Answer:
F(d) = 30 + 0.50d
Step-by-step explanation:
Given
Charges = P8.00 ---- first 4 km
Additional = P0.50
Required
Write a function to address the scenario.
Represent the whole distance covered with d.
First,we need to determine the total charges for the first four hours.
Charges = 8.00 * 4
Charges = 32.00
Next, we determine the charges for additional distance.
Charges = 0.50 * (d - 4)
d - 4 is the remaining distance after the first 4.
Charges = 0.50d - 2
The function is then written as;
F(d) = 32 + 0.50d - 2
F(d) = 32 - 2 + 0.50d
F(d) = 30 + 0.50d
Answer:
9·x² - 36·x = 4·y² + 24·y + 36 in standard form is;
(x - 2)²/2² - (y + 3)²/3² = 1
Step-by-step explanation:
The standard form of a hyperbola is given as follows;
(x - h)²/a² - (y - k)²/b² = 1 or (y - k)²/b² - (x - h)²/a² = 1
The given equation is presented as follows;
9·x² - 36·x = 4·y² + 24·y + 36
By completing the square, we get;
(3·x - 6)·(3·x - 6) - 36 = (2·y + 6)·(2·y + 6)
(3·x - 6)² - 36 = (2·y + 6)²
(3·x - 6)² - (2·y + 6)² = 36
(3·x - 6)²/36 - (2·y + 6)²/36 = 36/36 = 1
(3·x - 6)²/6² - (2·y + 6)²/6² = 1
3²·(x - 2)²/6² - 2²·(y + 3)²/6² = 1
(x - 2)²/2² - (y + 3)²/3² = 1
The equation of the hyperbola is (x - 2)²/2² - (y + 3)²/3² = 1.
