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LenaWriter [7]
3 years ago
13

The function f is defined by the following rule

Mathematics
1 answer:
swat323 years ago
3 0

Answer:-18

-6

0

3

9

Step-by-step explanation:

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3 years ago
I will give brainliest if you answer it!!
aleksandrvk [35]

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B, C and D all appear to be repeating decimals

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8 0
4 years ago
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Margaret is planning a wedding reception. She wants to spend less than her budget of $5,748, and Margaret has already spent $2,4
noname [10]

Answer:

She can invite 109 guests and be right on top of budget. If she wants to spend less than her budget though she can only invite 108 people.

Step-by-step explanation:

First you have to subtract 5,748 - 2,478 which is 3270. Next you have to divide 3270 by 30 because each person costs 30 dollars. 3270 divided by 30 is 109.

I hope this helps! Please mark me brainliest if I am correct! Have a nice day!

6 0
3 years ago
Find the absolute maximum and absolute minimum values of f on the given interval.
anyanavicka [17]

The question is missing parts. Here is the complete question.

Find the absolute maximum and absolute minimum values of f on the given interval.

f(x)=xe^{-\frac{x^{2}}{32} } , [ -2,8]

Answer: Absolute maximum: f(4) = 2.42;

              Absolute minimum: f(-2) = -1.76;

Step-by-step explanation: Some functions have absolute extrema: maxima and/or minima.

<u>Absolute</u> <u>maximum</u> is a point where the function has its greatest possible value.

<u>Absolute</u> <u>minimum</u> is a point where the function has its least possible value.

The method for finding absolute extrema points is

1) Derivate the function;

2) Find the values of x that makes f'(x) = 0;

3) Using the interval boundary values and the x found above, determine the function value of each of those points;

4) The highest value is maximum, while the lowest value is minimum;

For the function given, absolute maximum and minimum points are:

f(x)=xe^{-\frac{x^{2}}{32} }

Using the product rule, first derivative will be:

f'(x)=e^{-\frac{x^{2}}{32} }(1-\frac{x^{2}}{16} )

f'(x)=e^{-\frac{x^{2}}{32} }(1-\frac{x^{2}}{16} ) = 0

1-\frac{x^{2}}{16}=0

\frac{x^{2}}{16}=1

x^{2}=16

x = ±4

x can't be -4 because it is not in the interval [-2,8].

f(-2)=-2e^{-\frac{(-2)^{2}}{32} }=-1.76

f(4)=4e^{-\frac{4^{2}}{32} }=2.42

f(8)=8e^{-\frac{8^{2}}{32} }=1.08

Analysing each f(x), we noted when x = -2, f(-2) is minimum and when x = 4, f(4) is maximum.

Therefore, absolute maximum is f(4) = 2.42 and

absolute minimum is f(-2) = -1.76

8 0
3 years ago
21 points pls help
svp [43]

Answer:

D

Step-by-step explanation:

w = 6 m

L = 1/2 m

Area = 1/2 * 6

Area = 3 m^2

Are you sure you have the givens correct?  If you are hesitant about this answer, ask your instructor how it is done.

Edit

In that case (L = 12) then the area is

Area = 12 * 6

Area = 72

So the answer is D

5 0
3 years ago
Read 2 more answers
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