Answer:
x≤ −96
/23
Step 1: Simplify both sides of the inequality.
Step 2: Subtract 1/8x from both sides.
Step 3: Subtract 12 from both sides.
Step 4: Multiply both sides by 8/23.
Answer:
{(3,6),(2,1),(-8,4),(2,0)}
Step-by-step explanation:
when determining if a set of points is a function
determine if any of the x values or y values repeat. if they do then it is not a function because it will not pass the function like test. (which is where you can draw a line trough the function and it will hit more than once.
none if the x values repeat and none if the y values do either.
x values : 3,2,-8,2
y values: 6,1,4,0
Answer:
<h2>
V ≈ 628.3 cm³</h2>
Step-by-step explanation:
Volume of cylinder: V = π×r² × h
diameter = 20 cm ⇒ r = 10 cm
h = 2 cm
Volume:
V = π×(10)²×2 = 200π cm³ = 628.318.... cm³ ≈ 628.3 cm³
<h3>
Answer: x = 7 and y = 3</h3>
=====================================================
Explanation:
Apply the difference of squares rule
x² - 4y² = 13
x² - (2y)² = 13
(x - 2y)(x + 2y) = 13
Since x and y are positive integers, this means x-2y and x+2y are both integers as well.
The value 13 is prime. Its only factors are 1 and 13
Since the above equation shows 13 factoring into x-2y and x+2y, then we have two cases:
- A) x-2y = 1 and x+2y = 13
- B) x-2y = 13 and x+2y = 1
----------------
Let's consider case A
We have this system of equations

Add the equations straight down
- x+x becomes 2x
- -2y+2y becomes 0y = 0 which goes away
- 1+13 becomes 14
Therefore we have 2x = 14 solve to x = 7
From here, plug this into either equation to solve for y
x-2y = 1
7 - 2y = 1
-2y = 1-7
-2y = -6
y = -6/(-2)
y = 3
You should get the same result if you used x+2y = 13
----------------
Since we've found that x = 7 and y = 3, notice how case B is not possible
Example: x-2y = 13 becomes 7-2(3) = 13 which is false.
Also, x+2y = 1 would turn into 7+2(3) = 1 which is also false.
-----------------
Let's check those x and y values in the original equation
x² - 4y² = 13
7² - 4*(3)² = 13
49 - 4(9) = 13
49 - 36 = 13
13 = 13
The answer is confirmed.