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**Question 1**
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The pie chart above shows the distribution of subjects offered by students in SSS III level. If 80 students enrolled in the class. What is the size of the angle of the sector in economics?

**Answer Details**

xo + 36o + 54o + 28o + 90o + 120o = 360o

x + 328o = 360o

x = 360o - 328o

x = 32o

**Question 2**
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OGIVE is constructed using

**Answer Details**

An ogive is a graph used to show the cumulative frequency of a given dataset. It is constructed using a cumulative frequency table which contains the frequencies and the cumulative frequencies. The cumulative frequency is the total of the frequencies up to that point in the data set. Therefore, the correct answer is "Cumulative frequency table". The other options mentioned, third quartile range, semi-quartile range, and inter-quartile range, are statistical measures that can be used to describe the spread of a dataset, but they are not directly used in the construction of an ogive.

**Question 3**
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implify 1 - (17 $\frac{1}{7}$ x 3 12 $\frac{1}{2}$) รท $\u0e23\u0e17$ 34

**Answer Details**

1 - (17
$\frac{1}{7}$ x 3 12
$\frac{1}{2}$) รท
$\u0e23\u0e17$ 34
$\frac{3}{4}$

1 - (17
$\frac{1}{7}$ x 72
$\frac{7}{2}$) x 43
$\frac{4}{3}$

1 - 12
$\frac{1}{2}$ x 43
$\frac{4}{3}$

1 - 23
$\frac{2}{3}$ = 13

**Question 4**
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From the cyclic quadrilateral above, find < TVS

**Answer Details**

< TVS = 180โโ(80+20)
${180}^{\u0e42\x88\x98}\u0e42\x88\x92(80+20)$

= 180โ100=80โ

**Question 5**
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Find the mean of 10, 8, 5, 11, 12, 9, 6, 3, 15, and 23.

**Answer Details**

To find the mean of a set of numbers, you add up all the numbers and then divide by the total count of numbers. So, for the given set of numbers, we add them up: 10 + 8 + 5 + 11 + 12 + 9 + 6 + 3 + 15 + 23 = 102 There are 10 numbers in the set, so we divide the sum by 10: 102 รท 10 = 10.2 Therefore, the mean of the given set of numbers is 10.2. So the answer is 10.2.

**Question 6**
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If Q is a factor of 18 and T is prime numbers between 2 and 18. What is Qโฉ $\u0e42\x88\u0e09$T?

**Answer Details**

Given that Q is a factor of 18, the possible values of Q are 1, 2, 3, 6, 9, and 18. We also know that T is a prime number between 2 and 18. The prime numbers between 2 and 18 are 2, 3, 5, 7, 11, 13, and 17. To find the intersection of Q and T, we look for the values that are common to both sets. The only values of Q that are also in the set of prime numbers between 2 and 18 are 2 and 3. Therefore, the intersection of Q and T is (2, 3). So the answer is: (2, 3).

**Question 7**
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The nth term of the sequence 3, 9, 27, 81.... is

**Answer Details**

3,9,27,81,.....

r = 98
$\frac{9}{8}$ = 3

Tn = arn-1

Tn = 3(3)n-1 = 3(32)3

**Question 8**
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Make S the subject of the relation

p = s + sm2nr

**Answer Details**

p = s + sm2nr
$\frac{s{m}^{2}}{nr}$

p = s + ( 1 + m2nr
$\frac{{m}^{2}}{nr}$)

p = s (1 + nr+m2nr
$\frac{nr+{m}^{2}}{nr}$)

nr ร p = s (nr + m2)

s = nrpnr+m2

**Question 9**
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Calculate the perimeter of a sector of a circle of radius 9cm and angle 36o.

**Answer Details**

The perimeter of a sector of a circle is the sum of the arc length and the two radii. To find the perimeter, we need to first find the arc length using the formula: arc length = (angle/360) x 2 x π x radius In this case, the angle is given as 36°, the radius is given as 9cm, and π is approximately equal to 3.14. Substituting these values into the formula, we get: arc length = (36/360) x 2 x 3.14 x 9 = 2.83 cm (rounded to two decimal places) Now, we can find the perimeter by adding the two radii to the arc length: perimeter = 2 x 9 + 2.83 = 20.83 cm Therefore, the answer is (B) (18 + 9π/5) cm, rounded to two decimal places.

**Question 10**
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If x10 = 235. Find x

**Answer Details**

x10 = 235

x10 = 2 x 51 + 3 x 50

= 10 + 3

= x10 = 13

**Question 11**
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Calculate the perimeter, in cm, of a sector of a circle of radius 8cm and angle 45o

**Answer Details**

Perimeter = OP + OQ + PQ

= 8 + 8 + PQ

length PQ = ฮธ360ร2ฯr
$\frac{\mathrm{\u0e2e\u0e18}}{360\u0e23\x972\mathrm{\u0e2f\x80}r}$

= 45360
$\frac{45}{360}$ x 2 x ฯ
$\u0e2f\x80$ x 8

= 2ฯ
$\u0e2f\x80$

Perimeter of sector 2r + L

Where l = length of arc and r = radius

โด P = 2(8) + 2ฯ
$\u0e2f\x80$

= 16 + 2ฯ

**Question 12**
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The venn diagram above shows a class of 40 students with the games they play. How many of the students play two games only?

**Answer Details**

To find the number of students who play two games only, we need to add up the number of students in the regions of the Venn diagram that represent playing two games only. Looking at the diagram, we can see that there are two regions that represent playing two games only: the intersection of the football and basketball circles, and the intersection of the basketball and volleyball circles. The number of students who play two games only is the sum of the number of students in these two regions. From the diagram, we can see that there are 7 students in the football and basketball region, and 8 students in the basketball and volleyball region. Therefore, the total number of students who play two games only is: 7 + 8 = 15 So, 15 of the 40 students play two games only.

