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Sure, let's go through each of these questions step-by-step:
### Question 2
Which is the multiplicative identity for all rational numbers?
The multiplicative identity for all rational numbers is [tex]\(1\)[/tex]. This is because any rational number multiplied by 1 gives the rational number itself. [tex]\( \frac{a}{b} \times 1 = \frac{a}{b} \)[/tex].
### Question 3
Identify the property used in the following: [tex]\(-\frac{3}{4}+\frac{5}{2}=\frac{5}{2}+\left(-\frac{3}{4}\right)\)[/tex]
The property used here is the Commutative Property of Addition. This property states that the order in which two numbers are added does not affect the sum. So, [tex]\( a + b = b + a \)[/tex].
### Question 4
List five rational numbers between 1 and [tex]\(\frac{1}{5}\)[/tex].
Five rational numbers between 1 and [tex]\(\frac{1}{5}\)[/tex] are:
[tex]\[ 0.5, 0.3333333333333333, 0.25, 0.16666666666666666, 0.14285714285714285 \][/tex]
### Question 5
Show with the help of an example that subtraction of rational numbers is not associative.
Let's take rational numbers [tex]\(\frac{3}{4}\)[/tex], [tex]\(\frac{1}{2}\)[/tex], and [tex]\(\frac{1}{4}\)[/tex].
Calculate [tex]\((\frac{3}{4} - \frac{1}{2}) - \frac{1}{4}\)[/tex]:
1. [tex]\(\frac{3}{4} - \frac{1}{2} = \frac{3}{4} - \frac{2}{4} = \frac{1}{4}\)[/tex]
2. [tex]\(\frac{1}{4} - \frac{1}{4} = 0\)[/tex]
Now, calculate [tex]\(\frac{3}{4} - (\frac{1}{2} - \frac{1}{4})\)[/tex]:
1. [tex]\(\frac{1}{2} - \frac{1}{4} = \frac{2}{4} - \frac{1}{4} = \frac{1}{4}\)[/tex]
2. [tex]\(\frac{3}{4} - \frac{1}{4} = \frac{2}{4} = \frac{1}{2}\)[/tex]
Since [tex]\( 0 \neq \frac{1}{2} \)[/tex], subtraction of rational numbers is not associative.
### Question 6
Verify that [tex]\(-(-x)\)[/tex] is same as [tex]\(x\)[/tex] for [tex]\(x = \frac{15}{16}\)[/tex].
Let's substitute [tex]\(x = \frac{15}{16}\)[/tex]:
1. Calculate [tex]\(-(-\frac{15}{16})\)[/tex]
2. The double negative makes it positive: [tex]\(-(-\frac{15}{16}) = \frac{15}{16}\)[/tex]
Thus, [tex]\(-(-x) = x\)[/tex].
### Question 7
Find the additive inverse of the following:
(a) [tex]\(\left(-\frac{1}{3} x - \frac{2}{5}\right)\)[/tex]
The additive inverse of [tex]\(\left(-\frac{1}{3} x - \frac{2}{5}\right)\)[/tex] is what you add to it to get 0. This is simply changing all the signs:
[tex]\(\frac{1}{3} x + \frac{2}{5}\)[/tex].
(b) [tex]\(-3 \times \frac{4}{9}\)[/tex]
First, calculate [tex]\(-3 \times \frac{4}{9}\)[/tex]:
[tex]\[-3 \times \frac{4}{9} = -\frac{12}{9} = -\frac{4}{3}\][/tex]
The additive inverse of [tex]\(-\frac{4}{3}\)[/tex] is:
[tex]\(\frac{4}{3}\)[/tex].
### Question 8
Multiply [tex]\(\frac{7}{15}\)[/tex] by the reciprocal of [tex]\(-\frac{11}{30}\)[/tex] and add the result to the reciprocal of [tex]\(\frac{44}{25}\)[/tex]. Write the reciprocal of the result you get.
1. The reciprocal of [tex]\(-\frac{11}{30}\)[/tex] is [tex]\(-\frac{30}{11}\)[/tex].
2. Multiply [tex]\(\frac{7}{15}\)[/tex] by [tex]\(-\frac{30}{11}\)[/tex]:
[tex]\[ \frac{7}{15} \times -\frac{30}{11} = \frac{7 \times -30}{15 \times 11} = -\frac{210}{165} = -\frac{14}{11} \][/tex]
3. The reciprocal of [tex]\(\frac{44}{25}\)[/tex] is [tex]\(\frac{25}{44}\)[/tex].
4. Add [tex]\(-\frac{14}{11}\)[/tex] to [tex]\(\frac{25}{44}\)[/tex]:
First, find a common denominator (44).
[tex]\[ -\frac{14}{11} = -\frac{14 \times 4}{11 \times 4} = -\frac{56}{44} \][/tex]
Now add:
[tex]\[ -\frac{56}{44} + \frac{25}{44} = \frac{-56 + 25}{44} = \frac{-31}{44} \][/tex]
5. The reciprocal of [tex]\(\frac{-31}{44}\)[/tex] is:
[tex]\[ \frac{44}{-31} = -\frac{44}{31} \][/tex]
### Question 2
Which is the multiplicative identity for all rational numbers?
