59 |3x - 9| 0 60 |5x + 2| ≥ 0 63 \left|\frac{2 - 3x}{5}\ ight| = 2 64 \left|\frac{2x + 5}{3}\ ight| ≤ 1 65 \frac{3}{|5 - 2x|}

Question

59 |3x - 9| > 0
60 |5x + 2| ≥ 0
63 \left|\frac{2 - 3x}{5}\
ight| = 2
64 \left|\frac{2x + 5}{3}\
ight| ≤ 1
65 \frac{3}{|5 - 2x|} < 2
66 \frac{2}{|2x + 3|} ≥ 5
67 -2 < |x| < 4
68 1 < |x| < 5
69 1 < |x - 2| < 4
70 2 < |2x - 1| < 3
exer. 71–72: solve part (a) and use that answer to determine the answers to parts (b) and (c).
71 (a) |x + 5| = 3 (b) |x + 5| < 3 (c) |x + 5| > 3
72 (a) |x - 3| < 2 (b) |x - 3| = 2 (c) |x - 3| > 2
exer. 73–76: express the statement in terms of an inequality involving an absolute value.
73 the weight w of a wrestler must be within 2 pounds of 148 pounds.
74 the radius r of a ball bearing must be within 0.01 centimeter of 1 centimeter.
75 the difference of two temperatures t₁ and t₂ within a chemical mixture must be between 5°c and 10°c.
76 the arrival time t of train b must be at least 5 minutes different from the 4:00 pm arrival time of train a.
temperature scales: temperature readings on the fahrenheit and celsius scales are related by the formula ( c = \frac{5}{9}(f - 32) ). what values of f correspond to the values of c such that ( 30 ≤ c ≤ 40 )?

Accepted Answer

The image contains multiple math problems related to absolute value equations and inequalities, as well as applications of absolute value in real - world contexts (like weight, radius, temperature difference, arrival time) and a problem about the relationship between Fahrenheit and Celsius temperature scales. Since you haven't specified a particular problem, I'll solve a few of them. Let's start with problem 71 (a), (b), (c):

### Problem 71
#### Part (a): Solve $|x + 5|=3$
# Explanation:
## Step 1: Recall the definition of absolute value
If $|a| = b$ ($b\geq0$), then $a = b$ or $a=-b$. For $|x + 5|=3$, we have two cases:
Case 1: $x + 5=3$
## Step 2: Solve the first equation
Subtract 5 from both sides: $x=3 - 5=-2$
Case 2: $x + 5=-3$
## Step 3: Solve the second equation
Subtract 5 from both sides: $x=-3 - 5=-8$
# Answer: $x=-2$ or $x = - 8$

#### Part (b): Solve $|x + 5|\lt3$
# Explanation:
## Step 1: Recall the inequality property for absolute value
If $|a|\lt b$ ($b\gt0$), then $-b\lt a\lt b$. For $|x + 5|\lt3$, we get $-3\lt x + 5\lt3$
## Step 2: Solve the compound inequality
Subtract 5 from all parts: $-3-5\lt x+5 - 5\lt3 - 5$, which simplifies to $-8\lt x\lt - 2$
# Answer: $-8\lt x\lt - 2$ (in interval notation: $(-8,-2)$)

#### Part (c): Solve $|x + 5|\gt3$
# Explanation:
## Step 1: Recall the inequality property for absolute value
If $|a|\gt b$ ($b\gt0$), then $a\gt b$ or $a\lt - b$. For $|x + 5|\gt3$, we have two cases:
Case 1: $x + 5\gt3$
## Step 2: Solve the first inequality
Subtract 5 from both sides: $x\gt3 - 5=-2$
Case 2: $x + 5\lt - 3$
## Step 3: Solve the second inequality
Subtract 5 from both sides: $x\lt - 3 - 5=-8$
# Answer: $x\lt - 8$ or $x\gt - 2$ (in interval notation: $(-\infty,-8)\cup(-2,\infty)$)

