Make t the subject of formula S = ut + \(\frac{1}{2} at^2\)
\(\frac{1}{a}\) (-u + \(\sqrt{U^2 - 2as}\))
\(\frac{1}{a}\) {u \(\pm\) (U2 - 2as)}
\(\frac{1}{a}\) {u \(\pm\) \(\sqrt{2as}\)}
\(\frac{1}{a}\) {-u + \(\sqrt{( 2as)}\)}
Correct answer is A
Given S = ut + \(\frac{1}{2} at^2\)
S = ut + \(\frac{1}{2} at^2\)
∴ 2S = 2ut + at2
= at2 + 2ut - 2s = 0
t = \(\frac{-2u \pm 4u^2 + 2as}{2a}\)
= -2u \(\pi\) \(\frac{\sqrt{u^2 4u^2 + 2as}}{2a}\)
= \(\frac{1}{a}\) (-u + \(\sqrt{U^2 - 2as}\))
Solve the equation: \(y - 11\sqrt{y} + 24 = 0\)
8, 3
64, 9
6, 4
9, -8
Correct answer is B
\(y - 11\sqrt{y} + 24 = 0 \implies y + 24 = 11\sqrt{y}\)
Squaring both sides,
\(y^{2} + 48y + 576 = 121y\)
\(y^{2} + 48y - 121y + 576 = 0 \implies y^{2} - 73y + 576 = 0\)
\(y^{2} - 64y - 9y + 576 = 0\)
\(y(y - 64) - 9(y - 64) = 0\)
\((y - 9)(y - 64) = 0\)
\(\therefore \text{y = 64 or y = 9}\)
Factorize \(9p^2 - q^2 + 6qr - 9r^2\)
(3p - 3q + r)(3p - q - 3r)
(6p - 3q - 3r)(3p - q - 4r)
(3p - q + 3r)(3p + q - 3r)
(3q - p + 3r)(3q - p + 3r)
Correct answer is C
\(9p^{2} - q^{2} + 6qr - 9r^{2}\)
= \(9p^{2} - (q^{2} - 6qr + 9r^{2})\)
= \(9p^{2} - (q^{2} - 3qr - 3qr + 9r^{2})\)
= \(9p^{2} - (q(q - 3r) - 3r(q - 3r))\)
= \(9p^{2} - (q - 3r)^{2}\)
= \((3p + (q - 3r))(3p - (q - 3r))\)
= \((3p + q - 3r)(3p - q + 3r)\)
(0)
U
(8)
\(\phi\)
Correct answer is D
U = (1, 2, 3, 6, 7, 8, 9, 10)
E = (10, 4, 6, 8, 10)
F = (x : x\(^2\) = 2\(^6\), x is odd)
∴ F = \(\phi\) Since x\(^2\) = 2\(^6\) = 64
x = \(\pm 8\) which is even
∴ E ∩ F = \(\phi\) Since there are no common elements
(2, 4, 3, 5, 11) and (4)
(4, 3, 5, 11) and (3, 4)
(2, 5, 11) and (2)
(2, 3, 5, 11) and (2)
Correct answer is D
x = (all prime factors of 44) and y = (all prime factors of 60)
∴ x = (2, 11), y = (2, 3, 5)
X ∪ Y = (2, 3, 5, 11),
X ∩ Y = (2)