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My name is Ann [436]
2 years ago
9

[SCREENSHOT INCLUDED] Which of the following is the best estimate of f '(2) based on this table of values?

Mathematics
1 answer:
mel-nik [20]2 years ago
6 0

Using derivatives, it is found that the best estimate of f '(2) based on this table of values is of 10.

The rate of change <u>from x = 0 to x = 2</u> is given by:

r_1 = \frac{2 - (-16)}{2 - 0} = \frac{18}{2} = 9

From <u>x = 2 to x = 4</u>, it is given by:

r_2 = \frac{24 - 2}{4 - 2} = \frac{22}{2} = 11

The average of these rates is:

A = \frac{r_1 + r_2}{2} = \frac{9 + 11}{2} = 10

Hence, the best estimate of f '(2) based on this table of values is of 10.

To learn more about derivatives, brainly.com/question/18590720

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1 1/11 rewrite each number as a single fraction greater then one
mafiozo [28]

Answer:

12/11

Step-by-step explanation:

11/11=1

1/11=1/11

11/11+1/11=12/11

7 0
3 years ago
The rate (gallons per day) at which a pond loses water due to evaporation from one day after observation has started is give by
Andreyy89

Answer:  see table below

<u>Step-by-step explanation:</u>

\left\begin{array}{c|l}T&\qquad w(t)\\1&\dfrac{1}{1^2}=1\implies 1.00\\\\2&\dfrac{1}{2^2}=\dfrac{1}{4}\implies 0.25\\\\10&\dfrac{1}{10^2}=\dfrac{1}{100}\implies 0.01\\\\50&\dfrac{1}{50^2}=\dfrac{1}{2500}\implies 0.0004\\\\100&\dfrac{1}{100^2}=\dfrac{1}{10000}\implies 0.0001\\\\200&\dfrac{1}{200^2}=\dfrac{1}{40000}\implies 0.000025\\\\500&\dfrac{1}{500^2}=\dfrac{1}{250000}\implies 0.000004\\\\1000&\dfrac{1}{1000^2}=\dfrac{1}{100000}\implies 0.000001\\\end{array}\right

4 0
3 years ago
Please help! due in next 10min!?!
Paladinen [302]

Answer:

21.99

Step-by-step explanation:

This is simple just times 3.14 and 7 to get 21.99

7 0
3 years ago
Read 2 more answers
The value of a certain car decreases by 16% each year. What is the 1⁄2-life of the car?
svet-max [94.6K]

Answer:

The half life of the car is 3.98 years.

Step-by-step explanation:

The value of the car after t years is given by the following equation:

V(t) = V(0)(1-r)^{t}

In which V(0) is the initial value and r is the constant decay rate, as a decimal.

The value of a certain car decreases by 16% each year.

This means that r = 0.16

So

V(t) = V(0)(1-r)^{t}

V(t) = V(0)(1-0.16)^{t}

V(t) = V(0)(0.84)^{t}

What is the 1⁄2-life of the car?

This is t for which V(t) = 0.5V(0). So

V(t) = V(0)(0.84)^{t}

0.5V(0) = V(0)(0.84)^{t}

(0.84)^{t} = 0.5

\log{(0.84)^{t}} = \log{0.5}

t\log{0.84} = \log{0.5}

t = \frac{\log{0.5}}{\log{0.84}}

t = 3.98

The half life of the car is 3.98 years.

7 0
3 years ago
2<br> 2. Evaluate (a + y)+ 2y if a = 5 and y = -3.<br> F 58<br> G -2<br> H 70<br> J 10
NARA [144]
<h3>Answer:  G) -2</h3>

=======================================================

Explanation:

I'm assuming you meant to say (a+y)^2 + 2y

Replace each copy of 'a' with 5. Replace each copy of 'y' with -3. Use PEMDAS to simplify.

(a+y)^2 + 2y

(5 + (-3))^2 + 2(-3)

(5-3)^2 + 2(-3)

(2)^2 + 2(-3)

4 + 2(-3)

4 - 6

-2

So (a+y)^2 + 2y = -2 when a = 5 and y = -3.

3 0
3 years ago
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