PurpleWasp

(x ^ 3 + 8x ^ 2 + 21x + 18) can be factored into ( x + 2 )( x + 3 )( x + 3 ). We divide by ( x + 2 ), and they cancel. Expanding

( x + 3 ) ^ 2 gives us x ^ 2 + 6x + 9. A = 1, B = 6, and C = 9. D is the one value that x can't be. If x was -2, then we would be dividing by 0, which absolutely can't happen. So, D = -2. Adding them together, A + B + C + D = 1 + 6 + 9 + (-2) = 14.

The two youngest kids have four pieces each. That leaves 5 pieces to distribute to 5 kids. Each piece of candy can go to any of the 5 kids, so 5^5, or  5*5*5*5*5, or 3125 ways.

That's incorrect. Thanks, tho!

The new area of Square A is (2009 +x)^2, while the new area of Square B is (2009 - x)^2. The difference is \(\begin{align*} &(2009+x)^2-(2009-x)^2\\ &\qquad=(2009+x+2009-x)(2009+x-2009+x) \\ &\qquad=(2\cdot 2009)(2x) \end{align*}\)

For this to be at least as great as the area of a 2009 by 2009 square, \(2(2009)2(x)\geq 2009^2\Rightarrow x\geq \boxed{\frac{2009}{4}}.\)

Thanks so much!

Let a and b] be the roots of this polynomial. Therefore, a+b=13 and ab=4. Notice that a^2+ab+b^2. This means that the sum of the squares of the roots can be obtained by subtracting the term containing the product of a and b, so a^2+b^2=169-2(4)=161.

Expanding the desired expression, we get \((2a-3)(4b-6)=8ab-12a-12b+18=8ab-12(a+b)+18\). This implies that we need the sum and product of the roots of the given equation, which are \(10/2=5\) and \(5/2\), respectively. Thus, the desired expression equals \(\)\(\left(8\cdot \frac{5}{2}\right) - (12 \cdot 5) + 18 = \boxed{-22}\).

You have a missing number in your question, but I think that I can still solve it. If all the members are equally likely to be elected as President, then it doesn't matter if you are a freshman or sophomore. So, your missing number is that there are ___ times more sophomores than freshmen. If your missing number was 2, then there are twice as many sophomores then freshmen. So, the fraction of the team that is sophomores is 2/3. Does this help? If you can give me the missing number, then I can solve it more completely.

From \(x^2-mx+n=0\), we get k+t=m and kt=m. Since n is prime, one of k and t is n and the other is 1. K>t, so k=n and t=1. Then m=n+1. m is also prime, so we have two consecutive integers that are prime. Since one of every two consecutive integers is even, and the only even prime is 2, we must have n=2 and m=3. Therefore, \(m^n+n^m+k^t+t^k= 3^2+2^3+2^1+1^2=9+8+2+1=\boxed{20}\).

One way to deal with the restriction is to solve the problem as if it didn't exist and then subtract the solutions we can't have.

There are 6 children and 6 seats. The 1st child has 6 choices. For each of those 6 options, the 2nd child has 5 choices, because the 1st child occupies a chair. For each of those 5 options, the 3rd child has 4 choices, and so on. So, there are 6*5*4*3*2*1 options. One way to see this is to draw a counting tree. 6*5*4*3*2*1 is more commonly known has 6 factorial, or 6!. 6! = 720, which is the number of ways you can sit the children without restrictions. However, no two siblings can sit in front of their sibling. This removes 3 options, so the answer is 720 - 3 = \(\boxed{717}\).

The youngest kids both get 4 pieces, so we subtract eight pieces from the total and continue, ignoring this. There are 5 kids and 5 pieces. Each piece of candy can go to each kid and has 5 choices. If each piece has 5 options, then there are 5*5*5*5*5, or 5^5, ways to distribute the candy. 5^5 = \(\boxed{3125}\), which is the answer.

B) We can start the same way that we solved A). However, now we are dealing with the columns, and we have an extra counter. So, each column needs 3 counters. Instead of counting the number of ways we can place the counters, let's instead count the ways we can pick the empty square. We can pick each square in the column to be empty, so there are 4 ways. 4 ways to do this in each column, which is 4*4=16. So, the answer is \(\boxed{16}\).

A) The square has four columns and four rows. Each row and column has 4 squares. In each row, we have four squares to place the counters in each row, or 4 choose 2, which is \(\frac{(4\cdot3\cdot2\cdot1)}{(2\cdot1)}\). This equals 12. So, there are 12 ways to place 2 counters in each row. However, there are 4 rows, so we multiply by 4. 12 times 4 is \(\boxed{48}\).

0.82-0.03=0.79

6, 8, 10 is the only one that I found.

This is a system of equations problem. You can either solve it with elimination or substitution.

Let c = candy bar and b = bubble gum.

Elimination:

c + b = $1.50, and c - b = $1

Add the equations

c + b + c + (- b) = $1.50 + $1

2c = $2.50

c = $1.25

$1.25 + b = $1.50

b = $0.25

Substitution:

c + b = $1.50, and c - b = $1

c = $1.50 - b, so $1.50 - b - b = $1

$1.50 - 2b = $1

- 2b = - $0.50

b = $0.25

Since we don't actually have to solve for the candy bar, we can stop.