Given:
The function, f(x) = -2x^2 + x + 5
Quadratic equation: 0 = -2x^2 + x +5
where a = -2
b = 1
c = 5
The discriminate b^2 - 4ac = 41
To solve for the zeros of the quadratic function, use this formula:
x = ( -b +-√ (b^2 - 4ac) ) / 2a
x = ( 1 + √41 ) / 4 or 1.85
x = ( 1 - √41 ) / 4 or -1.35
Therefore, the zeros of the quadratic equation are 1.85 and -1.35.
Answer:
-3r-8
Step-by-step explanation:
Answer:
Blue line: y = -2
Red line: y = 2/3 x + 2
Perpendicular to red line: y = -3/2x + 2
Parallel to red line: y = 2/3 x + any real number
Answer:
the value of these two numbers are 1/2 and 3/2
Step-by-step explanation:
Given that:
x+y < 2 ---- (1)
y - x > 1 ---- (2)
From equation (2), let y > 1 + x then substitute it into equation (1)
x + 1 + x < 2
2x + 1 < 2
2x < 2 - 1
2x < 1
x < 1/2
From equation (2), replace the value of x to be 1/2
y - 1/2 > 1
y > 1 + 1/2
y > 3/2
X=8 my photo has the work done
