Question

1. Storing milk at temperatures colder than 35°F can affect its quality of taste. However, storing milk at temperatures warmer than 40°F is an unsafe food practice. Which of the inequalities below represent the union of these improper storage temperatures, t, in degrees Fahrenheit? A. T > 35 or t <40 B. T < 35 or t >40 C. T > 35 AND t < 40 D. T < 35 AND t > 40 2. What are the solution(s) to the absolute value equation below? 15| x-7|+4=10|x-7|+4 A. X= - 5\7 or x = 5\7 B. X=5 or x= -5 C. Only x=7 D. X= 7/5 or x= -7\5 3. Match the absolute value equation with the statement that describes the solutions to that equation. A. Two solutions B. Solution C. No solution |x+1|+5=2 |4x+12|=0 |3x|=9

Answer 1

Answer:

Part (A): The required inequality is T < 35 or t >40.

Part (B): The correct option is C) Only x=7.

Part (C) |x+1|+5=2 has no solution; |4x+12|=0 has one solution; |3x|=9 has two solution.

Step-by-step explanation:

Consider the provided information.

Part (A)

Storing milk at temperatures colder than 35°F can affect its quality of taste. However, storing milk at temperatures warmer than 40°F is an unsafe food practice.

The union is written as A∪B or “A or B”.

The intersection of two sets is written as A∩B or “A and B”

We need to determine the inequalities represent the union of these improper storage.

That means we will use A∪B or “A or B”.

The improper storage temperatures is when temperature is less than 35°F or greater than 40°F.

Hence, the required inequality is T < 35 or t >40.

Part (B) 15| x-7|+4=10|x-7|+4

Solve the inequality as shown below:

Subtract 4 from both sides.

$15| x-7|+4-4=10|x-7|+4-4$

Subtract 10|x-7| from both sides

$15\left|x-7\right|-10\left|x-7\right|=10\left|x-7\right|-10\left|x-7\right|$

$5\left|x-7\right|=0\\\left|x-7\right|=0\\x=7$

Hence, the correct option is C) Only x=7.

Part (C) Match the solution,

$\left|x+1\right|+5=2$

Subtract 2 from both sides.

$\left|x+1\right|=-3$

Absolute value cannot be less than 0.

Hence, |x+1|+5=2 has no solution.

$|4x+12|=0$

$4x+12=0$

$x=-3$

Hence, |4x+12|=0 has one solution.

$|3x|=9$

$\mathrm{If}\:|u|\:=\:a,\:a>0\:\mathrm{then}\:u\:=\:a\:\quad \mathrm{or}\quad \:u\:=\:-a$

$3x=-9\quad \mathrm{or}\quad \:3x=9\\x=-3\quad \mathrm{or}\quad \:x=3$

Hence, |3x|=9 has two solution.