Question

1. find all values of x which make this equation true (|2x - 11| = 3)

a ({7,4})
b ({7,-7})
c ({-4,-7})
d ({-4,4})

1. find the area of the circle in terms of (pi).

a. (30pi \text{in.}^2)
b. (900pi \text{in.}^2)
c. (60pi \text{in.}^2)
d. (225pi \text{in.}^2)
(image of a circle with diameter 30 in.)

Explanation:

Question 4

Step1: Recall absolute value definition

If \(|a| = b\) (\(b\geq0\)), then \(a = b\) or \(a=-b\). So for \(|2x - 11| = 3\), we have two cases:
Case 1: \(2x - 11 = 3\)

Step2: Solve Case 1

Add 11 to both sides: \(2x=3 + 11=14\)
Divide by 2: \(x=\frac{14}{2}=7\)
Case 2: \(2x - 11=-3\)

Step3: Solve Case 2

Add 11 to both sides: \(2x=-3 + 11 = 8\)
Divide by 2: \(x=\frac{8}{2}=4\)

Step1: Recall circle area formula

The area of a circle is \(A=\pi r^{2}\), where \(r\) is the radius. The diameter \(d = 30\) in, so radius \(r=\frac{d}{2}=\frac{30}{2}=15\) in.

Step2: Calculate area

Substitute \(r = 15\) into the formula: \(A=\pi(15)^{2}=\pi\times225 = 225\pi\) \(in^{2}\)

Answer:

A. \(\{7, 4\}\)

Question 6