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**Question 1**
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The probability of an event P is 34 $\frac{3}{4}$ while that of another event Q is 16 $\frac{1}{6}$. If the probability of both P and Q is 12 $\frac{1}{2}$. What is the probability of either P or Q.

**Answer Details**

**Question 2**
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Find n if 34n = 100112

**Answer Details**

To find n if 34n = 100112, convert both sides to base 10

= 3n + 4 = (1 x 24) + (0 x23) + (0 x 22) + (1 x 21) + 1 x 2o

= 3n + 4 = 16 + 0 + 0 + 2 + 1

3n + 4 = 19

3n = 15

n = 5

**Question 3**
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If y = x sin x, Find d2yd2x

**Answer Details**

To find the second derivative of y = x sin x, we need to differentiate the function twice with respect to x. First, let's find the first derivative: y' = (x cos x) + (sin x) Using the product rule and the derivative of sin x. Next, we can find the second derivative: y'' = [(x cos x) + (sin x)]' = (cos x - x sin x) + cos x Using the product rule and the derivative of cos x. Therefore, the second derivative of y = x sin x is y'' = 2 cos x - x sin x.

**Question 5**
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If two dice are thrown together, what is the probability of obtaining at least a score of 10?

**Answer Details**

The total sample space when two dice are thrown together is 6 x 6 = 36

1234561.1.11.21.31.41.51.622.12.22.32.42.52.633.13.23.33.43.53.644.14.24.34.44.54.655.15.25.35.45.55.666.16.26.36.46.56.6
$\begin{array}{ccccccc}& 1& 2& 3& 4& 5& 6\\ 1.& 1.1& 1.2& 1.3& 1.4& 1.5& 1.6\\ 2& 2.1& 2.2& 2.3& 2.4& 2.5& 2.6\\ 3& 3.1& 3.2& 3.3& 3.4& 3.5& 3.6\\ 4& 4.1& 4.2& 4.3& 4.4& 4.5& 4.6\\ 5& 5.1& 5.2& 5.3& 5.4& 5.5& 5.6\\ 6& 6.1& 6.2& 6.3& 6.4& 6.5& 6.6\end{array}$

At least 10 means 10 and above

P(at least 10) = 636
$\frac{6}{36}$

= 16

**Question 6**
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The locus of a point which is equidistant from two given fixed points is the

**Answer Details**

The locus of a point which is equidistant from two given fixed points is the perpendicular bisector of the straight line joining them. This means that any point on the perpendicular bisector of the straight line joining the two fixed points is equidistant from those points. To understand this concept, imagine two points on a plane and draw a straight line joining them. Now draw a perpendicular line bisecting the straight line joining them. Any point on this perpendicular line is equidistant from the two fixed points, since it is the same distance away from each of them. This is because the perpendicular line creates right angles with the straight line joining the two fixed points, and all points on a perpendicular line are equidistant from the two endpoints of the straight line it intersects.

**Question 7**
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Make t the subject of formula S = ut + 12at2

**Answer Details**

Given S = ut + 12at2
$\frac{1}{2}a{t}^{2}$

S = ut + 12at2
$\frac{1}{2}a{t}^{2}$

∴ 2S = 2ut + at2

= at2 + 2ut - 2s = 0

t = −2u±4u2+2as2a
$\frac{-2u\pm 4{u}^{2}+2as}{2a}$

= -2u π
$\pi $ √u24u2+2as2a
$\frac{\sqrt{{u}^{2}4{u}^{2}+2as}}{2a}$

= 1a $\frac{1}{a}$ (-u + √U2−2as $\sqrt{{U}^{2}-2as}$)

**Question 8**
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If the angles of quadrilateral are (P + 10)o(2P - 30)o(3P + 20)o and 4po, find p

**Answer Details**

**Question 9**
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In the figure, if XZ is 10cm, calculate RY in cm.

**Question 10**
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Obtain a maximum value of the function f(x) = x3 $3$ - 12x + 11.

**Answer Details**

To obtain the maximum value of the given function f(x) = x^3 - 12x + 11, we need to find its critical points. The critical points of a function are the points where the derivative of the function is either zero or undefined. So, let's find the derivative of the given function: f'(x) = 3x^2 - 12 Now, we'll set f'(x) = 0 and solve for x: 3x^2 - 12 = 0 x^2 - 4 = 0 (x - 2)(x + 2) = 0 So, the critical points of the function are x = 2 and x = -2. To determine whether these are maximum or minimum points, we'll take the second derivative of the function: f''(x) = 6x Now, we'll substitute the critical points into the second derivative: f''(2) = 12 f''(-2) = -12 Since f''(2) is positive, the critical point x = 2 is a minimum point. Since f''(-2) is negative, the critical point x = -2 is a maximum point. Therefore, the maximum value of the given function is obtained when x = -2. Now, we'll substitute x = -2 into the original function: f(-2) = (-2)^3 - 12(-2) + 11 f(-2) = -8 + 24 + 11 f(-2) = 27 So, the maximum value of the given function is 27, and the correct option is (D).

