In the quadratic equation ax² + bx + c = 0, a cannot be zero.
If a = 0, the equation becomes linear (bx + c = 0), and it's no longer a quadratic equation.
Given the equation x^2 + y^2 = 1, the range of y values is [-1, 1] since y ranges from -1 to 1 as x varies.
Using the identify 2 sin A cos A = sin 2A, so 2 sin 45° cos 45° = sin 90°.
Since sin 90° = 1, the expression equals 1.
Given ∫12x⁻⁵ √(-3x⁻⁴ + 1) dx
Let u = -3x⁻⁴ + 1
du/dx = 12x⁻⁵
du = 12x⁻⁵ dx
∫√u du = ∫u^(1/2) du
= (2/3)u^(3/2) + C
Substituting back u = -3x⁻⁴ + 1:
= (2/3)(-3x⁻⁴ + 1)^(3/2) + C
A binomial expression contains exactly two terms, and x² + y² fits that definition.
x²y² is a monomial (single term).
x + y + 1 is a trinomial (three terms).
To evaluate the expression, follow the order of operations:
(2²)³ = (2 × 2)³
= 4³
= 4 × 4 × 4
= 64
The derivative of y = x² is dy/dx = 2x.
At x = 2, the slope is 2(2) = 4.
The tangent line equation is y - 4 = 4(x - 2), which simplifies to y = 4x - 4, or 4x - y = 4.
x + y =8
if x=5 and y=3
5^2 - 3^2 =2 5–9 =16, condition satisfied.
x-y = 5–3
=2
ND27-5-2023
In ax² + bx + c = 0, if > 0, b > 0, and c > 0, the parabola opens upward and lies entirely above the x-axis unless discriminant D = b² - 4ac ≥ 0.
If real roots exist, they are negative because both b and c are positive.
