Answer :
Let's solve the given problem step-by-step.
### (a) What is the product of [tex]\(a^m\)[/tex] and [tex]\(a^n\)[/tex]? [1K]
To find the product of [tex]\(a^m\)[/tex] and [tex]\(a^n\)[/tex] where [tex]\(a\)[/tex] is a base and [tex]\(m\)[/tex], [tex]\(n\)[/tex] are exponents, we use the property of exponents:
[tex]\[ a^m \times a^n = a^{m+n} \][/tex]
This property states that when multiplying two expressions with the same base, we add the exponents.
### (b) Find the product. [2U]
We need to multiply [tex]\((3.4 \times 10^5)\)[/tex] and [tex]\((5.1 \times 10^{-7})\)[/tex]. To do this, we multiply the coefficients and add the exponents:
[tex]\[ 3.4 \times 5.1 = 17.34 \][/tex]
[tex]\[ 10^5 \times 10^{-7} = 10^{5 + (-7)} = 10^{-2} \][/tex]
Thus, the product is:
[tex]\[ 17.34 \times 10^{-2} = 0.1734 \][/tex]
### (c) Calculate the quotient. [1A]
Now we need to divide the product obtained in part (b) by [tex]\(1.7 \times 10^{-4}\)[/tex]:
[tex]\[ \frac{0.1734}{1.7 \times 10^{-4}} \][/tex]
First, divide the coefficients:
[tex]\[ \frac{0.1734}{1.7} = 0.102 \][/tex]
Then, adjust for the exponents since [tex]\(10^{0} / 10^{-4} = 10^{0 - (-4)} = 10^4\)[/tex]:
[tex]\[ 0.102 \times 10^4 = 1020 \][/tex]
So, the quotient is:
[tex]\[ 1020 \][/tex]
### (d) Convert 6800 into the quinary number system. [2U]
To convert 6800 to the quinary (base-5) number system, we repeatedly divide by 5 and record the remainders:
1. [tex]\(6800 \div 5 = 1360\)[/tex] remainder [tex]\(0\)[/tex]
2. [tex]\(1360 \div 5 = 272\)[/tex] remainder [tex]\(0\)[/tex]
3. [tex]\(272 \div 5 = 54\)[/tex] remainder [tex]\(2\)[/tex]
4. [tex]\(54 \div 5 = 10.8\)[/tex] remainder [tex]\(4\)[/tex]
5. [tex]\(10 \div 5 = 2\)[/tex] remainder [tex]\(0\)[/tex]
6. [tex]\(2 \div 5 = 0.4\)[/tex] remainder [tex]\(2\)[/tex]
Reading the remainders from bottom to top, we get the quinary number:
[tex]\[ 204200_5 \][/tex]
### (e) Convert 51 into the binary number system. [1U]
To convert 51 to the binary (base-2) number system, we repeatedly divide by 2 and record the remainders:
1. [tex]\(51 \div 2 = 25\)[/tex] remainder [tex]\(1\)[/tex]
2. [tex]\(25 \div 2 = 12\)[/tex] remainder [tex]\(1\)[/tex]
3. [tex]\(12 \div 2 = 6\)[/tex] remainder [tex]\(0\)[/tex]
4. [tex]\(6 \div 2 = 3\)[/tex] remainder [tex]\(0\)[/tex]
5. [tex]\(3 \div 2 = 1\)[/tex] remainder [tex]\(1\)[/tex]
6. [tex]\(1 \div 2 = 0.5\)[/tex] remainder [tex]\(1\)[/tex]
Reading the remainders from bottom to top, we get the binary number:
[tex]\[ 110011_2 \][/tex]
### Final Results:
1. (a) [tex]\(a^m \times a^n = a^{m+n}\)[/tex]
2. (b) [tex]\(0.1734\)[/tex]
3. (c) [tex]\(1020\)[/tex]
4. (d) [tex]\(204200_5\)[/tex]
5. (e) [tex]\(110011_2\)[/tex]
### (a) What is the product of [tex]\(a^m\)[/tex] and [tex]\(a^n\)[/tex]? [1K]
To find the product of [tex]\(a^m\)[/tex] and [tex]\(a^n\)[/tex] where [tex]\(a\)[/tex] is a base and [tex]\(m\)[/tex], [tex]\(n\)[/tex] are exponents, we use the property of exponents:
[tex]\[ a^m \times a^n = a^{m+n} \][/tex]
This property states that when multiplying two expressions with the same base, we add the exponents.
