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MAXImum [283]
2 years ago
10

The price of products may increase due to inflation and decrease due to depreciation. Marco is studying the change in the price

of two products, A and B, over time.
The price f(x), in dollars, of product A after x years is represented by the function below:

f(x) = 0.69(1.03)x

Part A: Is the price of product A increasing or decreasing and by what percentage per year? Justify your answer. (5 points)

Part B: The table below shows the price f(t), in dollars, of product B after t years:


t (number of years) 1 2 3 4
f(t) (price in dollars) 10,100 10,201 10,303.01 10,406.04


Which product recorded a greater percentage change in price over the previous year? Justify your answer. (5 points)
Mathematics
1 answer:
likoan [24]2 years ago
4 0

Answer:

A)  3%

B)  Product A

Step-by-step explanation:

<u>Exponential Function</u>

General form of an exponential function: y=ab^x

where:

  • a is the initial value (y-intercept)
  • b is the base (growth/decay factor) in decimal form
  • x is the independent variable
  • y is the dependent variable

If b > 1 then it is an increasing function

If 0 < b < 1 then it is a decreasing function

<u>Part A</u>

<u>Product A</u>

Assuming the function for Product A is <u>exponential</u>:

f(x) = 0.69(1.03)^x

The base (b) is 1.03.  As b > 1 then it is an <u>increasing function</u>.

To calculate the percentage increase/decrease, subtract 1 from the base:

⇒ 1.03 - 1 = 0.03 = 3%

Therefore, <u>product A is increasing by 3% each year.</u>

<u>Part B</u>

\sf percentage\:change=\dfrac{final\:value-initial\:value}{initial\:value} \times 100

To calculate the percentage change in Product B, use the percentage change formula with two consecutive values of f(t) from the given table:

\implies \sf percentage\:change=\dfrac{10201-10100}{10100}\times 100=1\%

Check using different two consecutive values of f(t):

\implies \sf percentage\:change=\dfrac{10303.01-10201}{10201}\times 100=1\%

Therefore, as 3% > 1%, <u>Product A recorded a greater percentage change</u> in price over the previous year.

Although the question has not asked, we can use the given information to easily create an exponential function for Product B.

Given:

  • a = 10,100
  • b = 1.01
  • n = t - 1 (as the initial value is for t = 1 not t = 0)

\implies f(t) = 10100(1.01)^{t-1}

To check this, substitute the values of t for 1 through 4 into the found function:

\implies f(1) = 10100(1.01)^{1-1}=10100

\implies f(2) = 10100(1.01)^{2-1}=10201

\implies f(3) = 10100(1.01)^{3-1}=10303.01

\implies f(4) = 10100(1.01)^{4-1}=10406.04

As these values match the values in the given table, this confirms that the found function for Product B is correct and that <u>Product B increases by 1% per year.</u>

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3 years ago
what’s the total area of a converter box buildings of 4 feet, width of 3 feet,and a height of 6 feet?
Katena32 [7]
The total will be 72
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4 years ago
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BlackZzzverrR [31]
The answer is A. 2, -2


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8 0
3 years ago
Math help!<br><br> Question 5
Rudiy27

Answer:

a. 4 teachers b. 58 students c. 62 people

Step-by-step explanation:

multiply 80 by 0.05, you get 4.

Multiply 1160 by 0.05, you get 58.

Add 80 and 1160 together, you get 1240. Multiply 1240 by 0.05, you get 62.

5 0
4 years ago
Simplify each only using positive exponents:<br> 2x^-3 • 4x^2<br> 2x^4 • 4x^-3<br> 2x^3y^-3 • 2x
erica [24]
<h2>Answer:</h2>

\frac{2}{x}

\frac{x}{2}

\frac{4x^4}{y^3}

<h2>Step-by-step explanation:</h2>

a. 2x^-3 • 4x^2

To solve this using only positive exponents, follow these steps:

i. Rewrite the expression in a clearer form

2x⁻³ . 4x²

ii. The position of the term with negative exponent is changed from denominator to numerator or numerator to denominator depending on its initial position. If it is at the numerator, it is moved to the denominator. If otherwise it is at the denominator, it is moved to the numerator. When this is done, the negative exponent is changed to positive.

In our case, the first term has a negative exponent and it is at the numerator. We therefore move it to the denominator and change the negative exponent to  positive as follows;

\frac{1}{2x^3} . 4x^2

iii. We then solve the result as follows;

\frac{1}{2x^3} . 4x^2 = \frac{2}{x}

Therefore, 2x⁻³ . 4x² = \frac{2}{x}

b. 2x^4 • 4x^-3

i. Rewrite as follows;

2x⁴ . 4x⁻³

ii. The second term has a negative exponent, therefore swap its position and change the negative exponent to a positive one.

2x^4 . \frac{1}{4x^3}

iii. Now solve by cancelling out common terms in the numerator and denominator. So we have;

\frac{x}{2}

Therefore, 2x⁴ . 4x⁻³ = \frac{x}{2}

c. 2x^3y^-3 • 2x

i. Rewrite as follows;

2x³y⁻³ . 2x

ii. Change position of terms with negative exponents;

2x^3.\frac{1}{y^3} .2x

iii. Now solve;

\frac{4x^4}{y^3}

Therefore, 2x³y⁻³ . 2x = \frac{4x^4}{y^3}

8 0
3 years ago
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