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harkovskaia [24]
4 years ago
5

Use the matrix tool to solve the system of equations. Enter the answer as an

Mathematics
1 answer:
Andreyy894 years ago
6 0

Answer:

(-1/2,2)

Step-by-step explanation:

2(4x+y=0)=8x+2y=0

8x+2y=0

-(8x-y=6)

3y=6

Divide both sides y=2

4x+2=0

4x=-2

divide both sides x=-1/2

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18 +2 = 20
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The price of products may increase due to inflation and decrease due to depreciation. Marco is studying the change in the price
inn [45]

Answer:

Product A has a greater percentage change in price.

Step-by-step explanation:

Part A:

The price f product A, f (<em>x</em>) after <em>x</em> years is given by: 

 f(x) = 0.69\cdot(1.03)^{x}

After <em>x</em> = 0 years, the price of product A is:

f(0) = 0.69\cdot(1.03)^{0}=0.69

After <em>x</em> = 1 years, the price of product A is:

f(1) = 0.69\cdot(1.03)^{1}=0.69\cdot (1+0.03)=0.69\cdot (1+3\%)

After 1 year, the price of product A is 3% times more than the original price.

This means that after one year, the new price is 103% of the original price, which means the price product A is increasing by 3%.

Again after <em>x</em> = 2 years, the price of product A is:

f(2) = 0.69\cdot(1.03)^{2}=[0.69\cdot (1+3\%)]\times (1.03)

This implies that after 2 years, the price of product A is 103% of the price after year 1.

This implies that the price of product A is 3% more than the previous year.

Thus, the price of product A is increasing each year by 3%.

Part B:

The data for Product B is as follows:

Time (t)          Price [f (t)]

   1                   10,100

   2                   10,201

   3                 10,303.01

   4                 10,406.04

Product B is clearly increasing in price.

Consider the changes in price of Product B in the following intervals of years:

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Price in year 1 = $10,100

Price in Year 2 = $10,201

Compute the increase percentage as follows:

\text{Increase}\%=\frac{10201-10100}{10100}=0.01=1\%

  • Year 2 - Year 3:

Price in Year 2 = $10,201

Price in year 3 = $10,303.01

Compute the increase percentage as follows:

\text{Increase}\%=\frac{10303.01-10201}{10201}=0.01=1\%

  • Year 3 - Year 4

Price in year 3 = $10,303.01

Price in Year 4 = $10,406.04

Compute the increase percentage as follows:

\text{Increase}\%=\frac{10,406.04-10303.01}{10303.01}=0.09999\approx 0.01=1\%

It is quite clear that the price of product B increases by 1% each year.

Thus, Product A has a greater percentage change in price.

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Answer:

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