Find the equation of a circle with centre (2, -3) and radius 2 units.
\(x^{2} + y^{2} - 4x + 6y + 9 = 0\)
\(x^{2} + y^{2} + 4x - 6y - 9 = 0\)
\(x^{2} + y^{2} + 4x + 6y - 9 = 0\)
\(x^{2} + y^{2} + 4x - 6y + 9 = 0\)
Correct answer is A
The equation of a circle with centre coordinate (a, b) and radius r is :
\((x - a)^{2} + (y - b)^{2} = r^{2}\)
Given centre = (2, -3) and radius r = 2 units
Equation = \((x - 2)^{2} + (y - (-3))^{2} = 2^{2}\)
\(x^{2} - 4x + 4 + y^{2} + 6y + 9 = 4\)
\(x^{2} + y^{2} - 4x + 6y + 4 + 9 - 4 = 0 \implies x^{2} + y^{2} - 4x + 6y + 9 = 0\)
3
2
-2
-3
Correct answer is D
The sum of deviations from the mean of a set of numbers equals 0.
\((k+3)^{2} + (k+7) + (-2) + k + (k+2)^{2} = 0\)
\((k^2 + 6k + 9) + (k+7) - 2 + k + (k^2 + 4k + 4) = 0\)
\(2k^{2} + 12k + 18 = 0\)
\(2k^{2} + 6k + 6k + 18 = 2k(k + 3) + 6(k + 3) = 0\)
\(k = -3 (twice)\)
\(-45(2i + \sqrt{2}j)\)
\(60(\sqrt{3}i + 7j)\)
\(30(7i + \sqrt{3}j)\)
\(-15(7i + \sqrt{3}j)\)
Correct answer is D
\(F = F\cos\theta + F\sin\theta\)
\(\implies 90N = 90\cos 120° + 90\sin 120°\)
\(120N = 120 \cos 240° + 120 \sin 240°\)
\(R = F_{1} + F_{2} \)
= \((90 \cos 120 + 120 \cos 240)i + (90\sin 120 + 120 \sin 240)j\)
= \(90(-0.5) + 120(-0.5))i + (90(\frac{\sqrt{3}}{2}) + (120(-\frac{\sqrt{3}}{2}))j\)
= \(-105i - 15\sqrt{3}j = -15(7i + \sqrt{3}j)\)
If a fair coin is tossed four times, what is the probability of obtaining at least one head?
\(\frac{1}{2}\)
\(\frac{1}{4}\)
\(\frac{13}{16}\)
\(\frac{15}{16}\)
Correct answer is D
P(at least one head) = 1 - P(4 tails)
Let \(p = \frac{1}{2}\) = probability of head and \(q = \frac{1}{2}\) = probability of tail.
\((p + q)^{4} = p^{4} + 4p^{3}q + 6p^{2}q^{2} + 4pq^{3} + q^{4}\)
P(4 tails) = \(q^{4} = (\frac{1}{2})^{4} = \frac{1}{16}\)
P(at least one head) = \(1 - \frac{1}{16} = \frac{15}{16}\)
Find the coefficient of \(x^3\) in the binomial expansion of \((3x + 4)^4\) in ascending powers of x
432
194
144
108
Correct answer is A
\((3x + 4)^{4} = ^{4}C_{0}(3x)^{0}(4)^{4} + ^{4}C_{1}(3x)^{1}(4)^{3} + ^{4}C_{2}(3x)^{2}(4)^{2} + ^{4}C_{3}(3x)^{3}(4)^{1} + ^{4}C_{4}(3x)^{4}(4)^{0}\)
\(x^{3} = ^{4}C_{3}(3x)^{3}(4) = \frac{4!}{3!1!} \times 3^{3} \times 4\)
= \(432x^{3}\)