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3 years ago

Which expression is equivalent

Mathematics

1 answer:

Likurg_2 [28]3 years ago

7 0

Answer:

${8}^{ \frac{3}{4} } • {3}^{ \frac{15}{4} }$ (third opti0n)

Step-by-step explanation:

$(8 • {3}^{5} {)}^{ \frac{3}{4} }$

${8}^{ \frac{3}{4} } • ( {3}^{5} {)}^{ \frac{3}{4} }$

$= {8}^{ \frac{3}{4} } • {3}^{ \frac{15}{4} }$

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A news item is spreading by word of mouth through a population of size 12,000 people. After t days, the number of people (in tho

Lilit [14]

Answer:

a) 2.303

b) $f'(t)=\frac{432e^{-0.48t}}{(1+75e^{-0.48t})^{2}}$

c) 1.3609

e) $\frac{dy}{dx}=0.04y(12-y)$

f) 4.5 thousand people(smaller)

7.5 thousand people (larger)

e) t=7.93 days (smaller)

t=10.06 days (larger)

g) 1.44

Step-by-step explanation:

The given function is $y=f(t)=\frac{12}{1+75e^{-0.48t}}$

To find how many thousand people have heard the news after 6 days, we substitute to get:

$f(6)=\frac{12}{1+75e^{-0.48*6}}=2.303$ thousands.

b) We rewrite to get: $f(t)=12(1+75e^{-0.48t})^{-1}$

We differentiate using the chain rule to obtain:

$f'(t)=-12(1+75e^{-0.48t})^{-2}\cdot 75e^{-0.48t}\cdot -0.48$

This simplifies to:

$f'(t)=432e^{-0.48t}(1+75e^{-0.48t})^{-2}$

We rewrite as positive index to get:

$f'(t)=\frac{432e^{-0.48t}}{(1+75e^{-0.48t})^{2}}$

c) To find the rate at which the news is spreading after 8 days, we substitute t=8 into $f'(t)=\frac{432e^{-0.48t}}{(1+75e^{-0.48t})^{2}}$ to get:

$f'(8)=\frac{432e^{-0.48\cdot 8}}{(1+75e^{-0.48\cdot 8})^{2}}=1.3609$

d) The solution to the logistic differential equation:

$\frac{dy}{dt}=ky(M-y)$

$y=\frac{M}{1+be^{-Mkt}}$

By comparison, M=12, and $Mk=0.48$

$12k=0.48\\k=0.04$

The differential equation is:

$\frac{dy}{dx}=0.04y(12-y)$

e) To how many people have heard the news when its rate of spread is 1.35 thousand per day, we equate the differential equation to 1.35 and so for t first.

$\frac{432e^{-0.48t}}{(1+75e^{-0.48t})^{2}}=1.35$

This gives us: t=10.06 ot t=7.93

We substitute the times into the function to get:

$f(7.93)=\frac{12}{1+75e^{-0.48\cdot7.93}}$ =4.5 thousand: smaller value

$f(10.06)=\frac{12}{1+75e^{-0.48\cdot10.06}}=7.5$ thousand: Larger value

f) The two times that the news is spreading at rate 1.35 thousand people per day are:

$t=7.93$ days: Smaller value

$t=10.06$ : Larger value

g) To find the fastest rate at which the news spread we take the second derivative and equate it to zero.

$f''(t)=0$

This corresponds to where the horizontal line is tangent to

$f'(t)=\frac{432e^{-0.48t}}{(1+75e^{-0.48t})^{2}}$

From the graph the point of tangency is:

(8.99,1.44)

Therefore the fastest rate at which the news spread is 1.44

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Step-by-step explanation:

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