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Question 1 Report
PQRST is a regular pentagon and PQVU is a rectangle with U and V lying on TS and SR respectively as shown in the diagram. Calculate TUP
Answer Details
Question 2 Report
Question 3 Report
Simplify \( \frac{1}{p} - \frac{1}{q} + \frac{p}{q} - \frac{q}{p} \)
Question 4 Report
Question 6 Report
Question 7 Report
simplify \( \frac{1}{\sqrt{3}-2} - \frac{1}{\sqrt{3}+2} \)
√3+2−√3+23−2√3+2√3−4
= 43−2
= 4−1
= -4
Question 8 Report
Question 9 Report
In the diagram, \(QP // ST\): \(PQR = 34^\circ\) \(QRS = 73^\circ\) and \(RS = RT\). Find \(SRT\)
R = 180∘ - 107∘
< p = 180∘ - (107∘ - 34∘ )
108 - 141∘ = 39∘
Angle < S = 39∘ (corr. Ang.) But in △ SRT
< S = < T = 39∘
SRT = 180 - (39∘ + 39∘ )
= 180∘ - 78∘
= 102∘
Question 10 Report
In the diagram, PQRs is a circle with 0 as centre and PQ/RT. If RTS = \(32^\circ\). Find PSQ
< RTS = < PQS = 32∘ (Alternative angle)
< PSQ = 90 - < PSQ = 90∘ - 32∘
= 58∘
Question 11 Report
Question 12 Report
Evaluate \( \frac{3524}{0.05} \) correct to 3 significant figures
Question 13 Report
Make x the subject of the relation \( \frac{1+ax}{1-ax}=\frac{p}{q} \)
Question 14 Report
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Question 15 Report
In the diagram, O is the centre of the circle and POQ a diameter. If POR = \(96^\circ\), find the value of ORQ.
< ROQ = 180 - 86 = 84?
? OQR = Isosceles
R = Q
R + Q + 84 = 180(angle in a ? )
2R = 96 since R = Q
R = 48?
ORQ = 48?
Question 16 Report
In the diagram above, |PQ| = |QR|, |PS| = |RS|, ∠PSR = 30o and ∠PQR = 80o. Find ∠SPQ.
Question 17 Report
| \(Weight(s)\) | \(0-10\) | \(10-20\) | \(20-30\) | \(40-50\) | |
| Number of coconuts | \(10\) | \(27\) | \(19\) | \(6\) | \(2\) |
Estimate the mode of the frequency distribution above.
Question 18 Report
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Question 27 Report
Calculate the length in cm. of the area of a circle of diameter 8cm which subtends an angle of \( \frac{1}{2} \)o at the centre of the circle
Question 28 Report
find the radius of a sphere whose surface area is 154cm2 \( \left(\pi=\frac{22}{7}\right) \)
Question 29 Report
Integrate 1−xx3 with respect to x
Question 30 Report
If x is negative, what is the range of values of x within which \( \frac{x+1}{3} > \frac{1}{X+3} \)
= x > 0, x < -3, x < -4 = x < -3(solution only)
Case 3 (-, +, -) = x < 0, x > -3, x < -4 = x < -0, -4 < x < 3(solutions)
Case 4 (-, -, +) = x < 0, x + 3 < 0, x + 4 > 0
= x < 0, x < -5, x > -4 = x < -0, -4 < x < -3(solution)
combining the solutions -4 < x < -3
Question 31 Report
If \(9\left(x - \frac{1}{2}\right)^3x^2\)
Question 32 Report
The chances of three independent events X, Y, Z occurring are \( \frac{1}{2} \), \( \frac{2}{3} \), \( \frac{1}{4} \) respectively. What are the chances of Y and Z only occurring?
Question 33 Report
From the figure, calculate TH in centimeters
TH5+QH = tan 30∘
TH = (b + QH) tan 30∘
QH = 56 (5 + QH) 1√3
QH(1 - 1√3 ) = 5√3
QH = 5√3√3−1√3
= 5√3−1
Question 34 Report
Solve without using tables \( \log_{5}(62.5) - \log_{5}\left(\frac{1}{2}\right) \)
Question 35 Report
Question 36 Report
If \( \sin \theta = \cos \theta \), find \( \theta \) between 0o and 360o
Question 37 Report
The bar chart shows the distribution of marks in a class test. How many students took the test?
Question 38 Report
In the diagram, QPS = SPR, PR = 9cm. PQ = 4cm and QS = 3cm, find SR.
QS/QP = SR/PR
3/4 = SR/g
4SR = 27
SR = 274
= 634 cm
Question 39 Report
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Question 40 Report
If \( \sqrt{x^2 + 9} = x + 1 \), solve for x
Question 41 Report
| Class | Frequency |
| 1−5 | 2 |
| 6−10 | 4 |
| 11−15 | 5 |
| 16−20 | 2 |
| 21−25 | 3 |
| 26−30 | 2 |
| 31−35 | 1 |
| 36−40 | 1 |
Find the median of the observation in the table given.
Question 42 Report
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Question 44 Report
A binary operation \( \ast \) is defined on a set of real numbers by \(x \ast y = x^y\) for all real values of \(x\) and \(y\). If \(x \ast 2 = x\). Find the possible values of \(x\)
Question 45 Report
Question 46 Report
The shaded portion in the Venn diagram is
Question 47 Report
In the figure, the line segment ST is tangent to two circles at S and T. O and Q are the centres of the circles with OS = 5cm. QT = 2cm and OR = 14cm. Find ST
SQ2 = 142 - 52
196 - 25 = 171
ST2 + TQ2 = SQ2
ST2 + 22 = 171
ST2 = 171 - 4
= 167
ST = √167
= 12.92 = 12.9cm
Question 48 Report
Evaluate \( \left(x + \frac{1}{x} + 1\right)^2 - \left(x + \frac{1}{x} + 1\right)^2 \)
Question 49 Report
Question 50 Report
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