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s2008m [1.1K]
3 years ago
14

Graphing Natural Exponential Functions In Exercise,sketch the graph of the function.See Example 1.

Mathematics
1 answer:
serg [7]3 years ago
5 0

Answer:

See the graph and explanation below.

Step-by-step explanation:

For this case we have the following function:

f(x) = e^{1-x}

We can calculate some points in order to see the tendency of the graph, we can select a set of points for example x =-2,-1.5,-1,0,1,1,5,2 and we can calculate the values for f(x) like this

x=-2

f(x=-2) =e^{1+2}= e^{3}=20.086

x=-1.5

f(x=-1.5) =e^{1+1.5}= e^{2.5}=12.182

x=-1

f(x=-1) =e^{1+1}= e^{2}=7.389

x=0

f(x=0) =e^{1-0}= e^{1}=2.718

This point correspond to the y intercept.

x=1

f(x=1) =e^{1-1}= e^{0}=1

x=2

f(x=2) =e^{1-2}= e^{-1}=0.368

We don't have x intercepts for this case since the functionnever crosses the x axis.

And then we can see the plot on the figure attached.

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And the 4 in 37,426 is 400
The answer is 100
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Step-by-step explanation:

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In the data set below, what is the mean absolute deviation? 43 25 43 30 55 10 81 If the answer is a decimal, round it to the nea
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3 years ago
Five minus three fourths
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4 years ago
There are 30 students in a class. 10 students have a pet dog, 13 students have a pet cat, and 7 students have a pet fish. 4 stud
tankabanditka [31]

Answer:

9 students have pets

Step-by-step explanation:

From the above question, we are given the following information

Total number of students = 30

Let Pet Dog = D

Pet Cat = C

Pet Fish = F

Number is students that have pet dog

(D) = 10 students

Number of students that have pet cat (C) = 13 students

Number of students that have pet fish (F) = 7 students

Number of students that have Pet dog and cat ( D and C) = 4 students

Number of students that have Pet cat and fish (C and F) = 6 students

Number of student that has pet dog and pet fish (D and F) = 2 students

1 student has all three = ( D and F and C)

Number of student that have a pet Dog only

= n(D) - [n( D and C) + n( D and F) - n(D and C and F)]

= 10 -( 4+ 2 -1)

= 10 - 5

= 5

Number of student that have Pet cat only

= n(C) -[ n( D and C) + n( C and F) - n( D and C and F)]

= 13 -( 4 + 6 - 1)

= 13 - 9

= 4

Number is student that have a pet fish only

= n(F) - [n (C and F) + n( D and F) - n( D and C and F)]

= 7 - [6 + 2 - 1]

= 7 - 7

= 0

The number of students that have pets is calculated as:

(Number of students that have dogs only + Number of student that have cats only + Number of students that have fish only)

= 5 + 4 + 0

= 9

Therefore only 9 students have pets.

4 0
3 years ago
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