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Let's solve each question step-by-step:
### Question 1: Solve the equation [tex]\(\frac{2+i}{1-i} = i(x + y i)\)[/tex]
1. Simplify the left-hand side:
[tex]\[ \frac{2+i}{1-i} \][/tex]
To simplify, we'll multiply both the numerator and the denominator by the conjugate of the denominator, which is [tex]\(1 + i\)[/tex]:
[tex]\[ \frac{(2+i)(1+i)}{(1-i)(1+i)} \][/tex]
2. Perform the multiplication:
[tex]\[ (2+i)(1+i) = 2 \cdot 1 + 2 \cdot i + i \cdot 1 + i \cdot i = 2 + 2i + i + i^2 = 2 + 3i + (-1) = 1 + 3i \][/tex]
[tex]\[ (1-i)(1+i) = 1 \cdot 1 + 1 \cdot i - i \cdot 1 - i \cdot i = 1 + i - i - i^2 = 1 - (-1) = 1 + 1 = 2 \][/tex]
3. Thus, the simplified form is:
[tex]\[ \frac{1 + 3i}{2} = \frac{1}{2} + \frac{3i}{2} = \frac{1}{2} + \frac{3}{2}i \][/tex]
4. Now equate this to the given right-hand side:
[tex]\[ \frac{1}{2} + \frac{3}{2}i = i(x + y i) \][/tex]
5. Divide both sides by [tex]\(i\)[/tex]:
[tex]\[ i(x + y i) = i \cdot x + i^2 \cdot y = ix - y \][/tex]
So,
[tex]\[ \frac{1}{2} + \frac{3}{2}i = -y + ix \][/tex]
6. Separate real and imaginary parts:
[tex]\[ \frac{1}{2} = -y \][/tex]
[tex]\[ \frac{3}{2} = x \][/tex]
7. Solve for [tex]\(x\)[/tex] and [tex]\(y\)[/tex]:
[tex]\[ x = \frac{3}{2} \][/tex]
[tex]\[ y = -\frac{1}{2} \][/tex]
Thus, the solutions are:
[tex]\[ x = \frac{3}{2} \quad \text{and} \quad y = -\frac{1}{2} \][/tex]
---
### Question 2: Use binomial theorem to expand [tex]\((1 + 2x)^6\)[/tex] and evaluate [tex]\((0.96)^6\)[/tex]
1. The binomial theorem states:
[tex]\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \][/tex]
2. Expand [tex]\((1 + 2x)^6\)[/tex]:
Here, [tex]\(a = 1\)[/tex], [tex]\(b = 2x\)[/tex], and [tex]\(n = 6\)[/tex]:
[tex]\[ (1 + 2x)^6 = \sum_{k=0}^{6} \binom{6}{k} 1^{6-k} (2x)^k \][/tex]
[tex]\[ = \sum_{k=0}^{6} \binom{6}{k} (2x)^k \][/tex]
Expanding:
[tex]\[ = \binom{6}{0} (2x)^0 + \binom{6}{1} (2x)^1 + \binom{6}{2} (2x)^2 + \binom{6}{3} (2x)^3 + \binom{6}{4} (2x)^4 + \binom{6}{5} (2x)^5 + \binom{6}{6} (2x)^6 \][/tex]
[tex]\[ = 1 + 6(2x) + 15(2x)^2 + 20(2x)^3 + 15(2x)^4 + 6(2x)^5 + (2x)^6 \][/tex]
[tex]\[ = 1 + 12x + 60x^2 + 160x^3 + 240x^4 + 192x^5 + 64x^6 \][/tex]
3. Evaluate [tex]\((0.96)^6\)[/tex] using the binomial expansion:
[tex]\[ 0.96 = 1 - 0.04 \][/tex]
Let's use [tex]\(y = -0.04\)[/tex],
[tex]\[ (0.96)^6 = (1 + (-0.04))^6 = \sum_{k=0}^{6} \binom{6}{k} (1)^{6-k} (-0.04)^k \][/tex]
[tex]\[ = 1 + 6(-0.04) + 15(-0.04)^2 + 20(-0.04)^3 + 15(-0.04)^4 + 6(-0.04)^5 + (-0.04)^6 \][/tex]
[tex]\[ = 1 - 0.24 + 15(0.0016) - 20(0.000064) + 15(0.00000256) - 6(0.0000001024) + 0.00000001024 \][/tex]