**Question 13**
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Find dydx $\frac{dy}{dx}$. If y = 3x3 + 2x2 + 3x + 1

**Answer Details**

y = 3x3 + 2x2 + 3x + 1

dydx
$\frac{dy}{dx}$ = 9x2 + 4x + 3

**Question 14**
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If the 2nd term of a G.P is 89 $\frac{8}{9}$ and the 6th term is 412 $\frac{1}{2}$. Find the common ratio.

**Answer Details**

T2 = 89
$\frac{8}{9}$, T6 = 412
$\frac{1}{2}$

ar = 89
$\frac{8}{9}$

ar5 = 92
$\frac{9}{2}$

arar5
$\frac{ar}{a{r}^{5}}$ = 89
$\frac{8}{9}$ x 29
$\frac{2}{9}$

1r4
$\frac{1}{{r}^{4}}$ = 1681
$\frac{16}{81}$

r = 4โ8116
$\sqrt[4]{\frac{{\textstyle 81}}{{\textstyle 16}}}$ = 32

**Question 15**
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In the cyclic quadrilateral above . Find < PRO

**Answer Details**

< PRO = 180โโ(100+50)
${180}^{\u0e42\x88\x98}\u0e42\x88\x92(100+50)$

= 180โ150=30โ

**Question 16**
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Evaluate 0.8ร0.43ร0.0310.05ร0.72ร0.021 $\frac{{\textstyle 0.8\u0e23\x970.43\u0e23\x970.031}}{{\textstyle 0.05\u0e23\x970.72\u0e23\x970.021}}$ correct to four significant figures.

**Answer Details**

0.8ร0.43ร0.0310.05ร0.72ร0.021
$\frac{{\textstyle 0.8\u0e23\x970.43\u0e23\x970.031}}{{\textstyle 0.05\u0e23\x970.72\u0e23\x970.021}}$ x 1010
$\frac{{\textstyle 10}}{{\textstyle 10}}$ = 14.11

**Question 17**
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From the diagram above, Find the value of < ROP

**Answer Details**

< ROP = 95โ ${95}^{\u0e42\x88\x98}$ (Exterior angle of cyclic quadrilateral)

**Question 18**
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If N = p2 $\frac{{\textstyle p}}{{\textstyle 2}}$(T1โT2T1) $(\frac{{\textstyle {T}_{1}\u0e42\x88\x92{T}_{2}}}{{\textstyle {T}_{1}}})$. Find P when N = 12, T1 = 27 and T2 = 24.

**Answer Details**

N = p2
$\frac{{\textstyle p}}{{\textstyle 2}}$(T1โT2T1)
$(\frac{{\textstyle {T}_{1}\u0e42\x88\x92{T}_{2}}}{{\textstyle {T}_{1}}})$

12 = p2
$\frac{{\textstyle p}}{{\textstyle 2}}$(27โ2427)
$(\frac{{\textstyle 27\u0e42\x88\x9224}}{{\textstyle 27}})$

24 = P(327)
$(\frac{{\textstyle 3}}{{\textstyle 27}})$

P = 24 x 9 = 216

**Question 19**
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Find dydx $\frac{dy}{dx}$, if y = 23 $\frac{2}{3}$ X3 - 4x

**Answer Details**

y = 23
$\frac{2}{3}$x3 - 4x
$\frac{4}{x}$

dydx
$\frac{dy}{dx}$ = 2x2 - Lx-2 = 2x2 + 4x2

**Question 20**
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In a school of 150 students, 80 offer French while 60 offer Arabic and 20 offer neither. How many students offer both subjects?

**Answer Details**

80 - x + x + 60 - x + 20 = 150

160 - x = 150

x = 160 - 150 = 10

**Question 21**
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From the diagram above, find the required roots/region of x.

**Answer Details**

The required roots/region of x is: - -1 โค x โค 4 The diagram shows a closed circle at -1 and an open circle at 4, indicating that -1 and 4 are included and excluded from the solution set, respectively. This means that any value of x that is greater than or equal to -1 and less than or equal to 4 is a solution to the inequality. Therefore, the required region of x is the interval between -1 and 4, including -1 and excluding 4.

**Question 22**
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Evaluate ฯ2โซ0
$\underset{0}{\overset{\frac{\mathrm{\u0e2f\x80}}{2}}{\u0e42\x88\u0e0b}}$ sin *xdx*

**Answer Details**

ฯ2โซ0
$\underset{0}{\overset{\frac{\mathrm{\u0e2f\x80}}{2}}{\u0e42\x88\u0e0b}}$ sin *xdx* = -cosx|ฯ2
$\frac{\mathrm{\u0e2f\x80}}{2}$

= -(cosฯ2
$\frac{\mathrm{\u0e2f\x80}}{2}$ - cos0) = -(0-1) = 1

**Question 23**
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Evaluate โซ $\u0e42\x88\u0e0b$(cos4x + sin3x)dx

**Question 25**
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Find the derivative of y = ( 13 $\frac{1}{3}$x + 6)

**Answer Details**

Y = ( 13
$\frac{1}{3}$x + 6)2

dydx
$\frac{dy}{dx}$ = 2(13
$\frac{1}{3}$x + 6) 13
$\frac{1}{3}$

= 23
$\frac{2}{3}$ ( 13
$\frac{1}{3}$x + 6)

**Question 26**
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