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Semenov [28]
3 years ago
9

A sum of $5000 is invested at an interest rate of 5% per year. Find the time required for the money to double if the interest is

compounded continually. A(t)=Pe^rt
Mathematics
1 answer:
Temka [501]3 years ago
4 0
5000e^.05t=10000 solve
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A car rental agency charges $40 per day for the first 7 days, and $35 a day for each day after that. How much would joe be charg
Leno4ka [110]

Answer:

$385

Step-by-step explanation:

7x$40=280

3x$35=105

280+105=385

8 0
3 years ago
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Need help ASAP!!!!!
svetlana [45]

Answer:

cos 45 degree = A/H

cos 45 degree =  2 square root 2 / x

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or

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3 0
3 years ago
You have 300 pieces of candy in a jar. 28% of the pieces are red. How many pieces in the jar are NOT Red?
Citrus2011 [14]

Answer:

216

Step-by-step explanation:

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5 0
3 years ago
Type the correct answer in the box. Use numerals Instead of words. If necessary, use / for the fraction bar.
Zielflug [23.3K]

Using an exponential function, it is found that the value of David's Investment will be at least $415 after a period of 20 years.

<h3>What is an exponential function?</h3>

An increasing exponential function is modeled by:

A(t) = A(0)(1 + r)^t

In which:

  • A(0) is the initial value.
  • r is the growth rate, as a decimal.

In this problem, the initial value and the growth rate for his investment are given by:

A(0) = 230, r = 0.03.

Hence, the value after t years is given by:

A(t) = 230(1.03)^t

Then, the value will be of at least $415 when:

A(t) \geq 415

230(1.03)^t \geq 415

(1.03)^t \geq \frac{415}{230}

\log{(1.03)^t} \geq \log{\frac{415}{230}}

t\log{1.03} \geq \log{\frac{415}{230}}

t \geq \frac{\log{\frac{415}{230}}}{\log{1.03}}

t \geq 20

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

#SPJ1

4 0
2 years ago
How long does it take to travel 240 km at a constant speed of 12 km/h?
timama [110]
All that is necessary in this question is to divide. 240 km divided by 12 is 20. So this means that it would take exactly 20 hrs for someone to drive 240 km in 12 km an hr<span />
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