 + k.

Find the constant k

Can I have a reason why?

If  b = -1/2  the both the x⁴ and x² terms cancel out leaving the x-term; the degree becomes 1.

Since a is a constant, the x⁴ term remains and there is no term with a larger exponent; the degree stays 4.

Is it better now?

Thank you!  

Thank you!

Thank you!

Try n = 120

Never mind, I found out how to do it, but it appears that your answer is wrong?

So I think we might use the hockey stick thereom thing...

it's C(n+k-1, k-1).

So if the twins each get 0, we have 6 candy, 6+(5-2)-1 = 8. 5-2-1=2. C(8,2).

We also get C(6,2), C(4,2), C(2,2), and c(0,2) with increasing values

 Adding these together we get 50.

p = q + 8     given

p = 5q         given

substitute into the first equation

5q = q + 8

4q = 8

q = 2             so p = 10        2 quarters + 10 pennies = 60 cents

and gcd(m,100) is a single digit number?

Sorry this is lagging.

Wait... I can't! Here is the real problem: How many integers m  are there such that 0<100

Oh yeah! You are right. I'm sorry... I will edit it.

Don't overthink this problem

What you should know about this rule is that the constant p will ALWAYS double no matter what m is. So, the percent increase is always 100%.

Write the two term on the left-hand side as one fraction, with denominator (x - 5):

     A / (x - 5) + B(x + 1)  =  A / (x - 5) + B(x + 1)(x - 5) / (x - 5)  =  [A + B(x + 1)(x - 5)] / (x - 5)

Multiplying out, and simplifying,  the numerator:

     [A + Bx² - 4Bx - 5B] / (x - 5)     =     [ Bx² - 4Bx +(A - 5B) ] / (x - 5)

Setting the numerator on the left side with the numerator on the right side:

     Bx² - 4Bx +(A - 5B)    =   -3x² + 12x + 22

Setting the two x-squared terms equal to each other:  Bx²  =  -3x²     --->     B  =  -3

Setting the two x terms equal to each other:  -4Bx  =  12x     --->     -4B  =  12     --->     B  =  -3

Setting the two constant terms equal to each other:  A - 5B  =  22     --->     A - 5(-3)  =  22

                                                                        --->     A + 15  =  22     --->     A  =  7

I think you can use "Ptolemy's Theorem" to solve this question!
{[14^2 + 6^2] + [10^2 + 6^2]} / 2 = 184
The length of the midway chord is:
Sqrt(184).

x = k y^3 /z^1/2         not sure how to answer this unless we assume   k y^3  is constant....which is what I'll do...

x = k / z^1/2 

3 = k /(12^1/2)    k = 10.3923

Now when z = 75

x = 10.3923 / (75^1/2) =  1.2

40 b  + 20 n = 130   and

8   b   + 4 n = 28         Multiply this equation by  - 5   then add it to the top one to solve for b, notebooks

                                           then substitute this value of 'b' into one of the equations to solve for 'n'

WAIT a minute !   THIS question makes no sense !     both of the quantities are reduced to 1/5th on the second day, but the amount is not decreased to 1/5th.....no answer .

I think that part of the problem has been omitted ...

a)  Colin's name can be chosen on the first try but not on the second:  (1/10) x (9/10)  =  9/100

     or his name can be chosen on the second try but not the first:         (9/10) x (1/10)  =  9/100

     or his name can be chosen on both tries:                                         (1/10) x (1/10)  =  1/100

     The answer is the result of adding these three values.

b)  Colin's name can be chosen on the first try but not on the second:  (1/10) x (9/9)  =  1/10

     or his name can be chosen on the second try but not the first:         (9/10) x (1/9)  =  1/10

     Again, the answer is the result of adding these two valus.