Calculate each term:
[tex]\[ = 1 - 0.24 + 0.024 - 0.00128 + 0.0000384 - 0.0000006144 + 0.00000001024 \][/tex]
Sum these up:
[tex]\[ \approx 0.78376 \][/tex]
Thus, [tex]\((0.96)^6 \approx 0.78376\)[/tex]
---
### Question 3: Divide [tex]\(5x^4 + 3x^3 - 2x + 1\)[/tex] by [tex]\(x - 3\)[/tex]
1. Perform polynomial long division:
[tex]\[ \frac{5x^4 + 3x^3 - 2x + 1}{x - 3} \][/tex]
2. Step-by-step division:
[tex]\[ 5x^4 \div x = 5x^3 \implies 5x^3(x - 3) = 5x^4 - 15x^3 \][/tex]
Subtract:
[tex]\[ (5x^4 + 3x^3 - 2x + 1) - (5x^4 - 15x^3) = 18x^3 - 2x + 1 \][/tex]
Next term:
[tex]\[ 18x^3 \div x = 18x^2 \implies 18x^2(x - 3) = 18x^3 - 54x^2 \][/tex]
Subtract:
[tex]\[ (18x^3 - 2x + 1) - (18x^3 - 54x^2) = 54x^2 - 2x + 1 \][/tex]
Next term:
[tex]\[ 54x^2 \div x = 54x \implies 54x(x - 3) = 54x^2 - 162x \][/tex]
Subtract:
[tex]\[ (54x^2 - 2x + 1) - (54x^2 - 162x) = 160x + 1 \][/tex]
Next term:
[tex]\[ 160x \div x = 160 \implies 160(x - 3) = 160x - 480 \][/tex]
Subtract:
[tex]\[ (160x + 1) - (160x - 480) = 481 \][/tex]
3. Thus, the quotient and remainder are:
Quotient:
[tex]\[ 5x^3 + 18x^2 + 54x + 160 \][/tex]
Remainder:
[tex]\[ 481 \][/tex]
Hence, the division yields:
[tex]\[ \frac{5x^4 + 3x^3 - 2x + 1}{x - 3} = 5x^3 + 18x^2 + 54x + 160 + \frac{481}{x - 3} \][/tex]
### Question 1: Solve the equation [tex]\(\frac{2+i}{1-i} = i(x + y i)\)[/tex]
1. Simplify the left-hand side:
[tex]\[ \frac{2+i}{1-i} \][/tex]
To simplify, we'll multiply both the numerator and the denominator by the conjugate of the denominator, which is [tex]\(1 + i\)[/tex]:
[tex]\[ \frac{(2+i)(1+i)}{(1-i)(1+i)} \][/tex]
2. Perform the multiplication:
[tex]\[ (2+i)(1+i) = 2 \cdot 1 + 2 \cdot i + i \cdot 1 + i \cdot i = 2 + 2i + i + i^2 = 2 + 3i + (-1) = 1 + 3i \][/tex]
[tex]\[ (1-i)(1+i) = 1 \cdot 1 + 1 \cdot i - i \cdot 1 - i \cdot i = 1 + i - i - i^2 = 1 - (-1) = 1 + 1 = 2 \][/tex]
3. Thus, the simplified form is:
[tex]\[ \frac{1 + 3i}{2} = \frac{1}{2} + \frac{3i}{2} = \frac{1}{2} + \frac{3}{2}i \][/tex]
4. Now equate this to the given right-hand side:
[tex]\[ \frac{1}{2} + \frac{3}{2}i = i(x + y i) \][/tex]
5. Divide both sides by [tex]\(i\)[/tex]:
[tex]\[ i(x + y i) = i \cdot x + i^2 \cdot y = ix - y \][/tex]
So,
[tex]\[ \frac{1}{2} + \frac{3}{2}i = -y + ix \][/tex]
6. Separate real and imaginary parts:
[tex]\[ \frac{1}{2} = -y \][/tex]
[tex]\[ \frac{3}{2} = x \][/tex]
7. Solve for [tex]\(x\)[/tex] and [tex]\(y\)[/tex]:
[tex]\[ x = \frac{3}{2} \][/tex]
[tex]\[ y = -\frac{1}{2} \][/tex]
Thus, the solutions are:
