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Scrat [10]
1 year ago
8

Frank invests $1,000 in simple interest investment account that pays 8% a year. After a number of years, he withdraws his balanc

e of $2,200. Using the formula I = P r t, how many years was his money invested? 2. 75 years 10 years 15 years 27. 5 years.
Mathematics
1 answer:
MAXImum [283]1 year ago
7 0

There are 15 years was his money invested.

<h3>What is simple interset?</h3>

Simple Interest is an easy method of calculating the interest for a loan/principal amount.

Given

Frank invests $1,000 in simple interest investment account that pays 8% a year.

After a number of years, he withdraws his balance of $2,200.

The number of years was his money inversted is;

\rm I= P\times r \times t

Where p is the principal amount, r is the rate of interest, I is the amount and  t is the time.

Substitute all the values in the formula;

\rm I= P\times r \times t\\\\2200-1000= 1000 \times .08 \times t\\\\1200 = 80t\\\\t = \dfrac{1200}{80}\\\\t=15

Hence, in the 15 years was his money invested.

To know more about investing click the link given below.

brainly.com/question/26677076

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3 years ago
If Amir is running at a faster pace; 6.7 mph, to the nearest hundredth of an hour , after how much time will Amir catch up to sa
pickupchik [31]

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7

Step-by-step explanation:

6.7 rounds up to 7

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3 years ago
Jim had some dimes, quarters, and nickels. First, he counted only the dimes and quarters and found that he had 9 coins. Then, he
Olegator [25]

Answer:

$1.90

Step-by-step explanation:

Represent the numbers of nickels, dimes and quarters by n, d and q.

Then: d + q = 9, or q = 9 - d.

Also, n + d = 10

Lastly, n + d + q = 15.

Let's substitute 9 - d for q in the equation immediately above:

n + d + (9 - d) = 15, or n + 9 = 15.  Then n must be 6.

In summary, Jim has 6 nickels, (10 - 6) dimes and (9 - 4) quarters, or:

                                    6 nickels, 4 dimes and 5 quarters

Thus, he has 5($0.05) + 4($0.10) + 5($0.25), or

                         $0.25   +   $0.40   + $1.25, or          $1.90

7 0
2 years ago
A particle moves according to a law of motion s = f(t), t ? 0, where t is measured in seconds and s in feet.
Usimov [2.4K]

Answer:

a) \frac{ds}{dt}= v(t) = 3t^2 -18t +15

b) v(t=3) = 3(3)^2 -18(3) +15=-12

c) t =1s, t=5s

d)  [0,1) \cup (5,\infty)

e) D = [1 -9 +15] +[(5^3 -9* (5^2)+ 15*5)-(1-9+15)]+ [(6^3 -9(6)^2 +15*6)-(5^3 -9(5)^2 +15*5)] =7+ |32|+7 =46

And we take the absolute value on the middle integral because the distance can't be negative.

f) a(t) = \frac{dv}{dt}= 6t -18

g) The particle is speeding up (1,3) \cup (5,\infty)

And would be slowing down from [0,1) \cup (3,5)

Step-by-step explanation:

For this case we have the following function given:

f(t) = s = t^3 -9t^2 +15 t

Part a: Find the velocity at time t.

For this case we just need to take the derivate of the position function respect to t like this:

\frac{ds}{dt}= v(t) = 3t^2 -18t +15

Part b: What is the velocity after 3 s?

For this case we just need to replace t=3 s into the velocity equation and we got:

v(t=3) = 3(3)^2 -18(3) +15=-12

Part c: When is the particle at rest?

The particle would be at rest when the velocity would be 0 so we need to solve the following equation:

3t^2 -18 t +15 =0

We can divide both sides of the equation by 3 and we got:

t^2 -6t +5=0

And if we factorize we need to find two numbers that added gives -6 and multiplied 5, so we got:

(t-5)*(t-1) =0

And for this case we got t =1s, t=5s

Part d: When is the particle moving in the positive direction? (Enter your answer in interval notation.)

For this case the particle is moving in the positive direction when the velocity is higher than 0:

t^2 -6t +5 >0

(t-5) *(t-1)>0

So then the intervals positive are [0,1) \cup (5,\infty)

Part e: Find the total distance traveled during the first 6 s.

We can calculate the total distance with the following integral:

D= \int_{0}^1 3t^2 -18t +15 dt + |\int_{1}^5 3t^2 -18t +15 dt| +\int_{5}^6 3t^2 -18t +15 dt= t^3 -9t^2 +15 t \Big|_0^1 + t^3 -9t^2 +15 t \Big|_1^5 + t^3 -9t^2 +15 t \Big|_5^6

And if we replace we got:

D = [1 -9 +15] +[(5^3 -9* (5^2)+ 15*5)-(1-9+15)]+ [(6^3 -9(6)^2 +15*6)-(5^3 -9(5)^2 +15*5)] =7+ |32|+7 =46

And we take the absolute value on the middle integral because the distance can't be negative.

Part f: Find the acceleration at time t.

For this case we ust need to take the derivate of the velocity respect to the time like this:

a(t) = \frac{dv}{dt}= 6t -18

Part g and h

The particle is speeding up (1,3) \cup (5,\infty)

And would be slowing down from [0,1) \cup (3,5)

5 0
3 years ago
What statement best describes the function?
xxMikexx [17]

Answer:

d

Step-by-step explanation:

3 0
2 years ago
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