(tco 6) using partial fractions method solve the initial value problem y"-4y'+4y=0 given y(0)= 2.1 and y'(0)=3.9
1 answer:
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Answer:
y-intercept = (0, ).
Step-by-step explanation:
Given the linear equation in standard form, 12x + 13y = 8:
We must transform this equation into slope-intercept form to make it easier to determine the coordinates of the y-intercept.
12x + 13y = 8
Subtract 12x from both sides:
12x + 13y - 12x = - 12x + 8
13y = - 12x + 8
Next, divide both sides of the equation by 13 to solve for y:
or
Next, to determine the y-intercept, we must set x = 0 (because the y-intercept is the point where the graph of the linear equation crosses the y-axis).
Let x = 0:
Therefore, the value of y when x = 0 is . This is the y-intercept, and its coordinate is (0,
).
9514 1404 393
Answer:
not a right triangle
Step-by-step explanation:
The perimeter is 22, so we have ...
AB +BC +CA = 22
8 + 5 + CA = 22
CA = 22 -13 = 9
If this is a right triangle, the sum of the squares of the short sides will be equal to the square of the long side.
AB² +BC² = 8² +5² = 64 +25 = 89
This is not the same as ...
CA² = 9² = 81
The triangle is not a right triangle.
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<em>Additional comment</em>
If you're familiar with a few Pythagorean triples, you know that a couple involving a side length 5 are (3, 4, 5) and (5, 12, 13). That is, (5, 8, 9) is NOT a Pythagorean triple, so you know immediately that those side lengths do not form a right triangle. (No calculation is necessary.)
These expressions are different in the way that they are reciprocals of each other.