2.5 review & refresh
in exercises 1 - 4, solve the equation. graph the solution(s), if possible.
1. $|5d|=-45$
2. $-3|6p + 7|=-12$
3. $|r + 5|=3.5$
4. $|\frac{1}{4}w + 5|=|w + 1|$
5. the mean admission price for six glassblowing workshops is $17.50. the data shows the admission prices for five of the glassblowing workshops. find the admission price for the sixth workshop.
$11, $13, $21, $12, $23
6. one - fourth of the female students in a choir sing alto and one - half of the female students sing soprano. eighteen female students sing soprano. how many female students are in the choir?
in exercises 7 and 8, write an inequality that represents the graph.
7.
8.

Question

Accepted Answer

### 1. Solve $|5d|=-45$
# Explanation:
## Step1: Recall absolute - value property
The absolute - value of a number is always non - negative. That is, for any real number $x$, $|x|\geq0$.
Since $|5d|\geq0$ for all real $d$, and the right - hand side is $-45\lt0$, there is no solution for the equation $|5d| = - 45$.
# Answer:
No solution

### 2. Solve $-3|6p + 7|=-12$
# Explanation:
## Step1: Isolate the absolute - value expression
Divide both sides of the equation by $-3$. We get $|6p + 7|=\frac{-12}{-3}=4$.
## Step2: Set up two equations
Based on the definition of absolute value ($|x| = a$ where $a\gt0$ gives $x=a$ or $x = - a$), we have $6p+7 = 4$ or $6p+7=-4$.
### Case 1: Solve $6p+7 = 4$
## Step3: Isolate the variable term
Subtract 7 from both sides: $6p=4 - 7=-3$.
## Step4: Solve for $p$
Divide both sides by 6: $p=\frac{-3}{6}=-\frac{1}{2}$.
### Case 2: Solve $6p+7=-4$
## Step5: Isolate the variable term
Subtract 7 from both sides: $6p=-4 - 7=-11$.
## Step6: Solve for $p$
Divide both sides by 6: $p=-\frac{11}{6}$.
# Answer:
$p =-\frac{1}{2}$ or $p=-\frac{11}{6}$

### 3. Solve $|r + 5|=3.5$
# Explanation:
## Step1: Set up two equations
Based on the definition of absolute value ($|x| = a$ where $a\gt0$ gives $x=a$ or $x=-a$), we have $r + 5=3.5$ or $r + 5=-3.5$.
### Case 1: Solve $r + 5=3.5$
## Step2: Isolate the variable
Subtract 5 from both sides: $r=3.5−5=-1.5$.
### Case 2: Solve $r + 5=-3.5$
## Step3: Isolate the variable
Subtract 5 from both sides: $r=-3.5−5=-8.5$.
# Answer:
$r=-1.5$ or $r=-8.5$

### 4. Solve $|\frac{1}{4}w + 5|=|w + 1|$
# Explanation:
## Step1: Set up two equations
Based on the property $|a| = |b|$ which gives $a = b$ or $a=-b$, we have $\frac{1}{4}w+5=w + 1$ or $\frac{1}{4}w+5=-(w + 1)$.
### Case 1: Solve $\frac{1}{4}w+5=w + 1$
## Step2: Get rid of the fraction and isolate the variable terms
Multiply through by 4 to get $w + 20=4w+4$. Then subtract $w$ from both sides: $20=3w + 4$. Subtract 4 from both sides: $16 = 3w$. Divide by 3: $w=\frac{16}{3}$.
### Case 2: Solve $\frac{1}{4}w+5=-(w + 1)$
## Step3: Expand and solve
Expand the right - hand side: $\frac{1}{4}w+5=-w - 1$. Multiply through by 4 to get $w + 20=-4w-4$. Add $4w$ to both sides: $5w+20=-4$. Subtract 20 from both sides: $5w=-24$. Divide by 5: $w=-\frac{24}{5}$.
# Answer:
$w=\frac{16}{3}$ or $w=-\frac{24}{5}$

### 5. Find the admission price of the sixth workshop
# Explanation:
## Step1: Recall the mean formula
The mean $\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}$. Here, $n = 6$, $\bar{x}=17.50$, and the five known values are $x_1 = 11$, $x_2 = 13$, $x_3 = 21$, $x_4 = 12$, $x_5 = 23$. Let the sixth value be $x_6$.
We know that $17.50=\frac{11 + 13+21+12+23+x_6}{6}$.
## Step2: Multiply both sides by 6
$17.50\times6=11 + 13+21+12+23+x_6$. So, $105=80 + x_6$.
## Step3: Solve for $x_6$
Subtract 80 from both sides: $x_6=105 - 80 = 25$.
# Answer:
$\$25$

### 6. Find the number of female students in the choir
# Explanation:
## Step1: Let the total number of female students be $x$
We know that the fraction of students who sing alto is $\frac{1}{4}x$, the fraction of students who sing soprano is $\frac{1}{2}x$, and the number of students who sing soprano is 18.
Since $\frac{1}{2}x=18$.
## Step2: Solve for $x$
Multiply both sides by 2: $x = 36$.
# Answer:
36 female students

### 7. Write the inequality for the graph
# Explanation:
## Step1: Analyze the graph
The graph has a closed circle at $x=-3$ and an open circle at $x = 2$, and the line is shaded between them. The inequality is $-3\leq x\lt2$.
# Answer:
$-3\leq x\lt2$

### 8. Write the inequality for the graph
# Explanation:
## Step1: Analyze the graph
The graph has an open circle at $x = 3$ and a closed circle at $x = 4$, and the line is shaded between them. The inequality is $3\lt x\leq4$.
# Answer:
$3\lt x\leq4$

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

1. Solve \(|5d|=-45\)

Step1: Recall absolute - value property

Step1: Isolate the absolute - value expression

Step2: Set up two equations

Case 1: Solve \(6p+7 = 4\)

Step3: Isolate the variable term

Step4: Solve for \(p\)

Case 2: Solve \(6p+7=-4\)

Step5: Isolate the variable term

Step6: Solve for \(p\)

Step1: Set up two equations

Case 1: Solve \(r + 5=3.5\)

Step2: Isolate the variable

Case 2: Solve \(r + 5=-3.5\)

Step3: Isolate the variable

Answer:

2. Solve \(-3|6p + 7|=-12\)