**Question 11**
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If x = 3 - √3 $3$, find x2 + 36x2

**Answer Details**

x = 3 - √3
$3$

x2 = (3 - √3
$3$)2

= 9 + 3 - 6√34
$34$

= 12 - 6√3
$3$

= 6(2 - √3
$3$)

∴ x2 + 36x2
$\frac{36}{{x}^{2}}$ = 6(2 - √3
$3$) + 366(2−√3)
$\frac{36}{6(2-\sqrt{3})}$

6(2 - √3
$3$) + 62−√3
$\frac{6}{2-\sqrt{3}}$ = 6(- √3
$3$) + 6(2+√3)(2−√3)(2+√3)
$\frac{6(2+\sqrt{3})}{(2-\sqrt{3})(2+\sqrt{3})}$

= 6(2 - √3
$3$) + 6(2+√3)4−3
$\frac{6(2+\sqrt{3})}{4-3}$

6(2 - √3
$3$) + 6(2 + √3
$3$) = 12 + 12

= 24

**Question 12**
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The angle between latitudes 30oS and 13oN is

**Answer Details**

The distance between two latitudes is always the same, and this distance is equal to 1/360th of the circumference of the earth. The circumference of the earth is approximately 40,000 km, so 1/360th of the circumference is approximately 111 km. To find the angle between latitudes 30°S and 13°N, we need to add the distance between these latitudes to the angle at the center of the earth. The distance between 30°S and the equator (0°) is 30°, and the distance between the equator and 13°N is 13°, for a total distance of 43°. Therefore, the angle between latitudes 30°S and 13°N is 43°. So, the answer is

43o

**Question 13**
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If x = (all prime factors of 44) and y = (all prime factors of 60), the elements of X ∪ Y and X ∩ Y respectively are

**Answer Details**

The prime factors of 44 are 2, 2, 11. The prime factors of 60 are 2, 2, 3, 5. To find X ∪ Y (the union of X and Y), we need to list all the distinct prime factors from both 44 and 60. These are 2, 3, 5, 11. Therefore, X ∪ Y is (2, 3, 5, 11). To find X ∩ Y (the intersection of X and Y), we need to list all the prime factors that are common to both 44 and 60. These are 2. Therefore, X ∩ Y is (2). Therefore, the answer is (2, 3, 5, 11) and (2), which corresponds to option D.

**Question 14**
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A cliff on the bank of a river is 300 meter high. if the angle of depression of a point on the opposite side of the river is 60∘ $\circ $, find the width of the river.

**Answer Details**

The problem involves finding the width of a river given the height of a cliff and the angle of depression of a point on the opposite side of the river. The angle of depression is the angle formed between the horizontal line and the line of sight from the point on the opposite side of the river to the top of the cliff. We can use trigonometry to solve the problem. Let x be the width of the river. Then we have a right triangle with the height of the cliff as the opposite side, x as the adjacent side, and the angle of depression as 60 degrees. Using the tangent function, we have: tan(60) = opposite/adjacent sqrt(3) = 300/x x = 300/sqrt(3) x = 100sqrt(3) Therefore, the width of the river is 100sqrt(3) meters. So, the correct option is: - 100 - 75√3 m - 100√3m - 200√3m (100√3m) is the correct answer.

**Question 15**
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Find all values of x satisfying the inequality -11 ≤ $\le $ 4 - 3x ≤ $\le $ 28

**Answer Details**

To solve the inequality -11 ≤ 4 - 3x ≤ 28, we need to isolate x in the middle of the inequality. First, we subtract 4 from all parts of the inequality: -15 ≤ -3x ≤ 24. Then, we divide by -3, remembering to flip the direction of the inequality signs when dividing by a negative number: 5 ≥ x ≥ -8. Therefore, the correct option is -8 ≤ x ≤ 5.

