Answer :
To solve the given question, we need to address each part step-by-step.
1. First Three Multiples of 20:
The first three multiples of 20 are:
- 20 (20 multiplied by 1)
- 40 (20 multiplied by 2)
- 60 (20 multiplied by 3)
2. Evaluate the Inequality:
To compare [tex]\( \sqrt{64 + 36} \)[/tex] with [tex]\( \sqrt[3]{27} \)[/tex]:
- Calculate the left side: [tex]\(\sqrt{64 + 36} = \sqrt{100} = 10\)[/tex]
- Calculate the right side: [tex]\(\sqrt[3]{27} = 3\)[/tex]
- Since 10 is greater than 3, the statement [tex]\(\sqrt{64+36} > \sqrt[3]{27}\)[/tex] is true.
3. Ordering Values:
The values given are in descending order:
- 0.3 (or 0,[tex]$3$[/tex])
- [tex]\( \sqrt[3]{0.001} \)[/tex]
- [tex]\( (0.2)^3 \)[/tex]
Let's find their approximate values:
- [tex]\( 0.3 = 0.3 \)[/tex]
- [tex]\( \sqrt[3]{0.001} = 0.1 \)[/tex]
- [tex]\( (0.2)^3 = 0.008 \)[/tex]
So, in descending order: 0.3, 0.1, 0.008
4. True/False Statements:
- Evaluate [tex]\(\frac{c}{d} < \frac{d}{c}\)[/tex] and other statements individually:
- No numerical values given for [tex]\(c\)[/tex] and [tex]\(d\)[/tex], so without specific context, this cannot be simplified further.
- Check the calculation [tex]\(2^3 + 2^2 = 4^5\)[/tex]:
- [tex]\(2^3 + 2^2 = 8 + 4 = 12\)[/tex]
- [tex]\(4^5 = 1024\)[/tex]
- Since 12 is not equal to 1024, this statement is false.
- For [tex]\(3x^5 \cdot 4x^2 = 12x^{10}\)[/tex]:
- This should be simplified to [tex]\(12x^7\)[/tex], and thus does not match the given solution. The multiplication is incorrect as per standard algebraic rules.
- Evaluate [tex]\((3ab)^2 = 6a^2b^2\)[/tex]:
- [tex]\((3ab)^2 = 9a^2b^2\)[/tex], not [tex]\(6a^2b^2\)[/tex], so this is false.
By breaking down each part, we have addressed the entire question step-by-step. If you have any more questions or need further clarification, feel free to ask!
1. First Three Multiples of 20:
The first three multiples of 20 are:
- 20 (20 multiplied by 1)
- 40 (20 multiplied by 2)
- 60 (20 multiplied by 3)
2. Evaluate the Inequality:
To compare [tex]\( \sqrt{64 + 36} \)[/tex] with [tex]\( \sqrt[3]{27} \)[/tex]:
- Calculate the left side: [tex]\(\sqrt{64 + 36} = \sqrt{100} = 10\)[/tex]
- Calculate the right side: [tex]\(\sqrt[3]{27} = 3\)[/tex]
- Since 10 is greater than 3, the statement [tex]\(\sqrt{64+36} > \sqrt[3]{27}\)[/tex] is true.
3. Ordering Values:
The values given are in descending order:
- 0.3 (or 0,[tex]$3$[/tex])
- [tex]\( \sqrt[3]{0.001} \)[/tex]
- [tex]\( (0.2)^3 \)[/tex]
Let's find their approximate values:
- [tex]\( 0.3 = 0.3 \)[/tex]
- [tex]\( \sqrt[3]{0.001} = 0.1 \)[/tex]
- [tex]\( (0.2)^3 = 0.008 \)[/tex]
So, in descending order: 0.3, 0.1, 0.008
4. True/False Statements:
- Evaluate [tex]\(\frac{c}{d} < \frac{d}{c}\)[/tex] and other statements individually:
- No numerical values given for [tex]\(c\)[/tex] and [tex]\(d\)[/tex], so without specific context, this cannot be simplified further.
- Check the calculation [tex]\(2^3 + 2^2 = 4^5\)[/tex]:
- [tex]\(2^3 + 2^2 = 8 + 4 = 12\)[/tex]
- [tex]\(4^5 = 1024\)[/tex]
- Since 12 is not equal to 1024, this statement is false.
- For [tex]\(3x^5 \cdot 4x^2 = 12x^{10}\)[/tex]:
- This should be simplified to [tex]\(12x^7\)[/tex], and thus does not match the given solution. The multiplication is incorrect as per standard algebraic rules.
- Evaluate [tex]\((3ab)^2 = 6a^2b^2\)[/tex]:
- [tex]\((3ab)^2 = 9a^2b^2\)[/tex], not [tex]\(6a^2b^2\)[/tex], so this is false.
By breaking down each part, we have addressed the entire question step-by-step. If you have any more questions or need further clarification, feel free to ask!
