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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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Goes through (1,4) is parallel to y=3x-7
user100 [1]

Answer:

y = 3x + 1

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

y = 3x - 7 ← is in slope- intercept form

with slope m = 3

Parallel lines have equal slopes, thus

y = 3x + c ← is the partial equation

To find c substitute (1, 4) into the partial equation

4 = 3 + c ⇒ c = 4 - 3 = 1

y = 3x + 1 ← equation of parallel line

8 0
3 years ago
Calculate the limit of the function with L'Hospital rule​
mr_godi [17]

Answer:

L=24

Step-by-step explanation:

L'Hopital's Rule for Evaluating Limits:

Rule is that if \lim_{x \to a} \frac{f(x)}{g(x)} takes \frac{0}{0} or \frac{\infty}{\infty} form, then,

\lim_{x \to a} \frac{f(x)}{g(x)}= \lim_{x \to a} \frac{f'(x)}{g'(x)}

where f'(x)=\frac{df(x)}{dx} and g'(x)=\frac{dg(x)}{dx}

Now coming to the problem,

L= \lim_{x \to \frac{\pi}{6} } \frac{cot^{3}x-3cotx}{cos(x+\frac{\pi}{3} )}

Here f(x)=cot^{3}x-3cotx and g(x)=cos(x+\frac{\pi}{3} )

Substituting x=\frac{\pi}{6} in f(x) and g(x),

f(\frac{\pi}{6})=cot^{3}\frac{\pi}{6}-3cot\frac{\pi}{6}\\=3\sqrt{3}-3\sqrt{3}\\ =0

g(\frac{\pi}{6})=cos(\frac{\pi}{6}+\frac{\pi}{3})\\=cos\frac{\pi}{2}\\=0

Since L takes the form \frac{0}{0}, using l'hopital's rule

L= \lim_{x \to \frac{\pi}{6}} \frac{cot^{3}x-3cotx}{cos(x+\frac{\pi}{3})}= \lim_{x \to \frac{\pi}{6}} \frac{3cot^{2}x(-cosec^{2}x)-3(-cosec^{2}x)}{-sin(x+\frac{\pi}{3})}

now substituting x=\frac{\pi}{6} ,

L= \lim_{x \to \frac{\pi}{6}} \frac{3cot^{2}\frac{\pi}{6}(-cosec^{2}\frac{\pi}{6})-3(-cosec^{2}\frac{\pi}{6})}{-sin(\frac{\pi}{6}+\frac{\pi}{3})}\\=\frac{3*3^{2}(-2^{2})+3(2^{2})}{-1}\\=24

6 0
3 years ago
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