The roots of a quadratic equation are -3 and 1. Find its equation.
\(x^{2} - 3x + 1 = 0\)
\(x^{2} - 2x + 1 = 0\)
\(x^{2} + 2x - 3 = 0\)
\(x^{2} + x - 3 = 0\)
Correct answer is C
Given the roots of an equation such that you can find the sum and product of the roots, the equation can be given as:
\(x^{2} - (\alpha + \beta)x + (\alpha \beta) = 0 \)
\(\alpha + \beta = -3 + 1 = -2\)
\(\alpha \beta = -3 \times 1 = -3\)
Equation: \(x^{2} - (-2)x + (-3) = 0 \implies x^{2} + 2x - 3 = 0\)
\(q \implies p\)
\(\sim q \implies p\)
\(\sim q \implies \sim p\)
\(\sim p \implies \sim q\)
Correct answer is C
No explanation has been provided for this answer.
\(\frac{-5}{7}\)
\(-\frac{2}{5}\)
\(\frac{2}{5}\)
\(\frac{5}{7}\)
Correct answer is B
\(S_{\infty} = \frac{a}{1 - r}\) (Sum to infinity of a GP)
\(250 = \frac{350}{1 - r} \implies 250(1 - r) = 350\)
\(350 = 250 - 250r \implies 350 - 250 = -250r\)
\(250r = -100 \implies r = \frac{-100}{250} = -\frac{2}{5}\)
12.6 cm
12.0 cm
9.6 cm
9.0 cm
Correct answer is D
Length of arc (in radians) = \(r \theta\)
\(10.8 = 1.2r\)
\(r = \frac{10.8}{1.2} = 9.0 cm\)
The gradient of point P on the curve \(y = 3x^{2} - x + 3\) is 5. Find the coordinates of P.
(1, 5)
(1, 7)
(1, 13)
(1, 17)
Correct answer is A
\(y = 3x^{2} - x + 3\)
\(\frac{\mathrm d y}{\mathrm d x} = 6x - 1 = 5\)
\(6x = 5 + 1 = 6 \implies x = 1\)
\(y = 3(1^{2}) - 1 + 3 = 3 - 1 + 3 = 5\)
\(P = (1, 5)\)