Question

1. which of the sets shown includes the elements of set z that are both negative numbers and multiples of 4?

z = {-34, -28, -16, -2, 4, 8, 12, 26}
a {-28, -16}
b {-28, -16, 4, 8, 12}
c {-28, -16, -2, 4, 8, 12}
d {-34, -28, -16, -2, 4, 8, 12, 26}

1. order the numbers from least to greatest.

\\(\frac{45}{11}\\), 4.5, \\(\sqrt{20}\\), \\(\sqrt{\frac{45}{4}}\\)

1. write \\(\sqrt3{10}\\) using rational exponents.

a \\(3^{10}\\) c \\(3^{\frac{1}{10}}\\)
b \\(10^3\\) d \\(10^{\frac{1}{3}}\\)

Explanation:

Question 28

Step1: Identify negative numbers in Z

Negative numbers in \( Z = \{-34, -28, -16, -2, 4, 8, 12, 26\} \) are \(-34, -28, -16, -2\).

Step2: Identify multiples of 4

Multiples of 4 from Z: \(-28\) (since \(-28\div4 = -7\)), \(-16\) ( \(-16\div4=-4\) ), \(4\) ( \(4\div4 = 1\) ), \(8\) ( \(8\div4 = 2\) ), \(12\) ( \(12\div4 = 3\) ). Now, from negative numbers, the multiples of 4 are \(-28, -16\) (as \(-34\div4=-8.5\) and \(-2\div4=-0.5\) are not integers).

Step3: Find the set with negative multiples of 4

So the set with elements of Z that are both negative and multiples of 4 is \(\{-28, -16\}\)? Wait, no, wait. Wait, step 2: multiples of 4 in Z: let's re - check. \(-28\) is a multiple of 4 (\(-28 = 4\times(-7)\)), \(-16=4\times(-4)\), \(4 = 4\times1\), \(8 = 4\times2\), \(12=4\times3\). But we need negative numbers and multiples of 4. So negative numbers that are multiples of 4: \(-28\) (negative, multiple of 4), \(-16\) (negative, multiple of 4). The other multiples of 4 (4, 8, 12) are positive. So the set should contain only negative multiples of 4. Let's check the options:

• Option A: \(\{-28, -16\}\) (both negative and multiples of 4)
• Option B: includes positive multiples (4, 8, 12)
• Option C: includes - 2 which is not a multiple of 4
• Option D: includes all elements, not just negative multiples of 4

Wait, maybe I made a mistake. Wait, the question is "elements of Set Z that are both negative numbers and multiples of 4". So negative numbers (so less than 0) and multiples of 4 (so divisible by 4). So:

Check each element in Z:

• \(-34\): negative, \(-34\div4=-8.5\) not integer (not multiple of 4)
• \(-28\): negative, \(-28\div4 = -7\) (multiple of 4)
• \(-16\): negative, \(-16\div4=-4\) (multiple of 4)
• \(-2\): negative, \(-2\div4=-0.5\) (not multiple of 4)
• \(4\): positive (not negative)
• \(8\): positive (not negative)
• \(12\): positive (not negative)
• \(26\): positive (not negative)

So the elements that are both negative and multiples of 4 are \(-28\) and \(-16\), which is option A.

Step1: Calculate the approximate values of each number

• For \(\frac{45}{11}\approx4.09\) (since \(11\times4 = 44\), \(45-44 = 1\), so \(\frac{45}{11}=4+\frac{1}{11}\approx4.09\))
• \(4.5\) is already in decimal form.
• For \(\sqrt{20}\), since \(4^2 = 16\) and \(5^2 = 25\), \(\sqrt{20}\approx4.47\)
• For \(\sqrt{\frac{45}{4}}\), first \(\frac{45}{4}=11.25\), and \(\sqrt{11.25}\approx3.35\) (since \(3^2 = 9\) and \(4^2 = 16\))

Step2: Order the numbers from least to greatest

Now we have the approximate values: \(\sqrt{\frac{45}{4}}\approx3.35\), \(\frac{45}{11}\approx4.09\), \(\sqrt{20}\approx4.47\), \(4.5\)

So the order from least to greatest is \(\sqrt{\frac{45}{4}},\frac{45}{11},\sqrt{20},4.5\)

Step1: Recall the formula for converting radicals to rational exponents

The formula is \(\sqrt[n]{a}=a^{\frac{1}{n}}\), where \(n\) is the index of the radical and \(a\) is the radicand.

Step2: Apply the formula to \(\sqrt[3]{10}\)

For \(\sqrt[3]{10}\), the index \(n = 3\) and the radicand \(a = 10\). Using the formula \(\sqrt[n]{a}=a^{\frac{1}{n}}\), we get \(10^{\frac{1}{3}}\)

Answer:

A. \(\{-28, -16\}\)

Question 29