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omeli [17]
3 years ago
12

2.

Mathematics
1 answer:
iren [92.7K]3 years ago
5 0

Answer:

  • Part A: The price of fuel A is decreasing by 12% per month.

  • Part B: Fuel A recorded a greater percentage change in price over the previous month.

Explanation:

<u>Part A:</u>

The function     f(x)=2.27(0.88)^x     calculates the price of fuel A each month by multiplying the price of the month before by 0.88.

Month       price, f(x)

1                 2.27 (0.88) = 1.9976 ≈ 2.00

2                2.27(0.88)² = 1.59808 ≈ 1.60

3                2.27(0.88)³ = 1.46063 ≈ 1.46

Then, the price of fuel A is decreasing.

The percentage per month is (1 - 0.88) × 100 = 12%, i.e. the price decreasing by 12% per month.

<u>Part B.</u>

<u>Table:</u>

m       price, g(m)

1          3.44

2         3.30

3          3.17

4         3.04

To find if the function decreases with a constant ration divide each pair con consecutive prices:

  • ratio = 3.30 / 3. 44 = 0.959 ≈ 0.96
  • ratio = 3.17 / 3.30 = 0.960 ≈ 0.96
  • ratio = 3.04 / 3.17 = 0.959 ≈ 0.96

Thus, the price of fuel B is decreasing by (1 - 0.96) × 100 =4%.

Hence, the fuel A recorded a greater percentage change in price over the previous month.

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Answer:

i) Since P(2), P(-1) and P(½) gives 0, then it's true that 2,-1 and 1⁄2 are the zeroes of the cubic polynomial.

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-the Sum of the products of roots where 2 are taken at the same time is same as the corresponding coefficient.

-the product of the zeros of the polynomial is same as the corresponding coefficient

Step-by-step explanation:

We are given the cubic polynomial;

p(x) = 2x³ - 3x² - 3x + 2

For us to verify that 2,-1 and 1⁄2 are the zeroes of the cubic polynomial, we will plug them into the equation and they must give a value of zero.

Thus;

P(2) = 2(2)³ - 3(2)² - 3(2) + 2 = 16 - 12 - 6 + 2 = 0

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P(½) = 2(½)³ - 3(½)² - 3(½) + 2 = ¼ - ¾ - 3/2 + 2 = -½ + ½ = 0

Since, P(2), P(-1) and P(½) gives 0,then it's true that 2,-1 and 1⁄2 are the zeroes of the cubic polynomial.

Now, let's verify the relationship between the zeros and the coefficients.

Let the zeros be as follows;

α = 2

β = -1

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The coefficients are;

a = 2

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c = -3

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So, the relationships are;

α + β + γ = -b/a

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αβγ = -d/a

Thus,

First relationship α + β + γ = -b/a gives;

2 - 1 + ½ = -(-3/2)

1½ = 3/2

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LHS = RHS; So, the sum of the zeros and the coefficients are the same

For the second relationship, αβ + βγ + γα = c/a it gives;

2(-1) + (-1)(½) + (½)(2) = -3/2

-2 - 1½ + 1 = -3/2

-1½ - 1½ = -3/2

-3/2 = - 3/2

LHS = RHS, so the Sum of the products of roots where 2 are taken at the same time is same as the coefficient

For the third relationship, αβγ = -d/a gives;

2 * -1 * ½ = -2/2

-1 = - 1

LHS = RHS, so the product of the zeros(roots) is same as the corresponding coefficient

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