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3 years ago

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Sử dụng phương pháp trung điểm, khi thu nhập bằng 7.500 đô la, thì độ co giãn của cầu theo giá trong khoảng từ 16 đô la đến 20 đ

ô la là bao nhiêu?

1 answer:

meriva3 years ago

8 0

Is what a drug does to your body. O a) Irrigation b) Pharmacodynamics c) Excretion d) Pharmacokinetics

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Ari decides to run 8.5 kilometers this weekend. He jogs 8.9 laps around the running track on Saturday. How many more kilometers

Marina86 [1]

Answer:

D. y = 8.9 ÷ 0.4

Step-by-step explanation:

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4 years ago

Can someone please help me!!!

tatiyna

Answer:

y = sin x/2

Step-by-step explanation:

y = a.sin.bx

a = y max= 2

y = a.sin.bx = 2.sin.bx

sin.bx = sin x/2

b=1/2

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y=sin x/2

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2 years ago

Martin has read 40% of a book. He has 30 more pages to finish. How many pages are there in the book?

victus00 [196]

Answer:

75

Step-by-step explanation:

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3 years ago

Read 2 more answers

A given field mouse population satisfies the differential equation dp dt = 0.5p − 410 where p is the number of mice and t is the

ohaa [14]

Answer:

a) $t = 2 *ln(\frac{82}{5}) =5.595$

b) $t = 2 *ln(-\frac{820}{p_0 -820})$

c) $p_0 = 820-\frac{820}{e^6}$

Step-by-step explanation:

For this case we have the following differential equation:

$\frac{dp}{dt}=\frac{1}{2} (p-820)$

And if we rewrite the expression we got:

$\frac{dp}{p-820}= \frac{1}{2} dt$

If we integrate both sides we have:

$ln|P-820|= \frac{1}{2}t +c$

Using exponential on both sides we got:

$P= 820 + P_o e^{1/2t}$

Part a

For this case we know that p(0) = 770 so we have this:

$770 = 820 + P_o e^0$

$P_o = -50$

So then our model would be given by:

$P(t) = -50e^{1/2t} +820$

And if we want to find at which time the population would be extinct we have:

$0=-50 e^{1/2 t} +820$

$\frac{820}{50} = e^{1/2 t}$

Using natural log on both sides we got:

$ln(\frac{82}{5}) = \frac{1}{2}t$

And solving for t we got:

$t = 2 *ln(\frac{82}{5}) =5.595$

Part b

For this case we know that p(0) = p0 so we have this:

$p_0 = 820 + P_o e^0$

$P_o = p_0 -820$

So then our model would be given by:

$P(t) = (p_o -820)e^{1/2t} +820$

And if we want to find at which time the population would be extinct we have:

$0=(p_o -820)e^{1/2 t} +820$

$-\frac{820}{p_0 -820} = e^{1/2 t}$

Using natural log on both sides we got:

$ln(-\frac{820}{p_0 -820}) = \frac{1}{2}t$

And solving for t we got:

$t = 2 *ln(-\frac{820}{p_0 -820})$

Part c

For this case we want to find the initial population if we know that the population become extinct in 1 year = 12 months. Using the equation founded on part b we got:

$12 = 2 *ln(\frac{820}{820-p_0})$

$6 = ln (\frac{820}{820-p_0})$

Using exponentials we got:

$e^6 = \frac{820}{820-p_0}$

$(820-p_0) e^6 = 820$

$820-p_0 = \frac{820}{e^6}$

$p_0 = 820-\frac{820}{e^6}$

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4 years ago

Find the value of .c in the triangle shown below.

ki77a [65]

Answer:

x=5.19615 or 3√3. √27

Step-by-step explanation:

3 0

3 years ago