[tex]\[ x = \frac{3}{2} \quad \text{and} \quad y = -\frac{1}{2} \][/tex]
---
### Question 2: Use binomial theorem to expand [tex]\((1 + 2x)^6\)[/tex] and evaluate [tex]\((0.96)^6\)[/tex]
1. The binomial theorem states:
[tex]\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \][/tex]
2. Expand [tex]\((1 + 2x)^6\)[/tex]:
Here, [tex]\(a = 1\)[/tex], [tex]\(b = 2x\)[/tex], and [tex]\(n = 6\)[/tex]:
[tex]\[ (1 + 2x)^6 = \sum_{k=0}^{6} \binom{6}{k} 1^{6-k} (2x)^k \][/tex]
[tex]\[ = \sum_{k=0}^{6} \binom{6}{k} (2x)^k \][/tex]
Expanding:
[tex]\[ = \binom{6}{0} (2x)^0 + \binom{6}{1} (2x)^1 + \binom{6}{2} (2x)^2 + \binom{6}{3} (2x)^3 + \binom{6}{4} (2x)^4 + \binom{6}{5} (2x)^5 + \binom{6}{6} (2x)^6 \][/tex]
[tex]\[ = 1 + 6(2x) + 15(2x)^2 + 20(2x)^3 + 15(2x)^4 + 6(2x)^5 + (2x)^6 \][/tex]
[tex]\[ = 1 + 12x + 60x^2 + 160x^3 + 240x^4 + 192x^5 + 64x^6 \][/tex]
3. Evaluate [tex]\((0.96)^6\)[/tex] using the binomial expansion:
[tex]\[ 0.96 = 1 - 0.04 \][/tex]
Let's use [tex]\(y = -0.04\)[/tex],
[tex]\[ (0.96)^6 = (1 + (-0.04))^6 = \sum_{k=0}^{6} \binom{6}{k} (1)^{6-k} (-0.04)^k \][/tex]
[tex]\[ = 1 + 6(-0.04) + 15(-0.04)^2 + 20(-0.04)^3 + 15(-0.04)^4 + 6(-0.04)^5 + (-0.04)^6 \][/tex]
[tex]\[ = 1 - 0.24 + 15(0.0016) - 20(0.000064) + 15(0.00000256) - 6(0.0000001024) + 0.00000001024 \][/tex]
Calculate each term:
[tex]\[ = 1 - 0.24 + 0.024 - 0.00128 + 0.0000384 - 0.0000006144 + 0.00000001024 \][/tex]
Sum these up:
[tex]\[ \approx 0.78376 \][/tex]
Thus, [tex]\((0.96)^6 \approx 0.78376\)[/tex]
---
### Question 3: Divide [tex]\(5x^4 + 3x^3 - 2x + 1\)[/tex] by [tex]\(x - 3\)[/tex]
1. Perform polynomial long division:
[tex]\[ \frac{5x^4 + 3x^3 - 2x + 1}{x - 3} \][/tex]
2. Step-by-step division:
[tex]\[ 5x^4 \div x = 5x^3 \implies 5x^3(x - 3) = 5x^4 - 15x^3 \][/tex]
Subtract:
[tex]\[ (5x^4 + 3x^3 - 2x + 1) - (5x^4 - 15x^3) = 18x^3 - 2x + 1 \][/tex]
Next term:
[tex]\[ 18x^3 \div x = 18x^2 \implies 18x^2(x - 3) = 18x^3 - 54x^2 \][/tex]
Subtract:
[tex]\[ (18x^3 - 2x + 1) - (18x^3 - 54x^2) = 54x^2 - 2x + 1 \][/tex]
Next term:
[tex]\[ 54x^2 \div x = 54x \implies 54x(x - 3) = 54x^2 - 162x \][/tex]
Subtract:
[tex]\[ (54x^2 - 2x + 1) - (54x^2 - 162x) = 160x + 1 \][/tex]
Next term:
[tex]\[ 160x \div x = 160 \implies 160(x - 3) = 160x - 480 \][/tex]
Subtract:
[tex]\[ (160x + 1) - (160x - 480) = 481 \][/tex]
3. Thus, the quotient and remainder are:
Quotient:
[tex]\[ 5x^3 + 18x^2 + 54x + 160 \][/tex]
Remainder:
[tex]\[ 481 \][/tex]
Hence, the division yields:
[tex]\[ \frac{5x^4 + 3x^3 - 2x + 1}{x - 3} = 5x^3 + 18x^2 + 54x + 160 + \frac{481}{x - 3} \][/tex]
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