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Tamiku [17]
2 years ago
13

Lily is a botanist who works for a garden that many tourists visit. The function f(s) = 2s + 30 represents the number of flowers

that bloomed, where s is the number of seeds she planted. The function s(w) = 40w represents the number of seeds she plants per week, where w represents the number of weeks.
Part A: Write a composite function that represents how many flowers Lily can expect to bloom over a certain number of weeks. (4 points)

Part B: What are the units of measurement for the composite function in Part A? (2 points)

Part C: Evaluate the composite function in Part A for 35 weeks. (4 points)
Mathematics
1 answer:
Vesnalui [34]2 years ago
7 0

Answer:

  A. b(w) = 80w +30

  B. input: weeks; output: flowers that bloomed

  C. 2830

Step-by-step explanation:

<h3>Part A:</h3>

For f(s) = 2s +30, and s(w) = 40w, the composite function f(s(w)) is ...

  b(w) = f(s(w)) = 2(40w) +30

  b(w) = 80w +30 . . . . . . blooms over w weeks

__

<h3>Part B:</h3>

The input units of f(s) are <em>seeds</em>. The output units are <em>flowers</em>.

The input units of s(w) are <em>weeks</em>. The output units are <em>seeds</em>.

Then the function b(w) above has input units of <em>weeks</em>, and output units of <em>flowers</em> (blooms).

__

<h3>Part C:</h3>

For 35 weeks, the number of flowers that bloomed is ...

  b(35) = 80(35) +30 = 2830 . . . . flowers bloomed over 35 weeks

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Answer:

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Step-by-step explanation:

4x^2+36=0

4x^2=-36

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One of the legs of a right triangle measures 7 cm and its hypotenuse measures 12 cm.
Oksana_A [137]

Answer:

AC ≈ 9,7sm

Step-by-step explanation:

AB=7  One side of a right triangle

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2 years ago
Solve y'' + 10y' + 25y = 0, y(0) = -2, y'(0) = 11 y(t) = Preview
svetlana [45]

Answer:  The required solution is

y=(-2+t)e^{-5t}.

Step-by-step explanation:   We are given to solve the following differential equation :

y^{\prime\prime}+10y^\prime+25y=0,~~~~~~~y(0)=-2,~~y^\prime(0)=11~~~~~~~~~~~~~~~~~~~~~~~~(i)

Let us consider that

y=e^{mt} be an auxiliary solution of equation (i).

Then, we have

y^prime=me^{mt},~~~~~y^{\prime\prime}=m^2e^{mt}.

Substituting these values in equation (i), we get

m^2e^{mt}+10me^{mt}+25e^{mt}=0\\\\\Rightarrow (m^2+10y+25)e^{mt}=0\\\\\Rightarrow m^2+10m+25=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mt}\neq0]\\\\\Rightarrow m^2+2\times m\times5+5^2=0\\\\\Rightarrow (m+5)^2=0\\\\\Rightarrow m=-5,-5.

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Differentiating with respect to t, we get

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According to the given conditions, we have

y(0)=-2\\\\\Rightarrow A=-2

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Thus, the required solution is

y(t)=(-2+1\times t)e^{-5t}\\\\\Rightarrow y(t)=(-2+t)e^{-5t}.

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