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Ivahew [28]
3 years ago
5

Which descriptions from the list below accurately describe the relationship between AQRS and ATUV? Check all that apply. 284 -28

12 12 76" 176 176 1764 6 V Q A. Same sides B. Same size C. Similar D. Congruent
Mathematics
1 answer:
ololo11 [35]3 years ago
3 0

Answer:

All A to D

Step-by-step explanation:

The attachment completes the question

From the attachment,

  • it can be seen that QRS and TUV are of the same i.e. triangles

Also, from the attachment, the sides of each are:

RQ = UT = 22

QS = TV = 30

RS = UV = 23

And the angles are:

\angle Q = \angle T = 50

\angle R = \angle U = 83

\angle S = \angle V = 47

The above show that QRS and TUV are:

  • similar
  • same sides
  • and they are congruent

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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)

is

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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Answer:

A is the correct option

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This is approximately $6.05

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