-\(\frac{1}{2} \leq x\) < 3
-\(\frac{1}{2} < x \leq 3\)
-\(\frac{1}{2} < x < 3\)
-\(\frac{1}{2} \leq x \leq 3\)
Correct answer is C
2x2 - 5x - 3 = 0
2x2 - 6x + x - 3 = 0
2x(x - 3) + 1(x - 3) = 0
(2x + 1)(x - 3) = 0
2x + 1 = 0
2x = -1
x = -\(\frac{1}{2}\)
x - 3 = 0
-\(\frac{1}{2}\) < x < 3
In the diagram, |SR| = |QR|. < SRP = 65o and < RPQ = 48o, find < PRQ
65o
45o
25o
19o
Correct answer is D
< RSQ = < RPQ = 48o (angle in the same segment)
< SQR < RSQ (Base angle of an isosceles \(\bigtriangleup\))
< SQR = 480
< QRS + < RSQ + < RSQ = 180o(sum of interior angles of a \(\bigtriangleup\))
i.e. < QRS + 48o + 48o = 180
< QRS = 180 - (48 + 48) = 180 - 96 = 84o
but < PRQ + < PRS = < QRS
< PRQ = < QRS - < PRS - 84 - 65
= 19o
2.32cm
1.84cm
0.62cm
0.26cm
Correct answer is C
No explanation has been provided for this answer.
The diagram is a circle with centre P. PRST are points on the circle. Find the value of < PRS
144o
72o
40o
36o
Correct answer is A
Reflex < POS = 2x (angle at centre is twice that at circumference)
reflex < POS + < POS = 350o(angles at a point)
i.e. 2x + 8x = 360o
10x = 360o
x = \(\frac{360}{10}\)
= 36o
< PRS = \(\frac{1}{2}\)
< POS(< at centre twice that circumference)
= \(\frac{1}{2}\) x 8x = 4x
4 x 36o
< PRS = 144
In the diagram, MN//PO, < PMN = 112o, < PNO = 129oo and < MPN = yo. Find the value of y
51o
54o
56o
68o
Correct answer is B
In \(\bigtriangleup\) NPO + PNO + PNO + < NOP = 180o(sum of interior angles of a \(\bigtriangleup\) )
i.e. NPO + 129 + 37 = 180
< NOP = 180 - (129 + 37) = 14o
< MNP = < NOP = 14o (alt. < s)
In \(\bigtriangleup\) MPN
< PMN + < MNP + y = 180(sum of interior angles of a \(\bigtriangleup\))
i.e. 112 + 14 + y = 180o
y = 180 - (112 + 14) = 180 - 126 = 54o
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