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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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Matt, a 180 pound (82 kg) male, goes to a party and is served beer in a 16-ounce (473 ml) red plastic cup. he has 3 cups of beer
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<h3>What is Unitary Method?</h3>

The unitary method is a method in which you find the value of a unit and then the value of a required number of units.

For example: 5 cars have price of $500. what will be the price of 10 cars.

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5 cars → $500

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The standard drink contains 12 - ounce of beer.

Here, we have given 1 red cup which has 16 - ounce of beer.

He has 3 red cups.

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3 years ago
A pond forms as water collects in a conical depression of radius a and depth h. Suppose that water flows in at a constant rate k
Scrat [10]

Answer:

a. dV/dt = K - ∝π(3a/πh)^⅔V^⅔

b. V = (hk^3/2)/[(∝^3/2.π^½.(3a))]

The small deviations from the equilibrium gives approximately the same solution, so the equilibrium is stable.

c. πa² ≥ k/∝

Step-by-step explanation:

a.

The rate of volume of water in the pond is calculated by

The rate of water entering - The rate of water leaving the pond.

Given

k = Rate of Water flows in

The surface of the pond and that's where evaporation occurs.

The area of a circle is πr² with ∝ as the coefficient of evaporation.

Rate of volume of water in pond with time = k - ∝πr²

dV/dt = k - ∝πr² ----- equation 1

The volume of the conical pond is calculated by πr²L/3

Where L = height of the cone

L = hr/a where h is the height of water in the pond

So, V = πr²(hr/a)/3

V = πr³h/3a ------ Make r the subject of formula

3aV = πr³h

r³ = 3aV/πh

r = ∛(3aV/πh)

Substitute ∛(3aV/πh) for r in equation 1

dV/dt = k - ∝π(∛(3aV/πh))²

dV/dt = k - ∝π((3aV/πh)^⅓)²

dV/dt = K - ∝π(3aV/πh)^⅔

dV/dt = K - ∝π(3a/πh)^⅔V^⅔

b. Equilibrium depth of water

The equilibrium depth of water is when the differential equation is 0

i.e. dV/dt = K - ∝π(3a/πh)^⅔V^⅔ = 0

k - ∝π(3a/πh)^⅔V^⅔ = 0

∝π(3a/πh)^⅔V^⅔ = k ------ make V the subject of formula

V^⅔ = k/∝π(3a/πh)^⅔ -------- find the 3/2th root of both sides

V^(⅔ * 3/2) = k^3/2 / [∝π(3a/πh)^⅔]^3/2

V = (k^3/2)/[(∝π.π^-⅔(3a/h)^⅔)]^3/2

V = (k^3/2)/[(∝π^⅓(3a/h)^⅔)]^3/2

V = (k^3/2)/[(∝^3/2.π^½.(3a/h))]

V = (hk^3/2)/[(∝^3/2.π^½.(3a))]

The small deviations from the equilibrium gives approximately the same solution, so the equilibrium is stable.

c. Condition that must be satisfied

If we continue adding water to the pond after the rate of water flow becomes 0, the pond will overflow.

i.e. dV/dt = k - ∝πr² but r = a and the rate is now ≤ 0.

So, we have

k - ∝πa² ≤ 0 ---- subtract k from both w

- ∝πa² ≤ -k divide both sides by - ∝

πa² ≥ k/∝

5 0
3 years ago
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