Question

kari and samantha have determined that their water - balloon launcher works best when they launch the ball an angle within 3 degrees of 45 degrees. which equation can be used to determine the minimum and maxi optimal angles of launch, and what is the minimum angle that is still optimal?
○ |x - 3| = 45; minimum angle: 42 degrees
○ |x - 3| = 45; minimum angle: 45 degrees
○ |x - 45| = 3; minimum angle: 42 degrees
○ |x - 45| = 3; minimum angle: 45 degrees

Explanation:

Step1: Understand absolute value equation for range

The problem is about an angle \( x \) that is within 3 degrees of 45 degrees. The absolute value equation for a value \( x \) within \( d \) units of \( a \) is \( |x - a|=d \). Here, \( a = 45 \) (the target angle) and \( d=3 \) (the range), so the equation should be \( |x - 45| = 3 \).

Step2: Solve for minimum angle

To find the minimum angle, we solve the absolute value equation \( |x - 45|=3 \). The absolute value equation \( |A|=b \) (where \( b\geq0 \)) has solutions \( A = b \) or \( A=-b \). So for \( |x - 45| = 3 \), we have two cases:

• Case 1: \( x - 45=3 \) gives \( x = 45 + 3=48 \) (maximum angle)
• Case 2: \( x - 45=- 3 \) gives \( x=45 - 3 = 42 \) (minimum angle)

Answer:

\( |x - 45| = 3 \); minimum angle: 42 degrees (corresponding to the option: \( |x - 45| = 3 \); minimum angle: 42 degrees)