Question

1. if x is the smaller of two consecutive odd integers, and their product must be at least 63, then this relationship can be represented by the inequality

a. $x^2 + 2x - 63 \geq 0$
b. $x^2 + 2x - 63 \leq 0$
c. $(x - 7)(x - 9) \leq 63$
d. $(x - 7)(x - 9) \geq 63$

1. the number line that illustrates the solution to $x^2 + 3x - 4 \leq 0$ is

a. number line with -4 and 1 marked, arrows both ways, line outside the segment
b. number line with -4 and 1 marked, arrows both ways, line outside the segment (maybe different style?)
c. number line with -4 and 1 marked, line between them, arrows right
d. number line with -4 and 1 marked, line between them, arrows right (maybe different style?)

1. a condition that satisfies $(x - r)(x - s) > 0$ is

a. $x > r$ and $x > s$
b. $x > r$ and $x < -s$
c. $x > -r$ and $x > s$
d. $x > -r$ and $x < s$

Explanation:

Question 8

Step1: Define the two integers

Let the smaller odd integer be $x$, the larger consecutive odd integer is $x+2$.

Step2: Set up product inequality

Their product is at least 63, so $x(x+2) \geq 63$.

Step3: Simplify the inequality

Expand and rearrange: $x^2 + 2x - 63 \geq 0$.

Step1: Factor the quadratic

Factor $x^2 + 3x - 4$: $(x+4)(x-1) \leq 0$.

Step2: Find critical points

Critical points are $x=-4$ and $x=1$.

Step3: Test solution interval

The quadratic opens upward, so the inequality holds between the roots, including the endpoints (solid dots).

Step1: Analyze the inequality rule

For $(x-r)(x-s) > 0$, the product is positive when both factors are positive or both are negative.

Step2: Match valid condition

If we assume $r < s$, both positive means $x > r$ and $x > s$ (equivalent to $x > s$), which matches option A.

Answer:

A. $x^2 + 2x - 63 \geq 0$

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Question 9