The truth set of \(8 + 2x - x^2\) = 0 is {p, q}. Evaluate p + q.
4
2
-6
-2
Correct answer is B
\(8 + 2x - x^2 = 0\)
\(8 + 4x - 2x - x^2 = 0\)
4(2 + x) - x(2 + x) = 0
(4 - x)(2 + x) = 0
4 - x = 0 or 2 + x = 0
x = 4 or x =- 2
So, p = 4 and q = -2
∴ p + q = 2
\(\frac{9}{11}\)
\(\frac{18}{77}\)
\(\frac{1}{2}\)
\(\frac{11}{12}\)
Correct answer is B
1st bag → 4 W, 3 B = 7 marbles
2nd bag → 5 R, 6 B = 11 marbles
The only possible way of picking marbles of the same colour is to pick a blue marble from each bag
therefore, Pr( same colour) = \(\frac{3}{7}\times \frac{6}{11} = \frac{18}{77}\)
y = \(\frac{1}{2}x - 10\)
y = \(\frac{-1}{2}x + 10\)
y = \(\frac{-1}{2}x - 10\)
y = \(\frac{1}{2}x +10\)
Correct answer is C
(7, 4) and (-3, 9) = (\( y_1 , x_1)(y_2 , x_2)\)
slope\( m_1\) of the points(7, 4) and (-3, 9) = \(\frac{ y_2 - y_1}{ x_2 - x_1}\)
\( m_1 = \frac{ 9 - 4}{ -3 - 7} = \frac{5}{-10} = \frac{-1}{2}\)
\( m_1 = m_2\) ( condition for parallelism)
Since the line L passes through the point (6, -13) and is parallel to the points (7, 4) and (-3, 9)
\(\frac{-1}{2} = \frac{ y - (- 13)}{x - 6}\)
= \(\frac{-1}{2} = \frac{ y + 13}{x - 6}\)
= \(\frac{-1}{2}( x - 6) = y + 13\)
\(\frac{-1}{2}x + 3 = y + 13\)
= \(\frac{-1}{2}x + 3 - 13 = y\)
∴ y = \(\frac{-1}{2}x - 10\)
John was facing S35°E. If he turned 90° in the anticlockwise direction, find his new direction.
S55°E.
S35°W.
N55°E.
N35°W.
Correct answer is C
Consider |NS|
θ = 180° - (90° + 35°) (sum of angles on a straight line is 180°)
= 180° - 90° - 35°
= 90° - 35°
= 55°
∴ His new bearing is N55°E.
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