Answer:
Step-by-step explanation
The relative error is the absolute error divided by the absolute value of p. for an approximation p*, the relative error is
r = |p*-p|/|p|
we want r to be at most 10⁻³, thus
|p*-p|/|p| ≤ 10⁻³
|p*-p| ≤ |p|* 10⁻³
therefore, p*-p should lie in the interval [ - |p| * 10⁻³ , |p| * 10⁻³ ], and as a consecuence, p* should be in the interval [p - |p| * 10⁻³ , p + |p| * 10⁻³ ]
Answer:
Step-by-step explanation:
Average rate of change is the same thing as the slope. Because this is parabolic, we cannot find the exact rate of change as we could if this were a linear function. But we can use the same idea. When t = 3, h(t) = 33, so the coordinate point is (3, 33). When t = 6, h(t) = 0, so the coordinate is (6, 0). Plug those values into the slope formula:
and
which is -11
From 3 to 6 seconds, the rocket is falling 11 yards per second.
Answer:
yes
Step-by-step explanation:
The line intersects each parabola in one point, so is tangent to both.
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For the first parabola, the point of intersection is ...
y^2 = 4(-y-1)
y^2 +4y +4 = 0
(y+2)^2 = 0
y = -2 . . . . . . . . one solution only
x = -(-2)-1 = 1
The point of intersection is (1, -2).
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For the second parabola, the equation is the same, but with x and y interchanged:
x^2 = 4(-x-1)
(x +2)^2 = 0
x = -2, y = 1 . . . . . one point of intersection only
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If the line is not parallel to the axis of symmetry, it is tangent if there is only one point of intersection. Here the line x+y+1=0 is tangent to both y^2=4x and x^2=4y.
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Another way to consider this is to look at the two parabolas as mirror images of each other across the line y=x. The given line is perpendicular to that line of reflection, so if it is tangent to one parabola, it is tangent to both.
