Tutor profile: Siva T.
Vertices of a quadrilateral ABCD are A(0, 0), B(4, 5), C(9, 9) and D(5, 4). What is the shape of the quadrilateral?
The lengths of the four sides, AB, BC, CD and DA are all equal to 41−−√41. Hence, the given quadrilateral is either a Rhombus or a Square. How to determine whether the quadrilateral is a square or a rhombus? The diagonals of a square are equal. The diagonals of a rhombus are unequal. Compute the lengths of the two diagonals AC and BD. The length of AC is 162−−−√162 and the length of BD is 2–√2. As the diagonals are not equal and the sides are equal, the given quadrilateral is a Rhombus.
Solve (x + 3)(x − 3).
=x(x − 3) + 3(x − 3) = x2 − 3x + 3x − 9 = x2 − 9
A golfer practising on a range with an elevated tee 4.9 m above the fairway is able to strike a ball so that it leaves the club with a horizontal velocity of 20 m s–1. (Assume the acceleration due to gravity is 9.80 m s–2, and the effects of air resistance may be ignored unless otherwise stated.) a How long after the ball leaves the club will it land on the fairway? b What horizontal distance will the ball travel before striking the fairway? c What is the acceleration of the ball 0.5 s after being hit? d Calculate the speed of the ball 0.80 s after it leaves the club. e With what speed will the ball strike the ground?
x = ut + 0.5at2 then 4.9 m = 0 + 0.5(9.8 m s–2)t 2 and t = 1.0 s b x = (average speed)(time) = (20 m s–1)(1.0 s) = 20 m c The acceleration of the ball is constant at any time during its flight, and is equal to the acceleration due to gravity = 9.8 m s–2 down d After 0.80 s, the ball has two components of velocity: vx = 20 m s–1 and vy = 0 + (9.8 m s–2)(0.80 s) = 7.84 m s–1 The speed of the ball at 0.80 s is given by: [(20 m s–1) 2 + (7.84 m s–1) 2 ] ½ = 21.5 m s–1 The ball will strike the ground 1.0 s after it is struck. Then vx = 20 m s–1 and vy = 0 + (9.8 m s–2)(1.0 s) = 9.8 m s–1 The speed of the ball at 1.0 s is given by: [(20 m s–1) 2 + (9.8 m s–1) 2 ] ½ = 22.3 m s–1
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