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NISA [10]
3 years ago
5

Can derivatives be used to measure the the slope of the tangent line at x=a

Mathematics
1 answer:
Pie3 years ago
6 0
Yes. In fact, that is basically the definition of a derivative. It is the instantaneous rate of change of a function. 

For example, picture the graph of the following function:

f(x) = x^2

The slope is constantly changing at every x-value, so to find  the slope at x=a, we find the  derivative of the  function.

f'(x)=2x

Once we have the derivative, simply plug in a for x to find the slope of the line tangent to f(x) at x=a.

For example, at x=5:

f'(5)=2(5)

The slope of f(x) at x=5 is 10.
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A triangle has ∠A, ∠B, and ∠C.<br> ∠B is 20° more than ∠A.<br> ∠C is double ∠B.<br> How big is ∠A?
san4es73 [151]

Answer:

30°

Step-by-step explanation:

∠A + ∠B + ∠C = 180

∠B = ∠A + 20

∠C = 2 * ∠B

<A + (∠A + 20) + (2 * ∠B) = 180

<A + (∠A + 20) + (2 * (∠A + 20)) = 180

<A + (∠A + 20) + 2∠A + 40 = 180

4∠A + 60 = 180

4∠A = 120

∠A = 30

5 0
3 years ago
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I need this rn 3×-7 equivalent expressions A. -21 B. 4 C. -4 D. 21
Drupady [299]

Answer: The answer is A. -21

Its simplified and i just multiplied 3 by -7 :)

8 0
2 years ago
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The amount of polonium-210 remaining, p(t), after t days in a sample can be modeled by the exponential function p(t) = 100e−0.00
Archy [21]

Answer:

% Po lost = 100[1 - e^(-0.005t)]  %; 73.0 g

Step-by-step explanation:

p(t) = 100e^(-0.005t)

Initial amount:        p(0) = 100

Amount remaining: p(t) = 100e^(-0.005t)

Amount lost: p(0) – p(t) = 100 - 100e^(-0.005t) = 100[1 - e^(-0.005t)]

% of Po lost  = amount lost/initial amount × 100 %

= [1 - e^(-0.005t)]  × 100 % = 100[1 - e^(-0.005t)]  %

p(63) = 100e^(-0.005 × 63) = 100e^(-0.315) = 100 × 0.730 = 73 g

The mass of polonium remaining after 63 days is 73 g.

4 0
3 years ago
What is ...1000.0000..000000km x 3,000?
aliya0001 [1]

Answer:

4

Step-by-step explanation:

duh

3 0
3 years ago
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11.447 is greater than 11.473
Stells [14]
No 11.473 is greater then 11.447
8 0
3 years ago
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