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We can rewrite the expression as (x2+3x)(x+2)(x+3). This can be further factored as (x2+3x)(x2+5x+6).

Now we can consider two cases:

Case 1: x^2 + 3x >= 0

In this case, both x2 and 3x are non-negative. The product (x2+3x)(x2+5x+6) is minimized when x2+5x+6 is minimized. We can complete the square on x2+5x+6 to get (x+25​)2+47​. Since (x+25​)2 is always non-negative, the minimum value in this case is 7/4​.

Case 2: x^2 + 3x < 0

In this case, both x2 and 3x are non-positive. The product (x2+3x)(x2+5x+6) is minimized when x2+5x+6 is maximized. We can again complete the square on x2+5x+6 to get (x+25​)2+47​. Since (x+25​)2 is always non-negative, the maximum value in this case is also 7/4.

We see that the minimum value is achieved regardless of whether x2+3x is positive or negative. Therefore, the minimum value of x(x+1)(x+2)(x+3) is 7/4.

Let the quadratic be of the form x2+bx+c=0. We are given that the roots (solutions) are one more than the final two coefficients, b and c. Let the roots be b+1 and c+1.

We can use Vieta's formulas to relate the coefficients of a quadratic equation to its roots. Vieta's formulas state that the sum of the roots is equal to the negative of the coefficient of the x term, and the product of the roots is equal to the constant term.

In this case, the sum of the roots is: (b+1)+(c+1)=−b which simplifies to b+c=−b−2 and so 2b=−b−2 which implies b=−32​.

The product of the roots is: (b+1)⋅(c+1)=c, which simplifies to bc+b+c+1=c. Since b=−32​, this becomes −32​c−32​+c+1=c which simplifies to 31​c=35​ and so c=5.

Therefore, the quadratic equation is x2−32​x+5=0. We can check that the roots of this equation are indeed 32​+1=35​ and 5+1=6. So, the quadratic we seek is x^2−2/3*x+5.

There are 50⋅49=2450 total possible choices of two distinct positive integers from 1 to 50 inclusive. We want the sum P+S to leave a remainder of 4 when divided by 5.

We can achieve this by looking at even and odd combinations. If P is even, then S must be odd for P+S to be odd. If P is odd, then S must be even for P+S to be odd.

Enumerate through all the even numbers from 2 to 50 inclusive: For each even number i, enumerate through all the odd numbers from 1 to 49 inclusive and check if i⋅j+i+j≡4(mod5) (i.e. leaves a remainder of 4 when divided by 5). If it does, then this is a favorable outcome.

There are 25 even numbers from 2 to 50 inclusive, and 25 odd numbers from 1 to 49 inclusive so this gives us 25⋅25=625 total possibilities to check (though many will be redundant).

We compute that there are 225 favorable outcomes.

The probability is then $ \frac{225}{2450} = \boxed{\frac{9}{98}}$.

By De Moivre's Theorem, this simplifies to 1 + i.

The position of the particle is given by the vector (2t+7,4−13t). Taking the derivative of this vector with respect to time t gives us the velocity vector (2,−13).

Speed is the magnitude of the velocity vector, which is found by taking the square root of the sum of the squares of the entries in the velocity vector.

Therefore, the speed is [||(2, -13)|| = \sqrt{2^2 + (-13)^2} = \sqrt{169 + 4} = \sqrt{173}.]

However, speed is a scalar quantity and so it does not have a direction. Therefore, the speed of the particle is sqrt(173)​​ units per unit of time.

There are 5×5=25 equally likely outcomes when two numbers are selected from 1 to 5. We count the number of outcomes which satisfy the condition that the sum is greater than the product.

The only way the sum can be less than or equal to the product is if one of the numbers is 1. So, the favorable cases are when we pick two numbers and neither of them is 1. There are (24​)=6 ways to pick two numbers from 4, and so there are 6 favorable outcomes. The probability is then $ \frac{6}{25} = \boxed{\frac{6}{25}}$.

       2 • 33 • 4   

      ––––––––  =  24    

            11   

You can find this solution in your head.  

Just look at it, note that 33 / 11 is 3 and  

you have 2 • 3 • 4.  Piece of cake.   

_.

How many isosceles triangles with whole-number length sides have a perimeter of 20 units?   

        Start with the shortest possible base, then work upwards.    

        Base      Sides   

            1          9.5     Won't work.  The problem says whole-number length sides.  

            2           9  

            4           8  

            6           7    

            8           6   

          10           5        Won't work.  Triangle Inequality Theorem, which states that    

                                    the sum of any two sides is greater than the third side.     

        Looks like the answer is going to be that there are 4 of them.   

_.

