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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}}$.

We can find the values of b and c using the relationship between the quadratic formula and the roots of the equation.

Roots and Quadratic Formula:

The quadratic formula relates the coefficients (a, b, and c) of the quadratic equation to its roots (r1 and r2):

x = (-b ± sqrt(b^2 - 4ac)) / (2a)

In this case, we know the roots (5 + 3i and 5 - 3i) and want to find b + c.

Properties of Complex Roots:

Complex numbers as roots of quadratic equations always come in conjugate pairs. This means that if one root is a + bi (where i is the imaginary unit), the other root will be a - bi.

In this case, our roots are 5 + 3i and 5 - 3i, which confirms they are complex conjugates.

Sum of Roots and Coefficients:

There's a useful relationship between the roots (r1 and r2) of a quadratic equation and its coefficients (a, b, and c):

Sum of roots (r1 + r2) = -b / a

Product of roots (r1 * r2) = c / a

Since we're looking for b + c, and the quadratic has a leading coefficient of 1 (a = 1), we can use these relationships directly.

Finding b + c:

Sum of Roots:

The sum of the roots (5 + 3i) and (5 - 3i) is:

(5 + 5) + (3i - 3i) = 10

Since the sum of roots is also -b / a (and a = 1), we have:

-b = 10

Therefore, b = -10

Product of Roots:

The product of the roots (5 + 3i) and (5 - 3i) is:

(5 + 3i) * (5 - 3i) = 25 + 25 = 50

Since the product of roots is also c / a (and a = 1), we have:

c = 50

b + c:

Finally, add b and c to find the desired value:

b + c = -10 + 50 = 40

Therefore, b + c = 40.

We can solve for x and y in this system of equations by using substitution or the method of completing the square. Here, we'll use substitution:

Solve one equation for one variable:

From the first equation, we can express one variable in terms of the other:

x = 10 - y (Equation 1)

Substitute into the second equation:

Substitute this expression for x in the second equation:

(10 - y)^2 + y^2 = 64

Expand and solve for y:

Expand the squared term:

100 - 20y + y^2 + y^2 = 64 Combine like terms: 2y^2 - 20y + 36 = 0 Factor the equation: 2(y^2 - 10y + 18) = 0

This factors further as: 2(y - 6)(y - 3) = 0

Therefore, the possible values for y are:

y = 6 or y = 3

Substitute y back to find x:

Substitute each value of y back into Equation (1) to find the corresponding value of x:

If y = 6:

x = 10 - 6 = 4

If y = 3:

x = 10 - 3 = 7

Solutions:

Therefore, the ordered pairs (x, y) that satisfy the system of equations are:

(x, y) = (4, 6)

(x, y) = (7, 3)