Answer :
Sure! Let's go through each of the questions step-by-step to fully understand and explain the solutions.
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1.1 Round off 2.09689 to two decimal places.
To round 2.09689 to two decimal places, we look at the third decimal place (6 in this case). Since it is greater than 5, we round up the second decimal place by 1.
Hence, [tex]\( 2.09689 \)[/tex] rounded to two decimal places is [tex]\( 2.10 \)[/tex].
[tex]\[ \boxed{2.10} \][/tex]
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1.2 Given the set [tex]\( B = \{-3, 0, \sqrt[3]{6}, 1, \sqrt{-2}\} \)[/tex]:
* 1.2.1 Identify Integer(s):
Integers are whole numbers that can be positive, negative, or zero. From the set [tex]\( B \)[/tex], the integers are:
[tex]\[ \boxed{\{ -3, 0, 1 \}} \][/tex]
* 1.2.2 Identify Natural number(s):
Natural numbers are positive integers starting from 1 and upwards. From the integers identified:
[tex]\[ \boxed{\{1\}} \][/tex]
* 1.2.3 Identify Non-real number(s):
Non-real numbers include complex numbers, which have an imaginary component. From the set, [tex]\(\sqrt{-2}\)[/tex] is a complex number because the square root of a negative number is imaginary.
[tex]\[ \boxed{\{ \sqrt{-2} \}} \][/tex]
Simplified this is approximately [tex]\(1.4142135623730951i\)[/tex].
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1.3 Factorize fully the following expressions:
* 1.3.1 [tex]\( 8y^3 - 1 \)[/tex]
This expression is a difference of cubes, which can be factored using the formula [tex]\( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \)[/tex]. In this case, [tex]\( a = 2y \)[/tex] and [tex]\( b = 1 \)[/tex].
[tex]\[ 8y^3 - 1 = (2y - 1)(4y^2 + 2y + 1) \][/tex]
So, the factorized form is:
[tex]\[ \boxed{(2y - 1)(4y^2 + 2y + 1)} \][/tex]
* 1.3.2 [tex]\( m(x - y) + (y - x) \)[/tex]
First, notice that [tex]\( y - x \)[/tex] is the same as [tex]\( -(x - y) \)[/tex]:
[tex]\[ m(x - y) + (y - x) = m(x - y) - (x - y) = (m - 1)(x - y) \][/tex]
So, the factorized form is:
[tex]\[ \boxed{(m - 1)(x - y)} \][/tex]
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1.4 Simplify the following expressions:
* 1.4.1 [tex]\(\frac{1}{10}\)[/tex] of [tex]\(10^{100}\)[/tex]
To find [tex]\(\frac{1}{10}\)[/tex] of [tex]\(10^{100}\)[/tex]:
[tex]\[ \frac{1}{10} \times 10^{100} = 10^{100 - 1} = 10^{99} \][/tex]
So,
[tex]\[ \boxed{10^{99}} \][/tex]
* 1.4.2 [tex]\(\frac{3^{x+4} - 6 \cdot 3^{x+1}}{7 \cdot 3^{x+2}}\)[/tex]
First, simplify the numerator:
[tex]\[ 3^{x+4} - 6 \cdot 3^{x+1} = 3^{x} \cdot 3^{4} - 6 \cdot 3^{x+1} = 3^{x} \cdot 81 - 6 \cdot 3 \cdot 3^{x} = 81 \cdot 3^{x} - 18 \cdot 3^{x} = 63 \cdot 3^{x} \][/tex]
Now, incorporate this into the original expression:
[tex]\[ \frac{63 \cdot 3^{x}}{7 \cdot 3^{x+2}} = \frac{63}{7 \cdot 3^2} = \frac{63}{7 \cdot 9} = \frac{63}{63} = 1 \][/tex]
So, the simplified form is:
[tex]\[ \boxed{1} \][/tex]
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1.1 Round off 2.09689 to two decimal places.
To round 2.09689 to two decimal places, we look at the third decimal place (6 in this case). Since it is greater than 5, we round up the second decimal place by 1.
Hence, [tex]\( 2.09689 \)[/tex] rounded to two decimal places is [tex]\( 2.10 \)[/tex].
[tex]\[ \boxed{2.10} \][/tex]
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1.2 Given the set [tex]\( B = \{-3, 0, \sqrt[3]{6}, 1, \sqrt{-2}\} \)[/tex]:
* 1.2.1 Identify Integer(s):
Integers are whole numbers that can be positive, negative, or zero. From the set [tex]\( B \)[/tex], the integers are:
[tex]\[ \boxed{\{ -3, 0, 1 \}} \][/tex]
* 1.2.2 Identify Natural number(s):
Natural numbers are positive integers starting from 1 and upwards. From the integers identified:
[tex]\[ \boxed{\{1\}} \][/tex]
* 1.2.3 Identify Non-real number(s):
Non-real numbers include complex numbers, which have an imaginary component. From the set, [tex]\(\sqrt{-2}\)[/tex] is a complex number because the square root of a negative number is imaginary.
[tex]\[ \boxed{\{ \sqrt{-2} \}} \][/tex]
Simplified this is approximately [tex]\(1.4142135623730951i\)[/tex].
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1.3 Factorize fully the following expressions:
* 1.3.1 [tex]\( 8y^3 - 1 \)[/tex]
This expression is a difference of cubes, which can be factored using the formula [tex]\( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \)[/tex]. In this case, [tex]\( a = 2y \)[/tex] and [tex]\( b = 1 \)[/tex].
[tex]\[ 8y^3 - 1 = (2y - 1)(4y^2 + 2y + 1) \][/tex]
So, the factorized form is:
[tex]\[ \boxed{(2y - 1)(4y^2 + 2y + 1)} \][/tex]
* 1.3.2 [tex]\( m(x - y) + (y - x) \)[/tex]
First, notice that [tex]\( y - x \)[/tex] is the same as [tex]\( -(x - y) \)[/tex]:
[tex]\[ m(x - y) + (y - x) = m(x - y) - (x - y) = (m - 1)(x - y) \][/tex]
So, the factorized form is:
[tex]\[ \boxed{(m - 1)(x - y)} \][/tex]
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1.4 Simplify the following expressions:
* 1.4.1 [tex]\(\frac{1}{10}\)[/tex] of [tex]\(10^{100}\)[/tex]
To find [tex]\(\frac{1}{10}\)[/tex] of [tex]\(10^{100}\)[/tex]:
[tex]\[ \frac{1}{10} \times 10^{100} = 10^{100 - 1} = 10^{99} \][/tex]
So,
[tex]\[ \boxed{10^{99}} \][/tex]
* 1.4.2 [tex]\(\frac{3^{x+4} - 6 \cdot 3^{x+1}}{7 \cdot 3^{x+2}}\)[/tex]
First, simplify the numerator:
[tex]\[ 3^{x+4} - 6 \cdot 3^{x+1} = 3^{x} \cdot 3^{4} - 6 \cdot 3^{x+1} = 3^{x} \cdot 81 - 6 \cdot 3 \cdot 3^{x} = 81 \cdot 3^{x} - 18 \cdot 3^{x} = 63 \cdot 3^{x} \][/tex]
Now, incorporate this into the original expression:
[tex]\[ \frac{63 \cdot 3^{x}}{7 \cdot 3^{x+2}} = \frac{63}{7 \cdot 3^2} = \frac{63}{7 \cdot 9} = \frac{63}{63} = 1 \][/tex]
So, the simplified form is:
[tex]\[ \boxed{1} \][/tex]
