Discover the answers to your questions at Westonci.ca, where experts share their knowledge and insights with you. Join our Q&A platform to connect with experts dedicated to providing precise answers to your questions in different areas. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.
Sagot :
Absolutely, let’s delve into solving the problems step by step:
### Problem 74
The question requires us to find the length of the perpendicular from the point \((0,0)\) to the line \(3x + 4y - 10 = 0\).
We can use the formula for the distance \(d\) from a point \((x_0, y_0)\) to a line \(ax + by + c = 0\), which is given by:
[tex]\[ d = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}} \][/tex]
1. Substitute \((x_0, y_0) = (0,0)\) into the formula:
[tex]\[ d = \frac{|3 \cdot 0 + 4 \cdot 0 - 10|}{\sqrt{3^2 + 4^2}} \][/tex]
2. Simplify the expression in the numerator:
[tex]\[ d = \frac{|0 + 0 - 10|}{\sqrt{9 + 16}} = \frac{|-10|}{\sqrt{25}} = \frac{10}{5} = 2 \][/tex]
So, the length of the perpendicular from \((0,0)\) to the line \(3x + 4y - 10 = 0\) is \(2\).
The correct answer is \(\boxed{2}\).
### Problem 75
We need to find the value of \(x\) for which the matrix
[tex]\[ A = \begin{pmatrix} 6 & x-2 \\ 3 & x \end{pmatrix} \][/tex]
has no inverse.
A matrix has no inverse if its determinant is zero. The determinant of matrix \(A\) can be found by:
[tex]\[ \text{det}(A) = 6 \cdot x - 3 \cdot (x - 2) \][/tex]
1. Calculate the determinant:
[tex]\[ \text{det}(A) = 6x - 3(x - 2) \][/tex]
Simplify the expression:
[tex]\[ \text{det}(A) = 6x - 3x + 6 = 3x + 6 \][/tex]
2. Set the determinant equal to zero:
[tex]\[ 3x + 6 = 0 \][/tex]
3. Solve for \(x\):
[tex]\[ 3x = -6 \implies x = -2 \][/tex]
So, the value of \(x\) for which the matrix has no inverse is \( -2 \).
The correct answer is \(\boxed{-2}\).
Each problem has been carefully broken down and solved with clear steps to arrive at the correct answers.
### Problem 74
The question requires us to find the length of the perpendicular from the point \((0,0)\) to the line \(3x + 4y - 10 = 0\).
We can use the formula for the distance \(d\) from a point \((x_0, y_0)\) to a line \(ax + by + c = 0\), which is given by:
[tex]\[ d = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}} \][/tex]
1. Substitute \((x_0, y_0) = (0,0)\) into the formula:
[tex]\[ d = \frac{|3 \cdot 0 + 4 \cdot 0 - 10|}{\sqrt{3^2 + 4^2}} \][/tex]
2. Simplify the expression in the numerator:
[tex]\[ d = \frac{|0 + 0 - 10|}{\sqrt{9 + 16}} = \frac{|-10|}{\sqrt{25}} = \frac{10}{5} = 2 \][/tex]
So, the length of the perpendicular from \((0,0)\) to the line \(3x + 4y - 10 = 0\) is \(2\).
The correct answer is \(\boxed{2}\).
### Problem 75
We need to find the value of \(x\) for which the matrix
[tex]\[ A = \begin{pmatrix} 6 & x-2 \\ 3 & x \end{pmatrix} \][/tex]
has no inverse.
A matrix has no inverse if its determinant is zero. The determinant of matrix \(A\) can be found by:
[tex]\[ \text{det}(A) = 6 \cdot x - 3 \cdot (x - 2) \][/tex]
1. Calculate the determinant:
[tex]\[ \text{det}(A) = 6x - 3(x - 2) \][/tex]
Simplify the expression:
[tex]\[ \text{det}(A) = 6x - 3x + 6 = 3x + 6 \][/tex]
2. Set the determinant equal to zero:
[tex]\[ 3x + 6 = 0 \][/tex]
3. Solve for \(x\):
[tex]\[ 3x = -6 \implies x = -2 \][/tex]
So, the value of \(x\) for which the matrix has no inverse is \( -2 \).
The correct answer is \(\boxed{-2}\).
Each problem has been carefully broken down and solved with clear steps to arrive at the correct answers.
We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.