Let that be
Two vertical asymptotes at -1 and 0
If we simply
- Denominator has degree 2
- Numerator should have degree as 2 and coefficient as 3 inorder to get horizontal asymptote y=3 means the quadratic equation should contain 3x²
- But there should be a x intercept at -3 so one zeros should be -3
Find a equation
Find zeros
Horizontal asymptote
So our equation is
Graph attached
Answer:
75
General Formulas and Concepts:
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
Equality Properties
- Multiplication Property of Equality
- Division Property of Equality
- Addition Property of Equality
- Subtraction Property of Equality<u>
</u>
<u>Geometry</u>
- Sum of Angles in a Quadrilateral: 360
Step-by-step explanation:
<u>Step 1: Set up</u>
75 + 120 + 90 + ∠4 = 360
<u>Step 2: Solve</u>
- Add: 285 + ∠4 = 360
- [Subtraction Property of Equality] Subtract 285 on both sides: ∠4 = 75
First, you should solve both equations for the same variable. Since the first one is already solved for y, solve the second equation for y as well.
6y = 2x + 6 Divide both sides by 6
y =
x + 1
You can see that both lines have a slope of
.
Lines that have the same slope are
parallel lines.
Answer: -2.1p + 1.7
2.4 - 6p - 0.7 + 3.9p
Subtract 6p from 3.9p. (-2.1p)
Subtract 0.7 from 2.4. (1.7)
