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jek_recluse [69]
3 years ago
10

suppose you go to work for a company that pays one penny the first day 2 cents on the second day 4 cents on the third day and so

on if the daily wage kepps doubling what will your total income be for working 31 days
Mathematics
1 answer:
Luda [366]3 years ago
7 0

If my calculations are correct the total income should be 1073741824 pennies, which is $10737418.24.

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A town in east Texas received 10 inches of rain in two DAYS. If it kept raining at this rate for a 31 day month, how much rain d
nydimaria [60]
<h3>Answer: 155 inches</h3>

This is equivalent to approximately 12.917 feet.

===================================================

Work Shown:

(10 inches)/(2 days) = (x inches)/(31 days)

10/2 = x/31

10*31 = 2*x ... cross multiply

310 = 2x

2x = 310

x = 310/2

x = 155

Keep in mind that this water is being drained on a fairly continuous basis. This means that not all of the water is sticking around (we'd hope not anyway) at the same moment in time.

Note: 155 inches = 155/12 = 12.917 feet approximately.

--------------------------

Alternative method:

If the town got 10 inches in 2 days, this means it rains at a unit rate of 10/2 = 5 inches per day.

Multiply that by 31 days and we get 5*31 = 155 inches of total rain.

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3 years ago
Finish the following statement: In the altitude to hypotenuse theorem, the ALTITUDE is the geometric mean...
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The hypotenuse is believe is what your looking for if not I do not understand because of how it’s worded
8 0
3 years ago
What is the answer to 3x-2=3x+2
Ostrovityanka [42]
This is an equality. There is no answer, by only using the given information, that can be found for "x".
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Given that f(x) = 19x2 + 152, solve the equation f(x) = 0
podryga [215]

Answer: 152

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19(0)^2 + 152

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The per capita growth rate of many species varies temporally for a variety of reasons, including seasonality and habitat destruc
Veronika [31]

Answer:

n(t)=n_0e^{(1-e^{-t }-t)}

Step-by-step explanation:

If n(t) represents the population size at time t, where n is measured in individuals and t is measured in years.

\frac{dn}{dt}=n(e^{-t }-1), n(0)=n_o

\frac{dn}{n}=(e^{-t }-1)dt

Taking the integral of both sides

\int\frac{dn}{n}=\int(e^{-t }-1)dt\\\int\frac{dn}{n}= \int e^{-t }dt-\int1dt

ln |n| = -e^{-t }-t+C

Where C is integration constant

Taking the exponential of both sides

n=e^{(-e^{-t }-t+C)}

n=e^{(-e^{-t }-t)}e^C\\n=Ke^{(-e^{-t }-t)} whee the exponential of a constant is a constant K.

When t=0, n(0)=n_o

n_0=Ke^{-1

Therefore:

n=n_0e^{1}e^{(-e^{-t }-t)}

n(t)=n_0e^{(1-e^{-t }-t)}

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