**Question 16**
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a student blows a balloon and its volume increases at a rate of π
$\pi $(20 - t2)cm3S-1 after t seconds. If the initial volume is 0 cm3, find the volume of the balloon after 2 seconds

**Answer Details**

The volume of the balloon increases at a rate of π(20-t^2) cm^3S^-1 after t seconds. If we want to find the volume of the balloon after 2 seconds, we need to integrate the given rate with respect to time from 0 to 2, since we want to know the change in volume from the initial volume of 0 cm^3 after 2 seconds. Integrating the given rate with respect to time, we get: ∫[0,2] π(20-t^2) dt = π[20t - (t^3/3)] from 0 to 2 Plugging in the values, we get: π[20(2) - (2^3/3)] - π[20(0) - (0^3/3)] = π[40 - 8/3] = π[120/3 - 8/3] = π[112/3] = 37.33π Therefore, the volume of the balloon after 2 seconds is approximately 37.33π cubic centimeters.

**Question 17**
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If U = (1, 2, 3, 6, 7, 8, 9, 10) is the universal set. E = (10, 4, 6, 8, 10) and F = is odd. Find (E ∩ F), where means x 1x2 = 26, x the complement of a set

**Answer Details**

U = (1, 2, 3, 6, 7, 8, 9, 10)

E = (10, 4, 6, 8, 10)

F = (x : x2 = 26, x is odd)

∴ F = ϕ
$\varphi $ Since x2 = 26 = 64

x = + which is even

∴ E ∩ F = ϕ
$\varphi $ Since there are no common elements

**Question 18**
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If x + 1 is a factor of x3 + 3x2 + kx + 4, find the value of k

**Answer Details**

To find the value of k, we need to use the factor theorem which states that if x - a is a factor of a polynomial, then the polynomial is zero when x = a. Since x + 1 is a factor of x^3 + 3x^2 + kx + 4, we know that x = -1 makes the polynomial zero. Substituting -1 for x in the polynomial, we get: (-1)^3 + 3(-1)^2 + k(-1) + 4 = 0 -1 + 3 - k + 4 = 0 6 - k = 0 Solving for k, we get: k = 6 Therefore, the value of k is 6.

**Question 19**
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Solve the equation y2−11y+24=0

**Answer Details**

To solve the equation: y^2 - 11y + 24 = 0 We can use the quadratic formula, which states that for an equation of the form ax^2 + bx + c = 0, the solutions are given by: x = (-b ± sqrt(b^2 - 4ac)) / (2a) In this case, we have a = 1, b = -11, and c = 24. Substituting these values into the quadratic formula, we get: y = (-(-11) ± sqrt((-11)^2 - 4(1)(24))) / (2(1)) Simplifying this expression, we get: y = (11 ± sqrt(121 - 96)) / 2 y = (11 ± sqrt(25)) / 2 y = (11 ± 5) / 2 So the solutions to the equation are: y = 8 or y = 3 Therefore, the equation y^2 - 11y + 24 = 0 has two solutions: y = 8 and y = 3.

**Question 20**
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Determine the distance on the earth's surface between two town P (latoN, Long 20oN) and Q(Lat 60oN, Long 25oW) (Radius of the earth = 6400km)

**Answer Details**

**Question 21**
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x2468f4y65 $\begin{array}{ccccc}x& 2& 4& 6& 8\\ f& 4& y& 6& 5\end{array}$

If the mean of the above frequency distribution is 5.2, find y

**Answer Details**

Mean ¯x
$\stackrel{\xaf}{x}$ = ∑fx∑f
$\frac{\sum fx}{\sum f}$

= 5.21
$\frac{5.2}{1}$

= 8+4y+36+404+y+6+5
$\frac{8+4y+36+40}{4+y+6+5}$

= 5.21
$\frac{5.2}{1}$

= 84+4y15+y
$\frac{84+4y}{15+y}$

= 5.2(15 + y)

= 84 + 4y

= 5.2 x 15 + 5.2y

= 84 + 4y

= 78 + 5.2y

= 84 = 4y

= 5.2y - 4y

= 84 - 78

1.2y = 6

y = 61.2
$\frac{6}{1.2}$

= 6012
$\frac{60}{12}$

= 5

**Question 22**
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For an arithmetical sequence, the first term is 2 and the common difference is 3. Find the sum of the first 11 terms

**Answer Details**

An arithmetic sequence is a sequence of numbers where the difference between any two consecutive terms is constant. In this case, the first term is 2 and the common difference is 3, so the sequence is: 2, 5, 8, 11, 14, 17, 20, 23, 26, 29, 32 To find the sum of the first 11 terms of this sequence, we can use the formula for the sum of an arithmetic series: Sn = n/2 * (a1 + an) where Sn is the sum of the first n terms, a1 is the first term, and an is the nth term. In this case, n = 11, a1 = 2, and an = 32 (the 11th term can be found by adding the common difference 3, 10 times to the first term 2). So, substituting the values in the formula: Sn = 11/2 * (2 + 32) Sn = 11/2 * 34 Sn = 187 Therefore, the sum of the first 11 terms is 187. So the correct answer is.

