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Ajụjụ 2 Ripọtì
Akọwa Nkọwa
Ajụjụ 3 Ripọtì
Akọwa Nkọwa
Ajụjụ 4 Ripọtì
Akọwa Nkọwa
Given P(2, -3) and Q(-5, 1)
Midpoint = (2+(−5)2,−3+12)
= (−32,−1)
Slope of the line PQ = 1−(−3)−5−2
= −47
The slope of the perpendicular line to PQ = −1−47
= 74
The equation of the perpendicular line: y=74x+b
Using a point on the line (in this case, the midpoint) to find the value of b (the intercept).
−1=(74)(−32)+b
−1+218=138=b
∴ The equation of the perpendicular bisector of the line PQ is y=74x+138
≡8y=14x+13⟹8y−14x−13=0
Ajụjụ 5 Ripọtì
Convert 2710 to another number in base three
Akọwa Nkọwa
Ajụjụ 6 Ripọtì
| Score | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| Frequency | 1 | 0 | 7 | 5 | 2 | 3 | 1 | 1 |
The table above shows the scores of 20 students in further mathematics test. What is the range of the distribution?
Akọwa Nkọwa
Ajụjụ 7 Ripọtì
Akọwa Nkọwa
Ajụjụ 8 Ripọtì
If \(P = \begin{vmatrix} 5 & 3 \\ 2 & 1 \end{vmatrix}\) and \(Q = \begin{vmatrix} 4 & 2 \\ 3 & 5 \end{vmatrix}\), find \(2P + Q\)
Akọwa Nkọwa
Ajụjụ 9 Ripọtì
Akọwa Nkọwa
Ajụjụ 10 Ripọtì
In triangle PQR, \(q = 8\text{ cm}\), \(r = 6\text{ cm}\) and \(\cos P = \frac{1}{12}\). Calculate the value of \(p\).
Akọwa Nkọwa
Using the cosine rule, we have
p2=q2+r2−2qrcosP
p2=82+62−2(8)(6)(112)
= 64+36−8
p2=92∴p=92−−√cm
Ajụjụ 11 Ripọtì
Akọwa Nkọwa
Ajụjụ 12 Ripọtì
Evaluate \( \frac{1.25 \times 0.025}{0.05} \), correct to 1 decimal place
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Ajụjụ 13 Ripọtì
Akọwa Nkọwa
Ajụjụ 14 Ripọtì
The nth term of the progression \( \frac{4}{2}, \frac{7}{3}, \frac{10}{4}, \frac{13}{5} \) is ..
Akọwa Nkọwa
Ajụjụ 15 Ripọtì
P, Q and R are subsets of the universal set U. The Venn diagram showing the relationship \( (P \cap Q) \cup R \) is
Ajụjụ 16 Ripọtì
| Age | 20 | 25 | 30 | 35 | 40 | 45 |
| Number of people | 3 | 5 | 1 | 1 | 2 | 3 |
Find the median age of the frequency distribution in the table above.
Akọwa Nkọwa
Ajụjụ 17 Ripọtì
The bar chart above shows the allotment of time(in minutes) per week for selected subjects in a certain school. What is the total time allocated to the six subjects per week?
