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maksim [4K]
3 years ago
5

A store manager begins each shift with the same total amount of money. She keeps $200 in a safe and distributes the rest equally

to the 5 cashiers in the store. This situation can be represented by the function y = (x − 200)5. What does the variable x represent in this situation?
Mathematics
2 answers:
lina2011 [118]3 years ago
8 0

Answer:

The answer would be 5

Volgvan3 years ago
8 0

Answer:

x=the total amount of money the manager had in the beginning

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Angelina_Jolie [31]

Answer:

475

Step-by-step explanation:

If each represents 100

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Hence,

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3 0
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I dont know what I'm doing! could you help me​
photoshop1234 [79]

Answer:

c = 24.34

Step-by-step explanation:

Here, we can use the cosine rule

Generally, we have this as:

a^2 = b^2 + c^2 - 2bcCos A

12^2 = 14^2 + c^2 - 2(14)Cos 19

144 = 196 + c^2 - 26.5c

c^2 - 26.5c + 196-144 = 0

c^2 - 26.5c + 52 = 0

We can use the quadratic formula here

and that is;

{-(-26.5) ± √(-26.5)^2 -4(1)(52)}/2

(26.5 + 22.23)/2 or (26.5 - 22.23)/2

24.37 or 2.135

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8 0
3 years ago
Determine whether
lora16 [44]

Answer:

(a) and (b) are not equivalent

(c) is equivalent

Step-by-step explanation:

Given

\frac{25^m}{5}

Required

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We have:

25^{m-1

Apply law of indices

25^{m-1} = \frac{25^m}{25}

<em>This is not equivalent to </em>\frac{25^m}{5}<em></em>

(b)\ 25^{2m - 1}

We have:

25^{2m - 1}

Apply law of indices

25^{2m - 1} = \frac{25^{2m}}{25}

<em>This is not equivalent to </em>\frac{25^m}{5}<em></em>

<em></em>

<em />(c)\ 5^{2m-1}<em />

We have:

<em />5^{2m-1}<em />

Apply law of indices

<em />5^{2m-1} = \frac{5^{2m}}{5^1}<em />

<em />5^{2m-1} = \frac{5^{2m}}{5}<em />

<em>Evaluate the numerator</em>

<em />5^{2m-1} = \frac{25m}{5}<em />

<em />

<em>This is an equivalent expression</em>

4 0
3 years ago
Which of the following comparisons is true?
nlexa [21]
D) 436.683 < 436.684
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3 years ago
Factor completely. (3.2² - 12x)(x2 – 2x + 1) =​
pychu [463]

Answer:

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Step-by-step explanation:

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