Help! I'm a little stumped when it comes to variables in matrices. An explanation would be super rad!
1 answer:
Answer:
30
Step-by-step explanation:
To find the determinant of a 3x3 matrix, you can use this method. (See picture.)
Start with the first number in the top row, and block off the row and column. A 2x2 matrix will be left. Find the determinant of this 2x2 matrix, and multiply it by the number in the top row.
Repeat for the other two numbers in the top row. Add the first result, subtract the second, and add the third.
det A = -2 [(3)(-5) − (a)(0)] − 2 [(0)(-5) − (a)(0)] + b [(0)(0) − (3)(0)]
det A = -2 (3)(-5) − 0 + 0
det A = 30
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Answer:
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Step-by-step explanation:
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Answer:
On a coordinate plane, a hyperbola is shown. One curve opens up and to the right in quadrant 1, and the other curve opens down and to the left in quadrant 3. A vertical asymptote is at x = 1, and the horizontal asymptote is at y = 4
Step-by-step explanation:
The given function is presented as follows;
From the given function, we have;
When x = 1, the denominator of the fraction, , which is (x - 1) = 0, and the function becomes,
therefore, the function in undefined at x = 1, and the line x = 1 is a vertical asymptote
Also we have that in the given function, as <em>x</em> increases, the fraction tends to 0, therefore as x increases, we have;
Therefore, as x increases, f(x) → 4, and 4 is a horizontal asymptote of the function, forming a curve that opens up and to the right in quadrant 1
When -∞ < x < 1, we also have that as <em>x</em> becomes more negative, f(x) → 4. When x = 0, . When <em>x</em> approaches 1 from the left, f(x) tends to -∞, forming a curve that opens down and to the left
Therefore, the correct option is on a coordinate plane, a hyperbola is shown. One curve opens up and to the right in quadrant 1, and the other curve opens down and to the left in quadrant 3. A vertical asymptote is at x = 1, and the horizontal asymptote is at y = 4.