**Question 23**
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What is the value of x satisfying the equation 42x43x $\frac{{4}^{2x}}{{4}^{3x}}$ = 2?

**Answer Details**

42x43x
$\frac{{4}^{2x}}{{4}^{3x}}$ = 2

42x - 3x = 2

4-x = 2

(22)-x

= 21

Equating coefficients: -2x = 1

x = -12

**Question 24**
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If the binary operation ∗ $\ast $ is defined by m ∗ $\ast $ n = mn + m + n for any real number m and n, find the identity of the elements under this operation

**Answer Details**

m ∗
$\ast $ n = mn + m + n

m ∗
$\ast $ e = me + m + e, e ∗
$\ast $ m = e

∴ me + m + e, m(e + 1)e - e = 0

e + 1 = 0

∴ e = -1

**Question 25**
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No. of children0123456No. of families71167753 $\begin{array}{cccccccc}\text{No. of children}& 0& 1& 2& 3& 4& 5& 6\\ \text{No. of families}& 7& 11& 6& 7& 7& 5& 3\end{array}$

Find the mode and median respectively of the distribution above

**Question 26**
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Ice forms on a refrigerator ice box at rate of (4 - 0.61)g per minute after 1 minute. If initially, there were 2g of ice, find the mass of ice formed in 5 minutes.

**Answer Details**

**Question 27**
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Resolve 3x2+x−2 $\frac{3}{{x}^{2}+x-2}$ into partial fractions

**Answer Details**

3x2+x−2
$\frac{3}{{x}^{2}+x-2}$ = 3(x−1)(x+1)
$\frac{3}{(x-1)(x+1)}$

Ax−1
$\frac{A}{x-1}$ + Bx+2
$\frac{B}{x+2}$

A(x + 2) + B(x - 1) = 3

when x = 1, 3A = 3 →
$\to $ a = 1

when x = -2, -3B = 3 →
$\to $ B = -1

3 = 1 - 1

= 1x−1+1x+2

**Question 28**
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Evaluate the integral ∫π4π12
${\int}_{\frac{\pi}{12}}^{\frac{\pi}{4}}$ 2 cos 2x dx

**Answer Details**

let I = ∫π4π12
${\int}_{\frac{\pi}{12}}^{\frac{\pi}{4}}$ 2 cos 2x dx

= 2(sin 2x)π4
$\frac{\pi}{4}$ (sin 2x)π4
$\frac{\pi}{4}$π12
$\frac{\pi}{12}$

(2)π12
$\frac{\pi}{12}$

= -1 - 12
$\frac{1}{2}$

= 12

**Question 29**
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p = ∣∣ ∣∣x302y3424∣∣ ∣∣ $\left|\begin{array}{ccc}x& 3& 0\\ 2& y& 3\\ 4& 2& 4\end{array}\right|$

q = ∣∣ ∣∣x2z3y203z∣∣ ∣∣ $\left|\begin{array}{ccc}x& 2& z\\ 3& y& 2\\ 0& 3& z\end{array}\right|$ Where pT is the transpose p calculate /pT/ when x = 0, y = 1 and z = 2

**Answer Details**

= ∣∣ ∣∣030213422∣∣ ∣∣
$\left|\begin{array}{ccc}0& 3& 0\\ 2& 1& 3\\ 4& 2& 2\end{array}\right|$

PT = ∣∣ ∣∣024213032∣∣ ∣∣
$\left|\begin{array}{ccc}0& 2& 4\\ 2& 1& 3\\ 0& 3& 2\end{array}\right|$

/pT/ = ∣∣ ∣∣024313032∣∣ ∣∣
$\left|\begin{array}{ccc}0& 2& 4\\ 3& 1& 3\\ 0& 3& 2\end{array}\right|$

= 0[2 - 6] - 2[6 - 0] + 4[9 - 0]

= 0 - 12 + 36 = 24

**Question 30**
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Simplify (1.25×10−4)×(2.0×10−1)(6.25×105)

**Answer Details**

(1.25×10−4)×(2.0×10−1)(6.25×105)
$\frac{(1.25\times {10}^{-4})\times (2.0\times {10}^{-1})}{(6.25\times {10}^{5})}$ = 1.25×26.25
$\frac{1.25\times 2}{6.25}$ x 104 - 1 - 5

2.506.25
$\frac{2.50}{6.25}$ x 10-2 = 250625
$\frac{250}{625}$ x 10-2

0.4 x 10-2 = 4.0 x 10-3

**Question 31**
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Given that θ $\theta $ is an acute angle and sin θ $\theta $ = mn $\frac{m}{n}$, find cos θ

**Question 32**