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nata0808 [166]
2 years ago
11

The graph of f(x) = |x| is reflected across the y-axis and translated to the left 5 units. Which statement about the domain and

range of each function is correct?
Mathematics
1 answer:
aksik [14]2 years ago
7 0

Both the domain and range of the transformed function are the same as those of the parent function.

Option A is the correct answer.

The complete question is

Both the domain and range of the transformed function are the same as those of the parent function.

Neither the domain nor the range of the transformed function are the same as those of the parent function.

The range but not the domain of the transformed function is the same as that of the parent function.

The domain but not the range of the transformed function is the same as that of the parent function

<h3>What is a function ?</h3>

A function is a mathematical statement which defines relationship between a dependent variable and an independent variable.

It is given that

f(x) = |x|

The graph is plotted

When the graph is is reflected across the y-axis and translated to the left 5 units.

When reflected across y axis , g(x) = |-x| = |x|

When translated 5 units to the left , g(x) = |x+5|

Both the graph is plotted and attached with the answer

The domain is the any value that can be an input in the function

domain for both the function is {x≥0}

The range is the value of the function that exists at any given time {y≥0}

From the option given

Both the domain and range of the transformed function are the same as those of the parent function.

Option A is the correct answer.

To know more about Function

brainly.com/question/12431044

#SPJ1

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c=35

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////////////

3.      3

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2. Suppose 27 blackberry plants started growing in a yard. Absent constraint, the blackberry plants will spread by 80% a month.
Marta_Voda [28]

Explanation

The question indicates we should use a logistic model to estimate the number of plants after 5 months.

This can be done using the formula below;

\begin{gathered} P(t)=\frac{K}{1+Ae^{-kt}};A=\frac{K-P_{0_{}}}{P_0}_{} \\ \text{From the question} \\ P_0=\text{ Initial Plants=27} \\ K=\text{Carrying capacity =140} \end{gathered}

Workings

Step 1: We would need to get the value of A using the carrying capacity and initial plants that started growing in the yard.

This gives;

\begin{gathered} A=\frac{140-27}{27} \\ A=\frac{113}{27} \end{gathered}

Step 2: Substitute the value of A into the formula.

P(t)=\frac{140}{1+\frac{113}{27}e^{-kt}}

Step 3: Find the value of the constant k

Kindly recall that we are told that the plants increase by 80% after each month. Therefore, after one month we would have;

\begin{gathered} P(1)=27+(\frac{80}{100}\times27) \\ P(1)=\frac{243}{5} \end{gathered}

We can then have that after t= 1month

\begin{gathered} \frac{140}{1+\frac{113}{27}e^{-k\times1}}=\frac{243}{5} \\ Flip\text{ the equation} \\ \frac{1+\frac{113}{27}e^{-k}}{140}=\frac{5}{243} \\ 243(1+\frac{113}{27}e^{-k})=700 \\ 243+1017e^{-k}=700 \\ 1017e^{-k}=700-243 \\ 1017e^{-k}=457 \\ e^{-k}=\frac{457}{1017} \\ -k=\ln (\frac{457}{1017}) \end{gathered}

Step 4: Substitute -k back into the initial formula.

\begin{gathered} P(t)=\frac{140}{1+\frac{113}{27}e^{\ln (\frac{457}{1017})t}} \\ =\frac{140}{1+\frac{113}{27}(e^{\ln (\frac{457}{1017})})^t} \\ P(t)=\frac{140}{1+\frac{113}{27}(\frac{457}{1017}^{})^t} \\  \end{gathered}

The above model is can be used to find the population at any time in the future.

Therefore after 5 months, we can estimate the model to be;

\begin{gathered} P(5)=\frac{140}{1+\frac{113}{27}(\frac{457}{1017}^{})^5} \\ P(5)=\frac{140}{1.07668} \\ P(5)=130.029\approx130 \end{gathered}

Answer: The estimated number of plants after 5 months is 130 plants.

8 0
1 year ago
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