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a_sh-v [17]
2 years ago
9

from 2001 to 2012 the sales for a local doughnut store increased by about the same percent each year. The sales, S, for year t c

an be modeled by, S=372,300(6/5)^t. Where t=0 corresponds to 2001. What were the sales in 2001​
Mathematics
1 answer:
Elenna [48]2 years ago
6 0

Using the given exponential function, it is found that the number of sales in 2001 was of 327,300.

<h3>What is the exponential function in this question?</h3>

It models the number of sales S in t years after 2001, given by:

S(t) = 372300\left(\frac{6}{5}\right)^t

2001 is year 0, hence the amount is given by S(0), that is:

S(0) = 372300\left(\frac{6}{5}\right)^0 = 372300

The number of sales in 2001 was of 327,300.

More can be learned about exponential functions at brainly.com/question/25537936

#SPJ1

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Answer: PART A : \frac{4.4}{x} = \frac{3}{4}

PART B : 5.9 FEET

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height of the second sign = 4 feet

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\frac{h_{1}}{h_{2} } = \frac{L_{1}}{L_{2} }

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4 years ago
1. What are foci? 2. What is the first step to take to write the equation of a hyperbola? 3. How do you represent parts of a hyp
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Answer:  see below

<u>Step-by-step explanation:</u>

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2) When determining the equation of a hyperbola you need the following:

    a)  does the hyperbola open up or to the right?

    b)  what is the center (h, k) of the hyperbola?

    c)  What is the slope of the asymptotes of the hyperbola?

3) The equation of a hyperbola is:

\dfrac{(x-h)^2}{a^2}-\dfrac{(y-k)^2}{b^2}=1\qquad or\qquad \dfrac{(y-k)^2}{b^2}-\dfrac{(x-h)^2}{a^2}=1

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