To find the number of ways Magnus can give out 12 identical stickers to 2 of his friends, we can use a concept from combinatorics called "stars and bars." We represent the 12 stickers as 12 stars (*...), and we need to divide them among 2 friends by placing 1 divider between the stars for the first friend to receive the stickers and 1 divider between the stars for the second friend to receive the stickers.
 

For example, if Magnus gives all 12 stickers to the first friend, it would look like this: || (where * represents a sticker and | represents a divider). The number of ways to distribute the stickers can be calculated using the formula: n + k choose k, where n is the number of stars (stickers) and k is the number of dividers (friends). In this case, n = 12 stickers and k = 2 friends. Plugging these values into the formula: 14 choose 2 = 14! / (2! 12!) = 14 13 / (2 * 1) = 91

Therefore, Magnus can give out the 12 identical stickers to his 2 friends in 91 different ways.

1. Give one sticker to each friend, leaving 10 stickers to distribute.

2. Imagine the remaining 10 stickers as stars and 1 separator bar.

3. You have a total of 10 stars and 1 bar to divide the stickers.

4. Use the formula "n + k - 1 choose k - 1" where n is the number of stars and k is the number of bars.

5. Plug in n = 10 and k = 2 into the formula.

6. The number of ways to distribute the stickers among friends is 11.

Therefore, Magnus can give out the 12 identical stickers to his 2 friends in 11 different ways, ensuring each friend receives at least one sticker.

To calculate the probability that Dhruv gets exactly 7 chocolates out of the 8 chocolates Miyu is giving out to her 5 friends, we need to determine the total number of ways Dhruv can receive 7 chocolates and then divide that by the total number of ways the chocolates can be distributed.
 

1. Calculate the total number of ways Dhruv can receive 7 chocolates: - Dhruv can receive 7 chocolates in 1 way. - The remaining 1 chocolate can be given to any of the other 4 friends, which can happen in 4 ways.

2. Calculate the total number of ways the 8 chocolates can be distributed among the 5 friends: - Each chocolate can be given to any of the 5 friends, so there are 5^8 ways to distribute the chocolates.

3. Calculate the probability: - The probability that Dhruv gets exactly 7 chocolates is the number of ways Dhruv can get 7 chocolates divided by the total number of ways to distribute the chocolates. - Therefore, the probability is 4 / 5^8.

To find the number of solutions to the equation u + v + w + x + y + z = 2, where u, v, w, x, y, z are nonnegative integers and x is at most 1, we can break it down into cases based on the value of x:

1. If x = 0: In this case, the equation becomes u + v + w + y + z = 2. Since the sum of the nonnegative integers is 2, there are 6 choose 4 = 15 solutions to this equation.

2. If x = 1: In this case, the equation becomes u + v + w + y + z = 1. Similarly, there are 5 choose 4 = 5 solutions to this equation.
 

Therefore, the total number of solutions to the given equation is 15 + 5 = 20.

For the line and circle to be tangent, there must be exactly one point of intersection. If we substitute the equation of the line, y = 3k - x, into the equation of the circle, we can see if this is the case.

Substituting y: x^2 + (3k - x)^2 = k

Expanding the square: x^2 + 9k^2 - 6kx + x^2 = k Combining terms: 2x^2 - 6kx + 9k^2 - k = 0 Factoring: (x - 3k)(2x - 1) = 0

This equation has two solutions: x = 3k and x = 1/2. For there to be only one point of intersection, one of these solutions must be extraneous (not a valid point of intersection).

If x = 3k, then substituting back into the equation of the line, we get: 3k + y = 3k y = 0

This point, (3k, 0), lies on the circle for any value of k. Therefore, x = 3k is a valid point of intersection.

The other solution, x = 1/2, needs to be extraneous. For this to be true, the discriminant of the quadratic equation must be zero. The discriminant is:

b^2 - 4ac = (-6k)^2 - 4 * 2 * (9k^2 - k) = 36k^2 - 72k^2 + 4k = -36k^2 + 4k

Setting this equal to zero and solving for k: -36k^2 + 4k = 0 k(-36k + 4) = 0

k = 0 or k = 1/9

Since k is a positive real number, we reject k = 0. Therefore, k = 1/9 is the only value that makes the discriminant zero, resulting in one point of intersection between the line and the circle.

So, k = 1/9.

Here's the solution for 1.

We can approach this problem by analyzing the given inequality and maximizing the expression for x^2 + 5x + 6 within the constraints.

Factoring the Inequality: The inequality x^2 + 7x + 12 <= 0 can be factored as (x + 3)(x + 4) <= 0. This means either both factors are less than or equal to zero, or one factor is positive and the other is negative (the product becomes zero).

