Question

N(t) = 100,000e−(t − 10)2/8

Answer 1

The approximation of the slope of the graph at t = 10 days using this secant lines are 11750 and -11750, respectively.

How to determine the slope of the secant lines?

(a) t = 10 days and a day earlier

This means that:

t = 10 and t = 9

The function is given as:

$N(t) = 100000e^{\frac{-(t - 10)^2}{8}}$

Calculate N(10)

$N(10) = 100000e^{\frac{-(10 - 10)^2}{8}}$

Evaluate

N(10) = 100000

Next, calculate N(9)

$N(9) = 100000e^{\frac{-(9 - 10)^2}{8}}$

Evaluate

N(9) = 88249.6902585

The slope is then calculated using"

$m = \frac{N(10) - N(9)}{10 - 9}$

This gives

$m = \frac{100000 - 88249.6902585}{10 - 9}$

Evaluate

m = 11750.3097415

Approximate

m = 11750

(b) t = 10 days and a day later

This means that:

t = 10 and t = 100

The function is given as:

$N(t) = 100000e^{\frac{-(t - 10)^2}{8}}$

In (a), we have:

N(10) = 100000

Next, calculate N(11)

$N(9) = 100000e^{\frac{-(11 - 10)^2}{8}}$

Evaluate

N(11) = 88249.6902585

The slope is then calculated using:

$m = \frac{N(11) - N(10)}{11 - 10}$

This gives

$m = \frac{88249.6902585 - 100000}{11 - 10}$

Evaluate

m = -11750.3097415

Approximate

m = -11750

Hence, the approximation of the slope of the graph at t = 10 days using this secant lines are 11750 and -11750, respectively.

Read more about secant lines at:

brainly.com/question/14438198

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