Question

warm - up: solve each inequality then graph its solution on the number line. 1. 4|6 - 2a|+8≤24 2. 3 + 4|3x + 7|≥ - 89 warm - up: determine the next two number in each sequence. 1. 6, 13, 27, 55,... 2. 4, 16, 36, 64, 100,... 3. 9|x - 2|-10 < - 73 4. 7|(1/3)n|-9 < 12

Explanation:

Tuesday - Inequality 1: Solve \(|4(6 - 2a)|+8\leq24\)

Step1: Isolate the absolute - value expression

Subtract 8 from both sides of the inequality:
\[|4(6 - 2a)|\leq24 - 8\]
\[|4(6 - 2a)|\leq16\]

Step2: Split the absolute - value inequality

We get two inequalities:
\(4(6 - 2a)\leq16\) and \(4(6 - 2a)\geq - 16\)

For \(4(6 - 2a)\leq16\):

Step3: Distribute and solve

Distribute the 4: \(24-8a\leq16\)
Subtract 24 from both sides: \(-8a\leq16 - 24=-8\)
Divide both sides by - 8. When dividing by a negative number, the inequality sign flips. So \(a\geq1\)

For \(4(6 - 2a)\geq - 16\):

Step4: Distribute and solve

Distribute the 4: \(24-8a\geq - 16\)
Subtract 24 from both sides: \(-8a\geq - 16 - 24=-40\)
Divide both sides by - 8. When dividing by a negative number, the inequality sign flips. So \(a\leq5\)
The solution is \(1\leq a\leq5\)

Tuesday - Inequality 2: Solve \(3 + 4|3x + 7|\geq - 89\)

Step1: Isolate the absolute - value expression

Subtract 3 from both sides: \(4|3x + 7|\geq - 89 - 3=-92\)
Divide both sides by 4: \(|3x + 7|\geq - 23\)
Since the absolute - value of any real number is always non - negative (\(|y|\geq0\) for all real \(y\)), the solution of \(|3x + 7|\geq - 23\) is all real numbers, \(x\in(-\infty,\infty)\)

Wednesday - Sequence 1: \(1,6,13,27,55,\cdots\)

Step1: Look for the pattern

Find the differences between consecutive terms:
\(6 - 1 = 5\), \(13 - 6 = 7\), \(27 - 13 = 14\), \(55 - 27 = 28\)
We can observe that the differences are not constant. But if we rewrite the sequence, we can see that \(a_{n + 1}=2a_{n}+(n)\) (a non - standard pattern). Another way is:
\(1\times2 + 4=6\), \(6\times2+1 = 13\), \(13\times2 + 1=27\), \(27\times2+1 = 55\)
The next term: \(55\times2+1 = 111\)
The term after that: \(111\times2+1 = 223\)

Wednesday - Sequence 2: \(4,16,36,64,100,\cdots\)

Step1: Identify the pattern

The terms can be written as \(2^{2},4^{2},6^{2},8^{2},10^{2},\cdots\)
The general term of the sequence is \((2n)^{2}\), where \(n = 1,2,3,\cdots\)
The next term (\(n = 6\)) is \((2\times6)^{2}=144\)
The term after that (\(n = 7\)) is \((2\times7)^{2}=196\)

Wednesday - Inequality 3: Solve \(9|x - 2|-10\lt - 73\)

Step1: Isolate the absolute - value expression

Add 10 to both sides: \(9|x - 2|\lt - 73 + 10=-63\)
Divide both sides by 9: \(|x - 2|\lt - 7\)
Since the absolute - value of any real number is non - negative (\(|y|\geq0\) for all real \(y\)), there is no solution for this inequality.

Wednesday - Inequality 4: Solve \(7|\frac{n}{3}|-9\lt12\)

Step1: Isolate the absolute - value expression

Add 9 to both sides: \(7|\frac{n}{3}|\lt12 + 9 = 21\)
Divide both sides by 7: \(|\frac{n}{3}|\lt3\)

Step2: Split the absolute - value inequality

We get two inequalities: \(\frac{n}{3}\lt3\) and \(\frac{n}{3}\gt - 3\)

For \(\frac{n}{3}\lt3\):

Multiply both sides by 3: \(n\lt9\)

For \(\frac{n}{3}\gt - 3\):

Multiply both sides by 3: \(n\gt - 9\)
The solution is \(-9\lt n\lt9\)

Answer:

Tuesday - Inequality 1: \(1\leq a\leq5\)
Tuesday - Inequality 2: \(x\in(-\infty,\infty)\)
Wednesday - Sequence 1: 111, 223
Wednesday - Sequence 2: 144, 196
Wednesday - Inequality 3: No solution
Wednesday - Inequality 4: \(-9\lt n\lt9\)