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Let's go through each of the questions and solve them step-by-step.
### Question 4
Which of the following is the simplified form of
[tex]\[ \left[x-\frac{x+3}{x-1}\right] \div \frac{x-3}{x^2-2x+1} \][/tex]
First, rewrite the original expression in a more workable form:
[tex]\[ \frac{x - \frac{x+3}{x-1}}{\frac{x-3}{x^2-2x+1}} \][/tex]
First, let's simplify the numerator:
[tex]\[ x - \frac{x+3}{x-1} = \frac{x(x-1) - (x+3)}{x-1} = \frac{x^2 - x - x - 3}{x-1} = \frac{x^2 - 2x - 3}{x-1} \][/tex]
Next, notice that [tex]\(x^2 - 2x + 1\)[/tex] can be factored:
[tex]\[ x^2 - 2x + 1 = (x-1)^2 \][/tex]
Thus the denominator becomes:
[tex]\[ \frac{x-3}{x^2 - 2x + 1} = \frac{x-3}{(x-1)^2} \][/tex]
Now, rewrite the original expression again:
[tex]\[ \frac{\frac{x^2 - 2x - 3}{x-1}}{\frac{x-3}{(x-1)^2}} = \frac{x^2 - 2x - 3}{x-1} \cdot \frac{(x-1)^2}{x-3} = \frac{(x^2 - 2x - 3) \cdot (x-1)}{(x-1) \cdot (x-3)} \][/tex]
Simplify the fraction:
[tex]\[ = \frac{(x-1)(x+1)}{1} = x^2-1 \][/tex]
So, the simplified form is:
[tex]\[ x^2-1, \quad x \neq 1, 3 \][/tex]
The answer is:
[tex]\[ \boxed{D. x^2-1,\, x \neq 1,3} \][/tex]
### Question 5
If [tex]\(w = i - \frac{3 + i}{5}\)[/tex], then which of the following is equal to the modulus of [tex]\(w\)[/tex]?
Let's simplify [tex]\(w\)[/tex]:
[tex]\[ w = i - \frac{3 + i}{5} = i - \left( \frac{3}{5} + \frac{i}{5} \right) = i - \frac{3}{5} - \frac{i}{5} = \left(1 - \frac{1}{5}\right)i - \frac{3}{5} = \frac{4i}{5} - \frac{3}{5} \][/tex]
The modulus of a complex number [tex]\(a + bi\)[/tex] is given by:
[tex]\[ |a + bi| = \sqrt{a^2 + b^2} \][/tex]
Here, [tex]\(a = -\frac{3}{5}\)[/tex] and [tex]\(b = \frac{4}{5}\)[/tex]:
[tex]\[ |w| = \sqrt{\left(-\frac{3}{5}\right)^2 + \left(\frac{4}{5}\right)^2} = \sqrt{\frac{9}{25} + \frac{16}{25}} = \sqrt{\frac{25}{25}} = \sqrt{1} = 1 \][/tex]
The answer is:
[tex]\[ \boxed{A. 1} \][/tex]
### Question 6
In a homogeneous population where there is a categorical difference, items are selected randomly in each category. Which of the following describes the sampling method stated above?
The method described involves categorizing the population and randomly selecting items within each category. This is known as:
\boxed{D. Stratified sampling}
### Question 7
The solution set of the system
[tex]\[ \left\{\begin{array}{l} 2x - y = 5 \\ 3y + 15 = 6x \end{array}\right. \][/tex]
First, rewrite the second equation:
[tex]\[ 3y + 15 = 6x \quad \Rightarrow \quad 3y = 6x - 15 \quad \Rightarrow \quad y = 2x - 5 \][/tex]
Substitute [tex]\( y = 2x - 5 \)[/tex] into the first equation:
[tex]\[ 2x - (2x - 5) = 5 \][/tex]
[tex]\[ 2x - 2x + 5 = 5 \][/tex]
[tex]\[ 5 = 5 \][/tex]
This is a true statement for all [tex]\(x\)[/tex]. Hence, [tex]\(x\)[/tex] can be any real number, and [tex]\(y = 2x - 5\)[/tex].
