1) 0.682 2) 3.317 3) -0.489 4) 0.569 5) 0.824 6) a) L ( x ) = 21.6 x − 48.6 b) 16.2216 c) f (3.001) = 16.2216108 error = .0000108
(
7) a) L ( x ) = 2 + 2 3 x −
π 3
)
b) 1.96535 c) g
f (0.02) = 1.030153544 error = 0.000154 9) a)
( π3 − .01) = 1.966 error = 0.000687
dy 16 x − 5 y = dx 5 x + 3 y 2
8) a) L ( x ) = 1 +
3 x b) 1.03 c) 2
3
b) y = 3 x − 13 c) -0.4 d) k + 21k = −7.88 k = -0.373 10) a)
f ( 0.9 ) ≈ 4.8 f (1.2) ≈ 5.4 b) The estimates are too large since f(x) is concave down so the tangent lines lie above the graph of the function, and since L(x) appears to be above the graph of f’(x) which is the actual change of f(x).
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a = acceleration m/s2 (metres per second squared). Unit Analysis: Inertial mass â the m used in the second law is correctly described as the inertial mass.
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lying instead on local Euclidean distances that can be mis- leading in image space. .... 0 means the pair is parallel (aligned) and 1 means the pair is orthogonal ...
Page 1 of 7. Newton's ThirdLaw. 'When you push on the edge o{ etable or desk with your hand, your palm. indents in a way that can only happen if something is ...
Ex. A force of 49 N causes a 7 m/s2. acceleration. Find the mass of the. object it was pulling. Variables: 49 N = F. 7 m/s2. = a. m = ? Equation: F = ma. Solve: Newton's 2nd Law tells us that when. you accelerate (stomp on the gas) or. decelerate qui
on which we induce a pseudo-Riemannian metric from the Sasaki metric. Kowalski-Sekizawa [1] have shown how the geometry of the tangent sphere bundle. TrM over a Riemannian manifold (M,g) depends on the radius r. We generalize a part of their results
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Apr 25, 1972 - IIA is a diagram of apparatus for a simulated ping>pong type game;. FIG. IIB is a sketch of a television screen illustrating the manner of play of ...
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