Answer:
7. RHS
8.RHS
9.SAS
10.SAS
11.RHS
12.ASA
Step-by-step explanation:
n=3; We need a third degree polynomials with the following given zero's: 2 and 5i are zeros; f(-1)=156.
Since these are solutions
x = 2 ; x = 5i. Since imaginaries travel in pairs, the other answer is x= -5i.
We have (x-2)(x-5i)(x+5i) = 0
Now,
f(-1) = (-1-2)(-1-5i)(-1+5i) = 156.
f(-1) = (-3)(26) = -78.
But -78 x -2 = 156, so our polynomial becomes
Y= -2x (<em>x</em> - 2 ) x (<em>x </em>to the power of 2 + 25) = 0
Answer:
11
General Formulas and Concepts:
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Algebra I</u>
Step-by-step explanation:
<u>Step 1: Define</u>
g(x) = x³ - 4
h(x) = x² + 2
h(-2) is x = -2
g(-1) is x = -1
<u>Step 2: Find h(-2) - g(-1)</u>
- Substitute: ((-2)² + 2) - ((-1)³ - 4)
- Exponents: (4 + 2) - (-1 - 4)
- Add/Subtract: 6 - (-5)
- Subtract: 11