The solution set is:
[tex]\[ \boxed{D. \{(t, 2t-5): t \in R\}} \][/tex]
### Question 4
Which of the following is the simplified form of
[tex]\[ \left[x-\frac{x+3}{x-1}\right] \div \frac{x-3}{x^2-2x+1} \][/tex]
First, rewrite the original expression in a more workable form:
[tex]\[ \frac{x - \frac{x+3}{x-1}}{\frac{x-3}{x^2-2x+1}} \][/tex]
First, let's simplify the numerator:
[tex]\[ x - \frac{x+3}{x-1} = \frac{x(x-1) - (x+3)}{x-1} = \frac{x^2 - x - x - 3}{x-1} = \frac{x^2 - 2x - 3}{x-1} \][/tex]
Next, notice that [tex]\(x^2 - 2x + 1\)[/tex] can be factored:
[tex]\[ x^2 - 2x + 1 = (x-1)^2 \][/tex]
Thus the denominator becomes:
[tex]\[ \frac{x-3}{x^2 - 2x + 1} = \frac{x-3}{(x-1)^2} \][/tex]
Now, rewrite the original expression again:
[tex]\[ \frac{\frac{x^2 - 2x - 3}{x-1}}{\frac{x-3}{(x-1)^2}} = \frac{x^2 - 2x - 3}{x-1} \cdot \frac{(x-1)^2}{x-3} = \frac{(x^2 - 2x - 3) \cdot (x-1)}{(x-1) \cdot (x-3)} \][/tex]
Simplify the fraction:
[tex]\[ = \frac{(x-1)(x+1)}{1} = x^2-1 \][/tex]
So, the simplified form is:
[tex]\[ x^2-1, \quad x \neq 1, 3 \][/tex]
The answer is:
[tex]\[ \boxed{D. x^2-1,\, x \neq 1,3} \][/tex]
### Question 5
If [tex]\(w = i - \frac{3 + i}{5}\)[/tex], then which of the following is equal to the modulus of [tex]\(w\)[/tex]?
Let's simplify [tex]\(w\)[/tex]:
[tex]\[ w = i - \frac{3 + i}{5} = i - \left( \frac{3}{5} + \frac{i}{5} \right) = i - \frac{3}{5} - \frac{i}{5} = \left(1 - \frac{1}{5}\right)i - \frac{3}{5} = \frac{4i}{5} - \frac{3}{5} \][/tex]
The modulus of a complex number [tex]\(a + bi\)[/tex] is given by:
[tex]\[ |a + bi| = \sqrt{a^2 + b^2} \][/tex]
Here, [tex]\(a = -\frac{3}{5}\)[/tex] and [tex]\(b = \frac{4}{5}\)[/tex]:
[tex]\[ |w| = \sqrt{\left(-\frac{3}{5}\right)^2 + \left(\frac{4}{5}\right)^2} = \sqrt{\frac{9}{25} + \frac{16}{25}} = \sqrt{\frac{25}{25}} = \sqrt{1} = 1 \][/tex]
The answer is:
[tex]\[ \boxed{A. 1} \][/tex]
### Question 6
In a homogeneous population where there is a categorical difference, items are selected randomly in each category. Which of the following describes the sampling method stated above?
The method described involves categorizing the population and randomly selecting items within each category. This is known as:
\boxed{D. Stratified sampling}
### Question 7
The solution set of the system
[tex]\[ \left\{\begin{array}{l} 2x - y = 5 \\ 3y + 15 = 6x \end{array}\right. \][/tex]
First, rewrite the second equation:
[tex]\[ 3y + 15 = 6x \quad \Rightarrow \quad 3y = 6x - 15 \quad \Rightarrow \quad y = 2x - 5 \][/tex]
Substitute [tex]\( y = 2x - 5 \)[/tex] into the first equation:
[tex]\[ 2x - (2x - 5) = 5 \][/tex]
[tex]\[ 2x - 2x + 5 = 5 \][/tex]
[tex]\[ 5 = 5 \][/tex]
This is a true statement for all [tex]\(x\)[/tex]. Hence, [tex]\(x\)[/tex] can be any real number, and [tex]\(y = 2x - 5\)[/tex].
The solution set is:
[tex]\[ \boxed{D. \{(t, 2t-5): t \in R\}} \][/tex]
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