Where’s the ratios that we can select? can you provide?
Answer:
The solutions to the quadratic equations are:

Step-by-step explanation:
Given the function

substitute y = 0 in the equation to determine the zeros

Switch sides

Add 4 to both sides

Simplify

Rewrite in the form (x+a)² = b
But, in order to rewrite in the form x²+2ax+a²
Solve for 'a'
2ax = -6x
a = -3
so add a² = (-3)² to both sides


Apply perfect square formula: (a-b)² = a²-2ab+b²


solve

Add 3 to both sides

Simplify

now solving

Add 3 to both sides

Simplify

Thus, the solutions to the quadratic equations are:

Answer:
Step-by-step explanation:
Answer:
x ≈ 18
General Formulas and Concepts:
<u>Pre-Algebra</u>
- Order of Operations: BPEMDAS
- Equality Properties
<u>Trigonometry</u>
Law of Cosines: a^2 = b^2 + c^2 - 2(b)(c)cosA
- a is a side length
- b is a side length
- c is a side length
- A is an angle corresponding with side a
Step-by-step explanation:
<u>Step 1: Define</u>
a = x
A = 30°
b = 16
c = 30
<u>Step 2: Solve for </u><em><u>x</u></em>
- Substitute: x² = 16² + 30² - 2(16)(30)cos30°
- Exponents: x² = 256 + 900 -2(16)(30)cos30°
- Evaluate: x² = 256 + 900 -2(16)(30)(√3/2)
- Multiply: x² = 256 + 900 - 480√3
- Add: x² = 1156 - 480√3
- Subtract: x² = 324.616
- Isolate <em>x</em>: x = √324.616
- Evaluate: x = 18.0171
- Round: x ≈ 18
Answer:
2 is the answer to your question