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noname [10]
3 years ago
5

1.) Some banks charge a fee on savings accounts that are left inactive for an extended period of time. The equation y = 5000 * (

0.98) ^ x represents the value, y of one account that was left inactive for a period of x years .
a. What was the initial amount of money that was left in this savings account?

b. What is the percent of change each year in this savings account?
Mathematics
1 answer:
oee [108]3 years ago
7 0

Answer:

a. The initial amount of money that was left in this savings account was of 5000.

b. Decrease of 2% each year.

Step-by-step explanation:

Exponential function:

An exponential function, with an initial value of A(0), and a decay rate of r, as a decimal, is given by:

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

In this question, we have:

y = 5000*(0.98)^x

a. What was the initial amount of money that was left in this savings account?

This is y(0) = 5000

The initial amount of money that was left in this savings account was of 5000.

b. What is the percent of change each year in this savings account?

First as a decimal.

1 - r = 0.98

r = 1 - 0.98

r = 0.02

So a decrease of 2% each year.

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3 years ago
To find the product of 42.12 and 10³, move the decimal point in 42.12 __ places to the right because 10³ has __ zeros.
Yakvenalex [24]
<h3>To find the product of 42.12 and 10^3,  move the decimal point in 42.12 3 places to the right because 10^3 has 3 zeros</h3>

<em><u>Solution:</u></em>

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\text{ product of } 42.12 \text{ and } 10^3

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3 years ago
A heavy rope, 50 ft long, weighs 0.6 lb/ft and hangs over the edge of a building 120 ft high. Approximate the required work by a
Anastasy [175]

Answer:

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Exercise (b)

The work done in pulling half the rope to the top of the building is 562.5 lb·ft

Step-by-step explanation:

Exercise (a)

The given parameters of the rope are;

The length of the rope = 50 ft.

The weight of the rope = 0.6 lb/ft.

The height of the building = 120 ft.

We have;

The work done in pulling a piece of the upper portion, ΔW₁ is given as follows;

ΔW₁ = 0.6Δx·x

The work done for the second half, ΔW₂, is given as follows;

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The total work done, W = W₁ + W₂ = 0.6Δx·x + 0.6Δx·x + 375

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The work done in pulling the rope to the top of the building, W = 750 lb·ft

Exercise (b)

The work done in pulling half the rope is given by W₂ as follows;

W_2 =  \int\limits^{25}_0 {0.6 \cdot x} \, dx + 375= \left[0.6 \cdot \dfrac{x^2}{2} \right]^{25}_0 + 375 = 562.5

The work done in pulling half the rope, W₂ = 562.5 lb·ft

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