Mathematics Questions & Answers - Free Online Practice

1,726.

If K is a constant, which of the following equations best describes the parabola?

y = kx²

x = y² - k

y = k - x²

x² = y² - k

y = (k - x)²

View answer & explanation

Correct answer is B

The parabola is best described by the equation x = y² - k because all other equations do not give the equation of the parabola in this position. C for example is the equation of a hyperbola facing downwards. D is the equation of a hyperbola A and E are equations of parabola facing upwards

1,727.

In the fiqure where PQRTU is a circle, ISTI = IRSI and angle TSR = 52^o. Find the angle marked m

128^o

52^o

104^o

64^o

116^o

View answer & explanation

Correct answer is E

< STR = \(\frac{180 - 52}{2}\) = \(\frac{128}{2}\) = 64^o

< PTR = 180 - < STR(angle on a straight line)

= 180 - 64 = 116^o

< PQR + < PTR = 180(Supplementary)

< PQR + 118 = 180

< PQR = 180 - 118

= 64

M = 180 - < PQR

= 180 - < PQR = 180 - 64

= 116^o

1,728.

The diagram is the distance time graph of a vehicle. Find its average speed in kilometers per hour during the journey

155km/hr

50km/hr

40km/hr

124km/hr

84km/hr

View answer & explanation

Correct answer is E

Distance = 155 - 50 = 105km

Time = 75mins

= \(\frac{75}{60}\)hr = \(\frac{5}{4}\)hr

Average speed = \(\frac{Distance}{time}\) = \(\frac{105}{\frac{5}{4}}\)

= \(\frac{105 \times 4}{5}\)

= 84km\h

1,729.

In the diagram, angle QPR = 90^o, angle PSR = 90^o and PR = 5 units. Find the length of QS.

5 tan 25^o sin 65^o

5 cos 25^o sin 65^o

5 tan 25^o cos 65^o

cos 25^o cos 65^o

5 cosec 25^o

View answer & explanation

Correct answer is C

From \(\bigtriangleup\)QPR, < R = 180^o - (25^o + 90^o)

180^o - 115^o = 65^o

From \(\bigtriangleup\)PSQ, Sin 65^o = \(\frac{QPR}{hyp}\) = \(\frac{PS}{5}\)

PS = 5 sin 65^o

From \(\bigtriangleup\)PSR, tan = \(\frac{OPP}{adj}\) = \(\frac{PS}{QS}\)

but PS = 5 sin 65^o

QS tan 25^o = PS

QS tan 25^o = 5 sin 65^o

QS = \(\frac{5 sin 65^o}{tan 25^o}\)

= 5 tan 25^o cos 65^o