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# math homework

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1. Let a,b,c, and d be real numbers such that a > b and c > d. Enter the letters of the statements below that must be true, separated by commas.(A) a + c > b + d.  (B) 2a + 3c > 2b + 3d. (C) a - c > b - d.  (D) ac > bd.  (E) a^2 + c^2 > b^2 + d^2.  (F) a^3 + c^3 > b^3 + d^3

2. Find the number of ordered pairs (x,y) of real numbers such that (2x^2 + 4x + 8)(3y^2 - 6y + 7) = 24.

3. Let f(a) = (13+a)/(3a+7) where a is restricted to positive integers. What is the maximum value of f(a)?

May 25, 2024

#1
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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.

May 25, 2024
#2
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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.

May 25, 2024
#4
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For problem 3, the answer is 3/2.

May 26, 2024