\(p \vee q\)
\(\sim p \vee q\)
\(p \wedge (\sim q)\)
\( p \wedge q\)
Correct answer is D
No explanation has been provided for this answer.
Given that \(x^{2} + 4x + k = (x + r)^{2} + 1\), find the value of k and r
k = 5, r = -1
k = 5, r = 2
k = 2, r = -5
k = -1, r = 5
Correct answer is B
\(x^{2} + 4x + k = (x + r)^{2} + 1\)
\(x^{2} + 4x + k = x^{2} + 2rx + r^{2} + 1\)
Comparing the LHS and RHS equations, we have
\(2r = 4 \implies r = 2\)
\(k = r^{2} + 1 = 2^{2} + 1 = 5\)
\(\frac{5}{4}\)
\(\frac{3}{4}\)
\(\frac{1}{4}\)
\(\frac{-3}{4}\)
Correct answer is A
Given, \(x^{2} + x - 2 = 0\), a = 1, b = 1 and c = -2.
\(\alpha + \beta = \frac{-b}{a} = \frac{-1}{1} = -1\)
\(\alpha\beta = \frac{c}{a} = \frac{-2}{1} = -2\)
\(\frac{1}{\alpha^{2}} + \frac{1}{\beta^{2}} = \frac{\beta^{2} + \alpha^{2}}{(\alpha\beta)^{2}}\)
\(\beta^{2} + \alpha^{2} = (\alpha + \beta)^{2} - 2\alpha\beta = (-1)^{2} - 2(-2) = 1 + 4 = 5\)
\(\frac{1}{\alpha^{2}} + \frac{1}{\beta^{2}} = \frac{5}{(-2)^{2}} = \frac{5}{4}\).
The gradient of a curve at the point (-2, 0) is \(3x^{2} - 4x\). Find the equation of the curve.
\(y = 6x - 4\)
\(y = 6x^{2} - 4x + 12\)
\(y = x^{3} - 2x^{2}\)
\(y = x^{3} - 2x^{2} + 16\)
Correct answer is D
The gradient of a curve is gotten by differentiating the equation of the curve. Therefore, given the gradient, integrate to get the equation of the curve back.
\(\frac{\mathrm d y}{\mathrm d x} = 3x^{2} - 4x\)
\(y = \int {(3x^{2} - 4x)} \mathrm {d} x = \frac{3x^{2+1}}{2+1} - \frac{4x^{1+1}}{1+1} + c\)
= \(x^{3} - 2x^{2} + c \)
To find c (the constant of integration), when x = -2, y = 0
\(0 = (-2^{3}) - 2(-2^{2}) + c\)
\(0 = -8 - 8 + c \implies c = 16\)
\(\therefore y = x^{3} - 2x^{2} + 16\)
If \(x = i - 3j\) and \(y = 6i + j\), calculate the angle between x and y
trong>
75°
81°
85°
Correct answer is C
\(\overrightarrow{x} . \overrightarrow{y} = |\overrightarrow{x}||\overrightarrow{y}|\cos\theta\)
\(\overrightarrow{x} . \overrightarrow{y} = (i - 3j) . (6i + j) = 6 - 3 = 3\)
\(|\overrightarrow{x}| = \sqrt{1^{2} + (-3)^{2}} = \sqrt{10}\)
\(|\overrightarrow{y}| = \sqrt{6^{2} + 1^{2}} = \sqrt{37}\)
\(\therefore 3 = (\sqrt{10})(\sqrt{37})\cos \theta\)
\(\cos\theta = \frac{3}{\sqrt{370}} = 0.1559\)
\(\theta = \cos^{-1} 0.1559 \approxeq 81°\)