Question

evaluate.

1. $(3^{2})^{2}$
2. $(1^{5})^{8}$

evaluate.

1. $sqrt{196}$
2. $sqrt{1.44}$

evaluate to the nearest tenth.

1. $sqrt{43}$
2. $sqrt{6.75}$
3. circle the composite numbers.

17 16 3 9 21 39

1. write the prime numbers between 30 and 80.

Explanation:

Step1: Evaluate \((3^{2})^{2}\)

Use the power - of - a - power rule \((a^{m})^{n}=a^{mn}\). Here \(a = 3\), \(m = 2\), \(n = 2\). So \((3^{2})^{2}=3^{2\times2}=3^{4}\). Then \(3^{4}=3\times3\times3\times3 = 81\).

Step2: Evaluate \((1^{5})^{8}\)

By the power - of - a - power rule \((a^{m})^{n}=a^{mn}\), with \(a = 1\), \(m = 5\), \(n = 8\). So \((1^{5})^{8}=1^{5\times8}=1^{40}\). Since \(1\) raised to any non - negative integer power is \(1\), \(1^{40}=1\).

Step3: Evaluate \(\sqrt{196}\)

We need to find a number \(x\) such that \(x^{2}=196\). Since \(14\times14 = 196\), \(\sqrt{196}=14\).

Step4: Evaluate \(\sqrt{1.44}\)

Find a number \(x\) such that \(x^{2}=1.44\). Since \(1.2\times1.2 = 1.44\), \(\sqrt{1.44}=1.2\).

Step5: Evaluate \(\sqrt{43}\) to the nearest tenth

We know that \(6^{2}=36\) and \(7^{2}=49\). Using a calculator, \(\sqrt{43}\approx6.557\). Rounding to the nearest tenth, we look at the hundredth digit. Since \(5\geq5\), we round up. So \(\sqrt{43}\approx6.6\).

Step6: Evaluate \(\sqrt{6.75}\) to the nearest tenth

We know that \(2^{2}=4\) and \(3^{2}=9\). Using a calculator, \(\sqrt{6.75}\approx2.6\).

Step7: Identify composite numbers

A composite number is a positive integer that has at least one positive divisor other than \(1\) and itself.

• For \(17\): Its divisors are \(1\) and \(17\), so it is prime.
• For \(16\): Divisors are \(1,2,4,8,16\), so it is composite.
• For \(3\): Divisors are \(1\) and \(3\), so it is prime.
• For \(9\): Divisors are \(1,3,9\), so it is composite.
• For \(21\): Divisors are \(1,3,7,21\), so it is composite.
• For \(39\): Divisors are \(1,3,13,39\), so it is composite.

Step8: Find prime numbers between \(30\) and \(80\)

A prime number is a number greater than \(1\) with only two distinct positive divisors: \(1\) and itself.
The prime numbers between \(30\) and \(80\) are \(31,37,41,43,47,53,59,61,67,71,73,79\).

Answer:

1. \(81\)
2. \(1\)
3. \(14\)
4. \(1.2\)
5. \(6.6\)
6. \(2.6\)
7. Composite numbers: \(16,9,21,39\)
8. Prime numbers: \(31,37,41,43,47,53,59,61,67,71,73,79\)