EnchantedLava68

Using the given information, we can write 4 equations.  Let Allan's paper clip count = a, Beth's = b, Charis' = c, and Dollah's = d

\(a=\frac{1}{2}(b+c+d)\)

\(b=\frac{1}{5}(a+c+d)\)

\(c=\frac{2}{7}(a+b+d)\)

\(d=145\)

Substituting d  as 145 into each of the 3 equations, we have 

\(a=\frac{1}{2}(b+c+145)\)

\(b=\frac{1}{5}(a+c+145)\)

\(c=\frac{2}{7}(a+b+145)\)

From here, it is simply substitution, elimination, to solve for each of the variables.  In the end, we have 

(a, b, c, d) = (174, 87, 116, 145), so Allan+Beth = 174+87 = 261

Let x = the number of stickers Ken had at first.  

Thus, using the word question's information, we have the number of stickers Ken has after giving stickers to his brother as x-111, and the number of stickers Ken has after using the remaining stickers on his birthday card as  5/8(x-111).  This expression equals 1/9 of the original, so we have \(\frac{5}{8}(x-111)=\frac{1}{9}x\).  

Solving, we have x = 135 original stickers.  

The equation for a circle is \((x-a)^2+(y-b)^2=r^2\), where (a, b) is the center, and r is the radius.  

Plugging in (x, y) = (-17, -2) into the above equation, we have \((-17-a)^2+(-2-b)^2=r^2\)

Plugging in (x, y) = (0, 15) into the above equation, we have \(a^2+(15-b)^2=r^2\)

Plugging in (x, y) = (9, -2) into the above equation, we have\( (9-a)^2+(-2-b)^2=r^2\).  

In this, we have 3 variables (a, b, and r), and 3 equations, and hence can solve for each of them.  A bit of algebra later,

a = -4, b = 2, and r = \(\sqrt{185}\).  

Substituting these values into the equation for the circle, 

The equation becomes \(\left(x+4\right)^{2}+\left(y-2\right)^{2}=185\)

If interest is compounded twice a year, it is compounded every 6 months.  Thus, after 12 months, interest will have compounded twice.  \(1.1^2*10,000\) = \(1.21*10,000 = $12,100\)

The maximum difference is 18, so the answer will naturally be between 1 and 18.  

It is impossible to represent an odd number difference, thus the differences must always be even.  Numbers are 2, 4, 6, 8, 10, 12, 14, 16, 18, for a total of 9 possibilities.  

Proof that you cannot get an odd difference with only odd numbers.  

If a is an odd number then it can be written as 2x+1, where x is an integer.  

If b is an odd number, then it can be written as 2y+1, where y is an integer.  

a-b (odd minus odd) = 2x-1-(2y+1) = 2x-2y.  Factoring, we have 2(x-y).  Because both x and y are integers, 2(x-y) will always be even, never odd.  

I believe that the answer is 18.  

Counting is a difficult mathematical topic for me, so make sure to check my work.  

With repeats being tricky to count, I honestly think it is easier to simply calculate it manually, as opposed to with complicated counting tricks.  

1.  Repeats A twice.  

There are 3 combinations, revolving around where to put the B

BAA

ABA

AAB

2.  Repeats B twice.  

3 combinations, revolving around where to put the A

ABB

BAB

BBA

3.  Does not repeat A or B.  

Because it is impossible to repeat a C, or a D, repeats are nonexistent here.  

First, we place the A, then B, resulting in 6 total combinations.  

After that, we have one space left for either a C or a D, making the total for this case as 6*2 or 12.  

ABc  ABd

BAc  BAd

AcB  AdB

BcA  BdA

cAB  dAB

cBA   dBA

In total, there are 12+3+3 combinations or 18 total.  

For this response, I will assume that abc = -27 instead of 27, as it is impossible for 3 negative numbers to multiply to a positive number

The minimum value is 64

Expanding (1-a)(1-b)(1-c), we have −abc+ab+ac+bc−a−b−c+1

Thus, to find what the minimum value for the desired expression is, we have to calculate the minimum of -a-b-c, and the minimum of ab+ac+bc.  

Since a, b, and c are negative, we cannot use the AM-GM inequality for 3 variables on it, but after multiplying each of the variables by -1, making them positive, we have \(\frac{-a-b-c}{3}≥ \sqrt[3]{-abc}\), simplifying to \(-(a+b+c)≥9\), so the minimum value of -(a+b+c) is 9.  

Since a, b, c are negative, ab, ca, and bc must all be positive, hence allowing for an easy AM-GM application.  \(\frac{ab+ac+bc}{3}≥\sqrt[3]{abc^2}\), simplifying to \(ab+ac+bc≥ 27\).  Both -(a+b+c) and ab+ac+bc reach the minimum value when the equality condition of the AM-GM inequality is met, namely, when -a=-b=-c, or when a=b=c, and when ab=ac=bc.  Both of these simplify to a=b=c.  Since abc=-27, a=b=c=-3.  Thus, the minimum value of −abc+ab+ac+bc−a−b−c+1 occurs when a=b=c=-3, rendering the minimum value of (1-a)(1-b)(1-c) as 64, as you assumed.  

Hopefully this answered your question, have a great day!

After reviewing my work, the next day after this question I found my mistake was not multiplying b and c by 3.  This was because f(x) had roots of -3, -3, -3, but it was multiplied by 3, hence giving -b/3=-9, and c/=27

From this, b=27 and c=81, so f(1) = 3+27+81+81 = 162+30=192

\(4x^6+7x^4−x^3−6x−1\)

The most fundamental definition of probability is \(\frac{cases that work}{total cases}\).  Keep this in mind while you work, it is the most important aspect of every probability question.