Standard quadratic equation .. y = a x^2 + b x + c
<span>parabola 'a' not equal to zero </span>
<span>a<0 parabola opens downward </span>
<span>a>0 parabola opens upward </span>
<span>when |a| >>0 the parabola is narrower </span>
<span>when |a| is close to zero , the parabola is flatter </span>
<span>when the constant is varied it only effects the vertical position of the parabola , the shape remains the same</span>
The two numbers are 30 and 11
<em><u>Solution:</u></em>
Given that we have to separate the number 41 into two parts
Let the second number be "x"
<em><u>Given that first number is eight more than twice the second number</u></em>
first number = eight more than twice the second number
first number = 8 + twice the "x"
first number = 8 + 2x
So we can say first number added with second number ends up in 41
first number + second number = 41
8 + 2x + x = 41
8 + 3x = 41
3x = 41 - 8
3x = 33
x = 11
first number = 8 + 2x = 8 + 2(11) = 8 + 22 = 30
Thus the two numbers are 30 and 11
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<h2>Answer</h2>
Or as ordered pairs: 
<h2>Explanation</h2>
Lets solve our system of equations step by step
equation (1)
equation (2)
1. Solve for 
in equation (2)
equation (3)
2. Replace equation (3) in equation (1) and solve for 
or 
3. Replace the values of in equation (3) and solve for 
- For
or 
- For
or 
So, the solutions of our system of equation are:
The answer is C and D
First we must make one side of the equation zero, so we subtract 7 on both sides of the equation, making the equation 9x^2 - 6x - 6 = 0
Then we will use the quadratic formula to find the answers.
using the Quadratic Formula where:
a = 9, b = -6, and c = -6
and the formula is 6 +- 
