Answer:
Write and solve an equation of ratios:
16 miles 20 min
----------- = -----------
12 miles x
Then 16x = 240, and x = 240/16 = 15.
It will take 15 minutes to travel 12 miles.
Answer:
The answer to your question is a = 16
Step-by-step explanation:
Polynomial
(y - 4) (y² + 4y + 16)
Process
1.- Multiply y by each term of the polynomial
y(y² + 4y + 16) = y³ + 4y² + 16y
2.- Multiply -4 by each term of the polynomial
-4(y² + 4y + 16) = -4y² - 16y - 64
3.- Write both results
y³ + 4y² + 16y - 4y² - 16y - 64
In bold we notice that a = 16
Answer:


Step-by-step explanation:
Given

Solving (a): An equivalent inequality
We have:

Multiply both sides by -1 (this changes the inequality)


Solving (b): Values of u from least to greatest
implies that u ends at -4, starting from negative infinity
So, the list is:

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.