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### Question 1: Quadratic Equation Roots Problem
Given the quadratic equation:
[tex]\[4x^2 - 2kx + k - 1 = 0\][/tex]
We know that the product of the roots is [tex]\(-\frac{3}{2}\)[/tex].
For a quadratic equation [tex]\(ax^2 + bx + c = 0\)[/tex]:
- The sum of the roots ([tex]\(\alpha + \beta\)[/tex]) is given by [tex]\(-\frac{b}{a}\)[/tex].
- The product of the roots ([tex]\(\alpha \beta\)[/tex]) is given by [tex]\(\frac{c}{a}\)[/tex].
Here, [tex]\(a = 4\)[/tex], [tex]\(b = -2k\)[/tex], and [tex]\(c = k - 1\)[/tex].
According to the problem, the product of the roots [tex]\(\alpha \beta = \frac{c}{a} = -\frac{3}{2}\)[/tex]:
[tex]\[ \frac{k - 1}{4} = -\frac{3}{2} \][/tex]
Solving for [tex]\(k\)[/tex]:
[tex]\[ k - 1 = -6 \implies k = -5 \][/tex]
With [tex]\(k = -5\)[/tex], we can find the sum of the roots:
[tex]\[ \alpha + \beta = -\frac{b}{a} = -\frac{-2k}{4} = -\frac{-2(-5)}{4} = \frac{10}{4} = 2.5 = \frac{5}{2} \][/tex]
So, the sum of the roots is [tex]\(\frac{5}{2}\)[/tex].
Answer: C. [tex]\(\frac{5}{2}\)[/tex]
### Question 2: Probability of Sum Less Than 5
Let's calculate the possible outcomes when a regular triangular pyramid labeled 1, 2, and 3 is thrown twice.
We will list all the possible outcomes:
- (1, 1), (1, 2), (1, 3)
- (2, 1), (2, 2), (2, 3)
- (3, 1), (3, 2), (3, 3)
Next, calculate the sum for each paired outcome and determine how many sums are less than 5:
- Sum = 2: (1, 1)
- Sum = 3: (1, 2), (2, 1)
- Sum = 4: (1, 3), (2, 2), (3, 1)
- Sum = 5: (2, 3), (3, 2)
- Sum = 6: (3, 3)
Outcomes with sums less than 5: (1, 1), (1, 2), (2, 1), (1, 3), (2, 2), (3, 1)
- There are 6 favorable outcomes.
- Total possible outcomes = 9.
Probability:
[tex]\[ \text{Probability} = \frac{6}{9} = \frac{2}{3} \][/tex]
Answer: A. [tex]\(\frac{2}{3}\)[/tex]
### Question 3: Value of k for Line Passing Through Points
Given points [tex]\(P\left(-\frac{1}{2}, 3\right)\)[/tex] and [tex]\(Q(4, -5)\)[/tex] on the line [tex]\(y = kx + b\)[/tex].
To find the slope [tex]\(k\)[/tex]:
[tex]\[ k = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]
Given [tex]\(P\left(x_1 = -\frac{1}{2}, y_1 = 3\right)\)[/tex] and [tex]\(Q(x_2 = 4, y_2 = -5)\)[/tex]:
[tex]\[ k = \frac{-5 - 3}{4 - (-\frac{1}{2})} = \frac{-8}{4 + \frac{1}{2}} = \frac{-8}{\frac{9}{2}} = -\frac{8 \times 2}{9} = -\frac{16}{9} \][/tex]
Answer: C. [tex]\(-\frac{16}{9}\)[/tex]
Given the quadratic equation:
[tex]\[4x^2 - 2kx + k - 1 = 0\][/tex]
We know that the product of the roots is [tex]\(-\frac{3}{2}\)[/tex].
For a quadratic equation [tex]\(ax^2 + bx + c = 0\)[/tex]:
- The sum of the roots ([tex]\(\alpha + \beta\)[/tex]) is given by [tex]\(-\frac{b}{a}\)[/tex].
- The product of the roots ([tex]\(\alpha \beta\)[/tex]) is given by [tex]\(\frac{c}{a}\)[/tex].
Here, [tex]\(a = 4\)[/tex], [tex]\(b = -2k\)[/tex], and [tex]\(c = k - 1\)[/tex].
According to the problem, the product of the roots [tex]\(\alpha \beta = \frac{c}{a} = -\frac{3}{2}\)[/tex]:
[tex]\[ \frac{k - 1}{4} = -\frac{3}{2} \][/tex]
Solving for [tex]\(k\)[/tex]:
[tex]\[ k - 1 = -6 \implies k = -5 \][/tex]
With [tex]\(k = -5\)[/tex], we can find the sum of the roots:
[tex]\[ \alpha + \beta = -\frac{b}{a} = -\frac{-2k}{4} = -\frac{-2(-5)}{4} = \frac{10}{4} = 2.5 = \frac{5}{2} \][/tex]
So, the sum of the roots is [tex]\(\frac{5}{2}\)[/tex].
Answer: C. [tex]\(\frac{5}{2}\)[/tex]
### Question 2: Probability of Sum Less Than 5
Let's calculate the possible outcomes when a regular triangular pyramid labeled 1, 2, and 3 is thrown twice.
We will list all the possible outcomes:
- (1, 1), (1, 2), (1, 3)
- (2, 1), (2, 2), (2, 3)
- (3, 1), (3, 2), (3, 3)
Next, calculate the sum for each paired outcome and determine how many sums are less than 5:
- Sum = 2: (1, 1)
- Sum = 3: (1, 2), (2, 1)
- Sum = 4: (1, 3), (2, 2), (3, 1)
- Sum = 5: (2, 3), (3, 2)
- Sum = 6: (3, 3)
Outcomes with sums less than 5: (1, 1), (1, 2), (2, 1), (1, 3), (2, 2), (3, 1)
- There are 6 favorable outcomes.
- Total possible outcomes = 9.
Probability:
[tex]\[ \text{Probability} = \frac{6}{9} = \frac{2}{3} \][/tex]
Answer: A. [tex]\(\frac{2}{3}\)[/tex]
### Question 3: Value of k for Line Passing Through Points
Given points [tex]\(P\left(-\frac{1}{2}, 3\right)\)[/tex] and [tex]\(Q(4, -5)\)[/tex] on the line [tex]\(y = kx + b\)[/tex].
To find the slope [tex]\(k\)[/tex]:
[tex]\[ k = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]
Given [tex]\(P\left(x_1 = -\frac{1}{2}, y_1 = 3\right)\)[/tex] and [tex]\(Q(x_2 = 4, y_2 = -5)\)[/tex]:
[tex]\[ k = \frac{-5 - 3}{4 - (-\frac{1}{2})} = \frac{-8}{4 + \frac{1}{2}} = \frac{-8}{\frac{9}{2}} = -\frac{8 \times 2}{9} = -\frac{16}{9} \][/tex]
Answer: C. [tex]\(-\frac{16}{9}\)[/tex]
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