### Problem 73: Express the statement "The weight $w$ of a wrestler must be within 2 pounds of 148 pounds" as an absolute value inequality
# Explanation:
## Step 1: Understand the meaning of "within"
"Within 2 pounds of 148 pounds" means that the difference between the weight $w$ and 148 pounds is less than or equal to 2 pounds. In absolute value terms, if we let $a = w$ and $b = 148$ and the tolerance $t = 2$, the absolute value inequality for $|a - b|\leq t$
## Step 2: Write the inequality
Substituting the values, we get $|w - 148|\leq2$
# Answer: $|w - 148|\leq2$

### Problem 74: Express the statement "The radius $r$ of a ball bearing must be within 0.01 centimeter of 1 centimeter" as an absolute value inequality
# Explanation:
## Step 1: Interpret "within"
"Within 0.01 centimeter of 1 centimeter" means the difference between the radius $r$ and 1 centimeter is less than or equal to 0.01 centimeter. Using the absolute value inequality form $|a - b|\leq t$ where $a=r$, $b = 1$ and $t=0.01$
## Step 2: Form the inequality
We have $|r - 1|\leq0.01$
# Answer: $|r - 1|\leq0.01$

### Problem 75: Express the statement "The difference of two temperatures $T_1$ and $T_2$ within a chemical mixture must be between $5^{\circ}C$ and $10^{\circ}C$" as an absolute value inequality
# Explanation:
## Step 1: Analyze the "between" condition
"Between $5^{\circ}C$ and $10^{\circ}C$" for the difference $|T_1 - T_2|$ means that $5\lt|T_1 - T_2|\lt10$
# Answer: $5\lt|T_1 - T_2|\lt10$

### Problem 76: Express the statement "The arrival time $t$ of train B must be at least 5 minutes different from the 4:00 PM arrival time of train A" as an absolute value inequality
# Explanation:
## Step 1: Understand "at least 5 minutes different"
If we consider the arrival time of train A as $t_A = 4:00$ PM (we can represent time $t$ in minutes since a certain reference, but for the inequality, we can think of the difference $|t - t_A|$. "At least 5 minutes different" means $|t - 4:00|\geq5$ (where we can consider $t$ and $4:00$ in minutes or just as a time value, the absolute value represents the difference in time)
# Answer: $|t - 4:00|\geq5$ (if $t$ is in minutes, we can let $t_A = 240$ minutes (4:00 PM is 240 minutes after 12:00 AM) and the inequality is $|t - 240|\geq5$)

### Problem (last part about temperature scales): Given $C=\frac{5}{9}(F - 32)$, find the values of $F$ corresponding to $30\leq C\leq40$
# Explanation:
## Step 1: Substitute the inequality for $C$ into the formula
We have $30\leq\frac{5}{9}(F - 32)\leq40$
## Step 2: Multiply all parts by $\frac{9}{5}$ to isolate the term with $F$
$30\times\frac{9}{5}\leq\frac{5}{9}(F - 32)\times\frac{9}{5}\leq40\times\frac{9}{5}$
$54\leq F - 32\leq72$
## Step 3: Add 32 to all parts
$54 + 32\leq F-32 + 32\leq72 + 32$
$86\leq F\leq104$
# Answer: $86\leq F\leq104$ (the Fahrenheit temperature $F$ is between 86 and 104 degrees)

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QUESTION IMAGE

Question

Explanation:

Problem 71

Part (a): Solve \(|x + 5|=3\)

Step 1: Recall the definition of absolute value

Step 2: Solve the first equation

Step 3: Solve the second equation

Step 1: Recall the inequality property for absolute value

Step 2: Solve the compound inequality

Step 1: Recall the inequality property for absolute value

Step 2: Solve the first inequality

Step 3: Solve the second inequality

Answer:

Part (b): Solve \(|x + 5|\lt3\)