The multiplicative identity for all rational numbers is [tex]\(1\)[/tex]. This is because any rational number multiplied by 1 gives the rational number itself. [tex]\( \frac{a}{b} \times 1 = \frac{a}{b} \)[/tex].
### Question 3
Identify the property used in the following: [tex]\(-\frac{3}{4}+\frac{5}{2}=\frac{5}{2}+\left(-\frac{3}{4}\right)\)[/tex]
The property used here is the Commutative Property of Addition. This property states that the order in which two numbers are added does not affect the sum. So, [tex]\( a + b = b + a \)[/tex].
### Question 4
List five rational numbers between 1 and [tex]\(\frac{1}{5}\)[/tex].
Five rational numbers between 1 and [tex]\(\frac{1}{5}\)[/tex] are:
[tex]\[ 0.5, 0.3333333333333333, 0.25, 0.16666666666666666, 0.14285714285714285 \][/tex]
### Question 5
Show with the help of an example that subtraction of rational numbers is not associative.
Let's take rational numbers [tex]\(\frac{3}{4}\)[/tex], [tex]\(\frac{1}{2}\)[/tex], and [tex]\(\frac{1}{4}\)[/tex].
Calculate [tex]\((\frac{3}{4} - \frac{1}{2}) - \frac{1}{4}\)[/tex]:
1. [tex]\(\frac{3}{4} - \frac{1}{2} = \frac{3}{4} - \frac{2}{4} = \frac{1}{4}\)[/tex]
2. [tex]\(\frac{1}{4} - \frac{1}{4} = 0\)[/tex]
Now, calculate [tex]\(\frac{3}{4} - (\frac{1}{2} - \frac{1}{4})\)[/tex]:
1. [tex]\(\frac{1}{2} - \frac{1}{4} = \frac{2}{4} - \frac{1}{4} = \frac{1}{4}\)[/tex]
2. [tex]\(\frac{3}{4} - \frac{1}{4} = \frac{2}{4} = \frac{1}{2}\)[/tex]
Since [tex]\( 0 \neq \frac{1}{2} \)[/tex], subtraction of rational numbers is not associative.
### Question 6
Verify that [tex]\(-(-x)\)[/tex] is same as [tex]\(x\)[/tex] for [tex]\(x = \frac{15}{16}\)[/tex].
Let's substitute [tex]\(x = \frac{15}{16}\)[/tex]:
1. Calculate [tex]\(-(-\frac{15}{16})\)[/tex]
2. The double negative makes it positive: [tex]\(-(-\frac{15}{16}) = \frac{15}{16}\)[/tex]
Thus, [tex]\(-(-x) = x\)[/tex].
### Question 7
Find the additive inverse of the following:
(a) [tex]\(\left(-\frac{1}{3} x - \frac{2}{5}\right)\)[/tex]
The additive inverse of [tex]\(\left(-\frac{1}{3} x - \frac{2}{5}\right)\)[/tex] is what you add to it to get 0. This is simply changing all the signs:
[tex]\(\frac{1}{3} x + \frac{2}{5}\)[/tex].
(b) [tex]\(-3 \times \frac{4}{9}\)[/tex]
First, calculate [tex]\(-3 \times \frac{4}{9}\)[/tex]:
[tex]\[-3 \times \frac{4}{9} = -\frac{12}{9} = -\frac{4}{3}\][/tex]
The additive inverse of [tex]\(-\frac{4}{3}\)[/tex] is:
[tex]\(\frac{4}{3}\)[/tex].
### Question 8
Multiply [tex]\(\frac{7}{15}\)[/tex] by the reciprocal of [tex]\(-\frac{11}{30}\)[/tex] and add the result to the reciprocal of [tex]\(\frac{44}{25}\)[/tex]. Write the reciprocal of the result you get.
1. The reciprocal of [tex]\(-\frac{11}{30}\)[/tex] is [tex]\(-\frac{30}{11}\)[/tex].
2. Multiply [tex]\(\frac{7}{15}\)[/tex] by [tex]\(-\frac{30}{11}\)[/tex]:
[tex]\[ \frac{7}{15} \times -\frac{30}{11} = \frac{7 \times -30}{15 \times 11} = -\frac{210}{165} = -\frac{14}{11} \][/tex]
3. The reciprocal of [tex]\(\frac{44}{25}\)[/tex] is [tex]\(\frac{25}{44}\)[/tex].
4. Add [tex]\(-\frac{14}{11}\)[/tex] to [tex]\(\frac{25}{44}\)[/tex]:
First, find a common denominator (44).
[tex]\[ -\frac{14}{11} = -\frac{14 \times 4}{11 \times 4} = -\frac{56}{44} \][/tex]
Now add:
[tex]\[ -\frac{56}{44} + \frac{25}{44} = \frac{-56 + 25}{44} = \frac{-31}{44} \][/tex]
5. The reciprocal of [tex]\(\frac{-31}{44}\)[/tex] is:
[tex]\[ \frac{44}{-31} = -\frac{44}{31} \][/tex]
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