### (b) Find the product. [2U]
We need to multiply [tex]\((3.4 \times 10^5)\)[/tex] and [tex]\((5.1 \times 10^{-7})\)[/tex]. To do this, we multiply the coefficients and add the exponents:
[tex]\[ 3.4 \times 5.1 = 17.34 \][/tex]
[tex]\[ 10^5 \times 10^{-7} = 10^{5 + (-7)} = 10^{-2} \][/tex]
Thus, the product is:
[tex]\[ 17.34 \times 10^{-2} = 0.1734 \][/tex]
### (c) Calculate the quotient. [1A]
Now we need to divide the product obtained in part (b) by [tex]\(1.7 \times 10^{-4}\)[/tex]:
[tex]\[ \frac{0.1734}{1.7 \times 10^{-4}} \][/tex]
First, divide the coefficients:
[tex]\[ \frac{0.1734}{1.7} = 0.102 \][/tex]
Then, adjust for the exponents since [tex]\(10^{0} / 10^{-4} = 10^{0 - (-4)} = 10^4\)[/tex]:
[tex]\[ 0.102 \times 10^4 = 1020 \][/tex]
So, the quotient is:
[tex]\[ 1020 \][/tex]
### (d) Convert 6800 into the quinary number system. [2U]
To convert 6800 to the quinary (base-5) number system, we repeatedly divide by 5 and record the remainders:
1. [tex]\(6800 \div 5 = 1360\)[/tex] remainder [tex]\(0\)[/tex]
2. [tex]\(1360 \div 5 = 272\)[/tex] remainder [tex]\(0\)[/tex]
3. [tex]\(272 \div 5 = 54\)[/tex] remainder [tex]\(2\)[/tex]
4. [tex]\(54 \div 5 = 10.8\)[/tex] remainder [tex]\(4\)[/tex]
5. [tex]\(10 \div 5 = 2\)[/tex] remainder [tex]\(0\)[/tex]
6. [tex]\(2 \div 5 = 0.4\)[/tex] remainder [tex]\(2\)[/tex]
Reading the remainders from bottom to top, we get the quinary number:
[tex]\[ 204200_5 \][/tex]
### (e) Convert 51 into the binary number system. [1U]
To convert 51 to the binary (base-2) number system, we repeatedly divide by 2 and record the remainders:
1. [tex]\(51 \div 2 = 25\)[/tex] remainder [tex]\(1\)[/tex]
2. [tex]\(25 \div 2 = 12\)[/tex] remainder [tex]\(1\)[/tex]
3. [tex]\(12 \div 2 = 6\)[/tex] remainder [tex]\(0\)[/tex]
4. [tex]\(6 \div 2 = 3\)[/tex] remainder [tex]\(0\)[/tex]
5. [tex]\(3 \div 2 = 1\)[/tex] remainder [tex]\(1\)[/tex]
6. [tex]\(1 \div 2 = 0.5\)[/tex] remainder [tex]\(1\)[/tex]
Reading the remainders from bottom to top, we get the binary number:
[tex]\[ 110011_2 \][/tex]
### Final Results:
1. (a) [tex]\(a^m \times a^n = a^{m+n}\)[/tex]
2. (b) [tex]\(0.1734\)[/tex]
3. (c) [tex]\(1020\)[/tex]
4. (d) [tex]\(204200_5\)[/tex]
5. (e) [tex]\(110011_2\)[/tex]
