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3 years ago

7

The local theater sells tickets to its plays at a rate of $38 each. Patrons who attend multiple shows are encouraged to buy an a

nnual membership for $80. Members are able to buy tickets at a discounted rate of $20 each. Roger expects to attend 6 shows this year. Grace expects to attend 4 shows. Who would benefit from buying a membership this year?
neither Grace nor Roger
only Grace
only Roger
both Grace and Roger

1 answer:

Mekhanik [1.2K]3 years ago

7 0

Answer:

Roger

Step-by-step explanation:

Since he is attending more shows, he will benefit. He will save a net amount of $40

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Answer:

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Step-by-step explanation:

Given

$N = -2F$ --- North Dakota

$C = 1F$ --- Cheyenne, Wyoming

Solving (a): Inequality to compare both temperatures

From the given temperatures, we can conclude that:

$C > N$ or $N < C$

Because

$1 > -2$ i.e. 1 is greater than -2

or

$-2 < 1$ i.e. -2 is less than 1

Solving (b): Inequality to compare the absolute values of the temperatures

We have:

$N = -2F$

$C = 1F$

The absolute values are:

$|N| = |-2|$

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$|C| = |1|$

$|C| = 1$

By comparison:

$|N| > |C|$ or $|C| < |N|$

Because

$2 > 1$ i.e. 2 is greater than 1

or

$1 < 2$ i.e. 1 is less than 2

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3 years ago

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3 years ago

Read 2 more answers

Each of these extreme value problems has a solution with both a maximum value and a minimum value. Use Lagrange multipliers to f

tino4ka555 [31]

Answer:

The minimum value of the given function is f(0) = 0

Step-by-step explanation:

Explanation:-

Extreme value :- f(a, b) is said to be an extreme value of given function 'f' , if it is a maximum or minimum value.

i) the necessary and sufficient condition for f(x) to have a maximum or minimum at given point.

ii) find first derivative $f^{l} (x)$ and equating zero

iii) solve and find 'x' values

iv) Find second derivative $f^{ll}(x) >0$ then find the minimum value at x=a

v) Find second derivative $f^{ll}(x)$ then find the maximum value at x=a

Problem:-

Given function is f(x) = log ( x^2 +1)

<u>step1:</u>- find first derivative $f^{l} (x)$ and equating zero

$f^{l}(x) = \frac{1}{x^2+1} \frac{d}{dx}(x^2+1)$

$f^{l}(x) = \frac{1}{x^2+1} (2x)$ ……………(1)

$f^{l}(x) = \frac{1}{x^2+1} (2x)=0$

the point is x=0

<u>step2:-</u>

Again differentiating with respective to 'x', we get

$f^{ll}(x)=\frac{x^2+1(2)-2x(2x)}{(x^2+1)^2}$

on simplification , we get

$f^{ll}(x) = \frac{-2x^2+2}{(x^2+1)^2}$

put x= 0 we get $f^{ll}(0) = \frac{2}{(1)^2}$ > 0

$f^{ll}(x) >0$ then find the minimum value at x=0

<u>Final answer</u>:-

The minimum value of the given function is f(0) = 0

5 0

3 years ago