Akọwa Nkọwa
Ajụjụ 18 Ripọtì
Akọwa Nkọwa
Length of chord = 2rsin(θ2)
= 2×8×sin(902)
= 16×2√2
= 82–√cm
Ajụjụ 19 Ripọtì
Simplify \( \frac{\sqrt{5}(\sqrt{147}-\sqrt{12})}{\sqrt{15}} \)
Akọwa Nkọwa
Ajụjụ 20 Ripọtì
The graph is shown is correctly represented by
Akọwa Nkọwa
Ajụjụ 21 Ripọtì
Akọwa Nkọwa
Ajụjụ 22 Ripọtì
P varies jointly as m and u, and varies inversely as q. Given that p = 4, m = 3 and u = 2 and q = 1, find the value of p when m = 6, u = 4 and q = \( \frac{8}{5} \)
Akọwa Nkọwa
Ajụjụ 23 Ripọtì
Akọwa Nkọwa
Ajụjụ 25 Ripọtì
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Ajụjụ 27 Ripọtì
If \(S = \sqrt{t^2 - 4t + 4}\), find t in terms of S
Akọwa Nkọwa
Ajụjụ 28 Ripọtì
In the diagram, find the size of the angle marked ao
Akọwa Nkọwa
S = 280o2
= 140
< O = 360 - 280 = 80o
60 + 80 + 140 + a = 360o
(< in a quad); 280 = a = 360
a = 360 - 280
a = 80o
Ajụjụ 30 Ripọtì
Akọwa Nkọwa
Ajụjụ 31 Ripọtì
Akọwa Nkọwa
Ajụjụ 32 Ripọtì
Evaluate \( \int_{0}^{\frac{\pi}{2}} \sin x \, dx \)
Akọwa Nkọwa
Ajụjụ 33 Ripọtì
What is the probability that an integer \( (1 \le x \le 25) \) chosen at random is divisible by both 2 and 3?
Akọwa Nkọwa
Ajụjụ 34 Ripọtì
Akọwa Nkọwa
Ajụjụ 35 Ripọtì
Akọwa Nkọwa
Ajụjụ 36 Ripọtì
Simplify \( \frac{3^{-5n}}{9^{1-n}} \times 27^{n+1} \)
Akọwa Nkọwa
3−5n−2(1−n)+3(n+1)
3−5n−2+2n+3n+3
3−5n+5n+3−2
31
= 3
Ajụjụ 37 Ripọtì
Ajụjụ 38 Ripọtì
The pie chart above shows the statistical distribution of 80 students in five subjects in an examination. Calculate how many student offer Mathematics.
Akọwa Nkọwa
360x∘ - 24 + 12 + 12 = 360∘
36x∘ = 360∘
x∘ = 360036
= 10∘
Thus, the angle of sector representing Mathematics is 5 x 10∘ = 50∘ . Hence the number of students who offer mathematics is
55o36×80≈11
Ajụjụ 39 Ripọtì
Akọwa Nkọwa
Ajụjụ 41 Ripọtì
Akọwa Nkọwa
Ajụjụ 42 Ripọtì
Find the inverse \( \begin{vmatrix} 5 & 3 \\ 6 & 4 \end{vmatrix} \)
Akọwa Nkọwa
Ajụjụ 43 Ripọtì
Akọwa Nkọwa
Ajụjụ 44 Ripọtì
Akọwa Nkọwa
Ajụjụ 45 Ripọtì
A chord of a circle subtends an angle of 120° at the centre of a circle of diameter \(4\sqrt{3}\,cm\). Calculate the area of the major sector.
Akọwa Nkọwa
Angle of major sector = 360° - 120° = 240°
Area of major sector : θ360×πr2
r = 43√2=23–√cm
Area : 240360×π×(23–√)2
= 8πcm2
Ajụjụ 46 Ripọtì
If \( \tan \theta = \frac{3}{4} \), find the value of \( \sin \theta + \cos \theta \).
Akọwa Nkọwa
tanθ=oppadj=34
hyp2=opp2+adj2
hyp=32+42−−−−−−√
= 5
sinθ=35;cosθ=45
sinθ+cosθ=35+45
= 75=125
Ajụjụ 47 Ripọtì
Akọwa Nkọwa
Ajụjụ 48 Ripọtì
Akọwa Nkọwa
In the diagram above, < CDE = < CED = 15o
(base < s of isos. △)
< ECD = 180o - (15 + 15)o
= 180o - 30o = 150o
But x + 110o = 150o
(Sum of opp. interior < s of a△ = opp. exterior < )
x = 150o - 110o = 40o
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