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horsena [70]
2 years ago
9

Solve for j-2.45j=2.05

Mathematics
1 answer:
Romashka-Z-Leto [24]2 years ago
5 0
0.87 is the answer for this math question
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Suppose it is known that for a given differentiable function y=f(x), its tangent line (local linearization) at the point where a
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Answer:

y(-4) = 5

y'(-4) = -7

Step-by-step explanation:

Hi!

Since the tangent line T and the curve y must coincide at x=-4

y(-4) = T(-4) = 5

On the other hand, the derivative of the curve evaluated at -4 y'(x=-4) must be the slope of the tangent line. Which inspecting the tangent line T(x) is -7

That is:

y'(-4) = -7

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