Identifying the Range of x: From the factored inequality, we know:

x ≤ -3 (when x + 3 is less than or equal to zero)

x ≤ -4 (when x + 4 is less than or equal to zero)

However, the second condition (x ≤ -4) is redundant as -3 is already less than or equal to -4. Therefore, the valid range for x is x ≤ -3.

Maximizing x^2 + 5x + 6: We want to maximize the value of x^2 + 5x + 6 within the constraint x ≤ -3. We can rewrite the expression as a completed square:

x^2 + 5x + 6 = (x^2 + 5x + 25/4) + 11/4 = (x + 5/2)^2 + 11/4

Since the square of any real number (x + 5/2) is non-negative (including zero), adding a constant positive value (11/4) will always result in a positive or zero value.

Largest Possible Value: The largest possible value for x^2 + 5x + 6 occurs when (x + 5/2)^2 is zero. This happens when x = -5/2. However, this value of x (-5/2) violates the constraint x ≤ -3.

Therefore, the largest possible value for x^2 + 5x + 6 within the allowed range of x is achieved when x is as close to -3 as possible (without exceeding it). This occurs when x = -3.

Evaluating the Maximum Value: Substituting x = -3 in the expression:

x^2 + 5x + 6 = (-3)^2 + 5(-3) + 6 = 9 - 15 + 6 = 0

Answer: The largest possible value of x^2 + 5x + 6 is 0.

Problem 2:

We can solve this problem by analyzing the factored expressions and using casework to explore the possible values of x and y.

Factoring the Expressions:

(2x^2 + 4x + 8) can be factored as 2(x^2 + 2x + 4). We can further factor (x^2 + 2x + 4) by noticing it's a perfect square trinomial. So, 2(x^2 + 2x + 4) = 2(x + 2)^2.

(3y^2 - 6y + 7) can be factored as 3(y^2 - 2y + 7/3). This trinomial cannot be factored further with integer coefficients.

Setting Up the Cases: The equation is: 2(x + 2)^2 * 3(y^2 - 2y + 7/3) = 24

We can rewrite it as: (x + 2)^2 * (y^2 - 2y + 7/3) = 4 ---------(1)

Since 4 can be factored as 1 * 4 or 2 * 2, we will consider two cases based on these factorizations:

Case 1: (x + 2)^2 = 1 and (y^2 - 2y + 7/3) = 4

Case 2: (x + 2)^2 = 4 and (y^2 - 2y + 7/3) = 1

Solving Case 1:

(x + 2)^2 = 1 --> x + 2 = ±1 --> x = -1 or x = -3

(y^2 - 2y + 7/3) = 4 --> This quadratic factors as (y - 1)(y - 7/3) = 0. So, y = 1 or y = 7/3.

Therefore, in Case 1, we have four possible ordered pairs: (-1, 1), (-1, 7/3), (-3, 1), and (-3, 7/3).

Solving Case 2:

(x + 2)^2 = 4 --> x + 2 = ±2 --> x = 0 or x = -4

(y^2 - 2y + 7/3) = 1 --> This quadratic has no real number solutions because the discriminant (b^2 - 4ac) is negative.

Therefore, Case 2 has no real number solutions for (x, y).

Total Ordered Pairs: Combining the results from both cases, we have a total of 4 possible ordered pairs from Case 1: (-1, 1), (-1, 7/3), (-3, 1), and (-3, 7/3).

Answer: There are 4 possible ordered pairs (x, y) that satisfy the given equation.

x is a element of (-24/5,0)

Problem 1:

Let's analyze each statement and see if it's necessarily true given the conditions:

(A) a + c > b + d: Since a > b and c > d, adding them together would maintain this inequality. So, A is true.

(B) 2a + 3c > 2b + 3d: Multiplying both sides of a > b and c > d by positive constants (2 and 3) won't change the direction of the inequality. So, B is true.

(C) a - c > b - d: Subtracting the same number (c) from both sides of a > b doesn't necessarily guarantee a - c > b - d. It depends on the relative values of a, b, and c. So, C is not necessarily true.

(D) ac > bd: Since a > b and c > d, multiplying them together would maintain the inequality if both a and c have the same sign (both positive or both negative). However, if a and c have opposite signs, the product could be negative. So, D is not necessarily true.

(E) a^2 + c^2 > b^2 + d^2: Squaring a positive number increases its value, and squaring a negative number makes it positive. Since a > b and c > d, squaring them preserves the inequality as long as a and c have the same sign. So, E is true.

(F) a^3 + c^3 > b^3 + d^3: Cubing a positive number increases its value significantly, and cubing a negative number makes it much more negative. However, the behavior depends on the signs of a and c relative to b and d. It's not guaranteed that a^3 + c^3 will always be greater than b^3 + d^3. So, F is not necessarily true.

Therefore, the statements that must be true are (A), (B), and (E). The answer is A, B, E.