Answer:
(3 ± √23 * i) /4
Step-by-step explanation:
To solve this, we can apply the Quadratic Equation.
In an equation of form ax²+bx+c = 0, we can solve for x by applying the Quadratic Equation, or x = (-b ± √(b²-4ac))/(2a)
Matching up values, a is what's multiplied by x², b is what's multiplied by x, and c is the constant, so a = 2, b = -3, and c = 4
Plugging these values into our equation, we get
x = (-b ± √(b²-4ac))/(2a)
x = (-(-3) ± √(3²-4(2)(4)))/(2(2))
= (3 ± √(9-32))/4
= (3 ± √(-23))/4
= (3 ± √23 * i) /4
Answer:
D
Step-by-step explanation:
Answer:
(-19, 55)
General Formulas and Concepts:
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
Equality Properties
<u>Algebra I</u>
- Solving systems of equations using substitution/elimination
Step-by-step explanation:
<u>Step 1: Define Systems</u>
y = -3x - 2
5x + 2y = 15
<u>Step 2: Solve for </u><em><u>x</u></em>
<em>Substitution</em>
- Substitute in <em>y</em>: 5x + 2(-3x - 2) = 15
- Distribute 2: 5x - 6x - 4 = 15
- Combine like terms: -x - 4 = 15
- Isolate <em>x</em> term: -x = 19
- Isolate <em>x</em>: x = -19
<u>Step 3: Solve for </u><em><u>y</u></em>
- Define original equation: y = -3x - 2
- Substitute in <em>x</em>: y = -3(-19) - 2
- Multiply: y = 57 - 2
- Subtract: y = 55
Answer:
by stating what the time was before the zero point came.
Step-by-step explanation:
Answer: 30.8
Step-by-step explanation:
