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**Question 1**
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If P(2,3) and Q)2, 5) are points on a graph, calculate the length PQ

**Answer Details**

To calculate the length of PQ, we need to use the distance formula, which is: distance = square root of [(x2 - x1)^2 + (y2 - y1)^2] Here, P is at (2,3) and Q is at (2,5). So, x1=2, y1=3, x2=2, and y2=5. Plugging these values into the formula, we get: distance = square root of [(2 - 2)^2 + (5 - 3)^2] distance = square root of [0 + 4] distance = square root of 4 distance = 2 Therefore, the length of PQ is 2 units.

**Question 2**
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The dimensions of water tank are 13cm, 10cm and 70cm. If it is half-filled with water, calculate the volume of water in litres

**Answer Details**

To calculate the volume of water in the tank, we need to first calculate the total volume of the tank and then divide it by 2 (since the tank is half-filled with water). The volume of the tank can be calculated by multiplying its length, width, and height. Therefore, the total volume of the tank is: Volume of tank = Length × Width × Height = 13cm × 10cm × 70cm = 9100 cubic centimeters (or cm³) Since we want to find the volume of water in litres, we need to convert the volume from cubic centimeters to litres. There are 1000 cubic centimeters in one litre, so we can convert the volume by dividing it by 1000. Volume of tank in litres = Volume of tank in cubic centimeters ÷ 1000 = 9100 ÷ 1000 = 9.1 litres Finally, since the tank is half-filled with water, we need to divide the total volume by 2 to get the volume of water in the tank: Volume of water in tank = Volume of tank ÷ 2 = 9.1 ÷ 2 = 4.55 litres Therefore, the volume of water in the tank is 4.55 litres, which is option A.

**Question 3**
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The graph of y = \(ax^2 + bx + c\) is shown oon the diagram. Find the minimum value of y

**Question 4**
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A stationary boat is observed from a height of 100m. If the horizontal distance between the observer and the boat is 80m, calculate, correct to two decimal places, the angles of depression of the boat from point of observation

**Question 5**
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Marks | 0 | 1 | 2 | 3 | 4 | 5 |

Frequency | 7 | 4 | 18 | 12 | 8 | 11 |

The table gives the distribution of marks obtained by a number of pupils in a class test. Using this information, find the first quartile

**Question 6**
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A circular pond of radius 4m has a path of width 2.5m round it. Find, correct to two decimal places, the area of the path. [Take\(\frac{22}{7}\)]

**Answer Details**

To find the area of the path around the circular pond, we need to subtract the area of the inner circle from the area of the outer circle. The radius of the outer circle is the sum of the radius of the pond and the width of the path. Therefore, the radius of the outer circle is: 4m + 2.5m = 6.5m The area of the outer circle is given by: A_outer = π * r_outer^2 A_outer = π * (6.5m)^2 A_outer = 132.73\(m^2\) The area of the inner circle is simply: A_inner = π * r_inner^2 A_inner = π * (4m)^2 A_inner = 50.27\(m^2\) Therefore, the area of the path is: A_path = A_outer - A_inner A_path = 132.73\(m^2\) - 50.27\(m^2\) A_path = 82.46\(m^2\) Rounding to two decimal places, the area of the path is approximately 82.50\(m^2\). Therefore, the correct option is: - 82.50\(m^2\)

**Question 7**
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The diagram shows a circle O. If < ZYW = 33\(^o\) , find < ZWX

**Question 8**
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Make x the subject of the relation d = \(\sqrt{\frac{6}{x} - \frac{y}{2}}\)

**Question 9**
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In a class of 45 students, 28 offer chemistry and 25 offer Biology. If each student offers at least one of the two subjects, calculate the probability that a student selected at random from the class the class offers chemistry only.

**Question 10**
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In the diagram, PQ and PS are tangents to the circle O. If PSQ = m, <SPQ = n and <SQR = 33\(^o\), find the value of (m + n)

**Question 11**
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In the diagram, XY is a straight line. <POX = <POQ and <ROY = <QOR. Find the value of <POQ + <ROY.

**Question 12**
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In what number base was the addition 1 + nn = 100, where n > 0, done?

**Question 13**
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If x varies inversely as y and y varies directly as z, what is the relationship between x and z?

**Answer Details**

The given problem states that x varies inversely with y, which means that as y increases, x decreases and vice versa. Similarly, it also states that y varies directly with z, which means that as z increases, y increases and vice versa. Now, to find the relationship between x and z, we need to combine these two statements. Since y is the common variable, we can rewrite the second statement as y \(\alpha\) z. Substituting this value of y in the first statement, we get x \(\alpha\) \(\frac{1}{y}\), which can be further simplified as x \(\alpha\) \(\frac{1}{z}\). Hence, the relationship between x and z is x \(\alpha\) \(\frac{1}{z}\). is the correct answer.

**Question 14**
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Simplify; \(\sqrt{2}(\sqrt{6} + 2\sqrt{2}) - 2\sqrt{3}\)

**Answer Details**

To simplify this expression, we can use the distributive property of multiplication and simplify the terms inside the parentheses first: \(\sqrt{2}(\sqrt{6} + 2\sqrt{2}) - 2\sqrt{3}\) = \(\sqrt{2} \times \sqrt{6} + \sqrt{2} \times 2\sqrt{2} - 2\sqrt{3}\) = \(\sqrt{12} + 2\sqrt{4} - 2\sqrt{3}\) = \(2\sqrt{3} + 4 - 2\sqrt{3}\) = \(4\) Therefore, the simplified expression is 4. In summary, we can simplify this expression by first using the distributive property to multiply the \(\sqrt{2}\) by the terms inside the parentheses, then simplifying the resulting expression by combining like terms.

**Question 15**
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Water flows out of a pipe at a rate of 40\(\pi cm^2\) per seconds into an empty cylinder container of base radius 4cm. Find the height of water in the container after 4 seconds.

**Question 17**
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The ages of Tunde and Ola are in the ratio 1:2. If the ratio of Ola's age to Musa's age is 4:5, what is the ratio of Tunde's age to Musa's age?

**Question 18**
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Evaluate \(\frac{3\frac{1}{4} \times 1\frac{3}{5}}{11\frac{1}{3} - 5 \frac{1}{3}}\)

**Answer Details**

To evaluate this expression, we need to follow the order of operations, which is a set of rules for the order in which we perform arithmetic operations. The order of operations is: 1. Parentheses 2. Exponents 3. Multiplication and division (performed from left to right) 4. Addition and subtraction (performed from left to right) In this expression, there are no parentheses or exponents, so we can start by performing the multiplication and division. First, we need to convert the mixed numbers to improper fractions, so that we can easily multiply and divide them. \(\frac{3\frac{1}{4} \times 1\frac{3}{5}}{11\frac{1}{3} - 5 \frac{1}{3}} = \frac{\frac{13}{4} \times \frac{8}{5}}{\frac{32}{3}}\) Next, we can simplify the expression by canceling out common factors. \(\frac{\frac{13}{4} \times \frac{8}{5}}{\frac{32}{3}} = \frac{13 \times 2}{4 \times 5} = \frac{26}{20} = \frac{13}{10}\) Therefore, the answer is \(\frac{13}{10}\), which is equivalent to \(\frac{26}{20}\) and can be simplified to \(\frac{13}{15}\). So the correct option is: \(\frac{13}{15}\).

**Question 19**
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Two bottles are drawn with replacement from a crate containing 8 coke, 12 and 4 sprite bottles. What is the probability that the first is coke and the second is not coke?

**Answer Details**

There are three types of bottles in the crate: Coke, Sprite, and not Coke (which includes Sprite bottles). To find the probability that the first bottle is Coke and the second bottle is not Coke, we need to multiply two probabilities: the probability of selecting a Coke bottle first and the probability of selecting a not Coke bottle (i.e., a Sprite bottle) second. The probability of selecting a Coke bottle first is 8/24 because there are 8 Coke bottles in the crate out of a total of 24 bottles (8 Coke + 12 Sprite + 4 Sprite = 24). After selecting the first bottle, there will be 23 bottles left in the crate. If the first bottle was a Coke bottle, then there will be 7 Coke bottles and 12 Sprite bottles left in the crate. Therefore, the probability of selecting a not Coke (i.e., a Sprite) bottle second is 12/23. Multiplying these probabilities, we get: (8/24) * (12/23) = 96/552 ≈ 0.174 Therefore, the probability that the first bottle is Coke and the second bottle is not Coke is approximately 0.174, which is closest to option (C) 2/9.

**Question 20**
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Three exterior angles of a polygon are 30\(^o\), 40\(^o\) and 60\(^o\). If the remaining exterior angles are 46\(^o\) each, name the polygon.

**Answer Details**

The sum of all exterior angles of a polygon is 360 degrees. Exterior angles are formed by extending one side of a polygon to the next. So, the sum of the three exterior angles that we know is 30 + 40 + 60 = 130 degrees. And, if each of the remaining exterior angles measures 46 degrees, then there must be 360 - 130 = 230 degrees left. Dividing this by 46, we get 230/46 = 5. So, the polygon has a total of 5 + 3 = 8 sides. Therefore, the polygon is an octagon.

**Question 21**
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If the total surface area of a solid hemisphere is equal to its volume, find the radius

**Question 22**
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Fig. 1 and Fig. 2 are the addition and multiplication tables respectively in modulo 5. Use these tables to solve the equation (n \(\oplus 4\))

**Question 23**
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Calculate the gradient (slope) of the joining points (-1, 1) and (2, -2)

**Answer Details**

The gradient or slope of a line is a measure of how steep it is, and is defined as the ratio of the vertical change between two points to the horizontal change between the same two points. To find the gradient of the line that passes through the points (-1, 1) and (2, -2), we need to calculate the difference in the y-coordinates (the vertical change) divided by the difference in the x-coordinates (the horizontal change). So, the gradient is: (y2 - y1) / (x2 - x1) = (-2 - 1) / (2 - (-1)) = -3 / 3 = -1. Therefore, the gradient of the line that passes through the points (-1, 1) and (2, -2) is -1.

**Question 24**
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Given that a = log 7 and b = \(\log\) 2, express log 35 in terms of a and b.

**Answer Details**

We can use the logarithmic identities to simplify the expression for log 35 in terms of a and b. Firstly, we can write 35 as the product of 7 and 5: 35 = 7 x 5 Next, we can use the logarithmic identity: log (a x b) = log a + log b to express log 35 in terms of log 7 and log 5 as follows: log 35 = log (7 x 5) = log 7 + log 5 Now, we need to express log 5 in terms of a and b. We can write 5 as the product of 2 and 2.5: 5 = 2 x 2.5 Using the logarithmic identity, we get: log 5 = log (2 x 2.5) = log 2 + log 2.5 We can express log 2.5 in terms of log 10 (which is equal to 1) and log 2 as follows: log 2.5 = log (2.5/1) = log (5/2) = log 5 - log 2 Substituting this into our expression for log 35, we get: log 35 = log 7 + log 5 = log 7 + (log 5 - log 2) Finally, we can substitute the given values of a and b into the expression above to obtain: log 35 = a + (log 5 - b) Simplifying further, we get: log 35 = a + log 5 - b Therefore, the answer is (c) a - b + 1.

**Question 25**
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Calculate the variance of 2, 3, 3, 4, 5, 5, 5, 7, 7 and 9

**Answer Details**

To calculate the variance of a set of data, you need to follow these steps: 1. Find the mean of the data set. 2. For each data point, subtract the mean and square the result. 3. Add up all the squared differences. 4. Divide the sum of squared differences by the total number of data points minus one. Here are the steps for the given data set: 1. Find the mean: (2 + 3 + 3 + 4 + 5 + 5 + 5 + 7 + 7 + 9) / 10 = 5 2. Subtract the mean and square the result for each data point: (2 - 5)^2 = 9 (3 - 5)^2 = 4 (3 - 5)^2 = 4 (4 - 5)^2 = 1 (5 - 5)^2 = 0 (5 - 5)^2 = 0 (5 - 5)^2 = 0 (7 - 5)^2 = 4 (7 - 5)^2 = 4 (9 - 5)^2 = 16 3. Add up all the squared differences: 9 + 4 + 4 + 1 + 0 + 0 + 0 + 4 + 4 + 16 = 42 4. Divide the sum of squared differences by the total number of data points minus one: 42 / (10 - 1) = 4.67 Therefore, the variance of the given data set is approximately 4.67. The closest option is 4.2.

**Question 26**
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Find the 6th term of the sequence \(\frac{2}{3} \frac{7}{15} \frac{4}{15}\),...

**Question 27**
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The roots of a quadratic equation are \(\frac{-1}{2}\) and \(\frac{2}{3}\). Find the equation.

**Question 28**
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Which of the following is true about parallelogram?

**Question 29**
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The diagram shows a circle centre O. if <STR = 29 and <RST = 45, calculate the value

**Question 32**
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Marks | 0 | 1 | 2 | 3 | 4 | 5 |

Frequency | 7 | 4 | 18 | 12 | 8 | 11 |

The table gives the distribution of marks obtained by a number of pupils in a class test. Using this information, Find the median of the distribution

**Question 33**
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If the simple interest on a certain amount of money saved in a bank for 5 years at 2\(\frac{1}{2}\)% annum is N500.00, calculate the total amount due after 6 years at the same rate

**Answer Details**

The simple interest on a certain amount of money saved in a bank for 5 years at 2\(\frac{1}{2}\)% per annum is N500.00. We can use the formula for simple interest to find the principal amount, which is the initial amount of money saved. Simple Interest (SI) = (Principal * Rate * Time) / 100 In this case, we know that the rate is 2\(\frac{1}{2}\)% per annum, which is equivalent to 0.025 as a decimal. We also know that the time is 5 years, and the simple interest is N500.00. We can plug these values into the formula and solve for the principal amount: 500 = (P * 0.025 * 5) / 100 P = 500 * 100 / (0.025 * 5) P = 4000 So the principal amount is N4,000.00. Now we need to calculate the total amount due after 6 years at the same rate. We can again use the formula for simple interest, but this time the time is 6 years instead of 5. Simple Interest (SI) = (Principal * Rate * Time) / 100 SI = (4000 * 0.025 * 6) / 100 SI = N600.00 So the simple interest for 6 years is N600.00. To find the total amount due after 6 years, we need to add the simple interest to the principal amount: Total amount due = Principal + Simple Interest Total amount due = 4000 + 600 Total amount due = N4,600.00 Therefore, the total amount due after 6 years at the same rate is N4,600.00.

**Question 34**
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An arc of a circle of radius 7.5cm is 7.5cm long. Find, correct to the nearest degree, the angle which the arc subtends at the centre of the circle. [Take \(\pi = \frac{22}{7}\)]

**Question 35**
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Consider the statements: p it is hot, q: it is raining

Which of the following symbols correctly represents the statement "It is raining if and only if it it is cold"?

**Answer Details**

The statement "It is raining if and only if it is cold" can be written as "q if and only if \(\sim\)p". This is because the "if and only if" connector implies that both statements are true, so it is equivalent to saying "if it is raining, then it is cold" and "if it is cold, then it is raining". We can represent this statement using the "if and only if" symbol (\(\iff\)) as: q \(\iff\) \(\sim\)p Therefore, the correct option is (D) q \(\iff\) \(\sim\)p.

**Question 36**
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Given that t = \(2 ^{-x}\), find \(2 ^{x + 1}\) in terms of t.

**Answer Details**

We can start by using the laws of exponents to rewrite \(2^{x+1}\) in terms of \(2^{-x}\), which is given by \(t\). Recall that \(a^{m+n} = a^m \times a^n\) and \(a^{-n} = \frac{1}{a^n}\). Therefore, we have: $$2^{x+1} = 2^x \times 2^1 = 2^x \times 2 = 2 \times 2^x$$ Next, we can substitute \(2^{-x} = t\) into this expression: $$2 \times 2^x = 2 \times \frac{1}{t} = \frac{2}{t}$$ So the answer is: \(\frac{2}{t}\).

**Question 37**
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If M = {x : 3 \(\leq\) x < 8} and N = {x : 8 < x \(\leq\) 12}, which of the following is true?

i. 8 \(\in\) M \(\cap\) N

ii. 8 \(\in\) M \(\cup\) N

iii. M \(\cap\) N = \(\varnothing\)

**Answer Details**

Let's first understand what M and N represent. M is a set of numbers x that are greater than or equal to 3 but less than 8, while N is a set of numbers x that are greater than 8 but less than or equal to 12. i. 8 is not in the set M because M only includes numbers less than 8. Similarly, 8 is not in the set N because N only includes numbers greater than 8. Therefore, 8 is not in the intersection of M and N, making option i false. ii. The union of M and N includes all the numbers in both sets. Therefore, the union of M and N would include all numbers greater than or equal to 3 but less than or equal to 12. 8 is included in this range, so it must also be in the union of M and N. Therefore, option ii is true. iii. The intersection of M and N includes all the numbers that are in both sets. However, M and N do not overlap since there are no numbers that are both greater than or equal to 3 and less than or equal to 12. Therefore, the intersection of M and N is an empty set, making option iii true. Therefore, the correct answer is option iii only.

**Question 38**
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A sum of N18,100 was shared among 5 boys and 4 girls with each boy taking N20.00 more than each girl. Find a boy's share.

**Answer Details**

The total amount of money shared among 5 boys and 4 girls is N18,100.00. If each boy takes N20.00 more than each girl, then the difference between a boy's share and a girl's share is N20.00. We can use this information to find the total share for both the boys and girls. Let's call the girl's share "x". If a boy takes N20.00 more than a girl, then the boy's share is "x + N20.00". The total amount of money for the girls is 4 * x, and the total amount of money for the boys is 5 * (x + N20.00). We can use this information to set up an equation: 4 * x + 5 * (x + N20.00) = N18,100.00 Expanding the right side of the equation: 4 * x + 5 * x + 5 * N20.00 = N18,100.00 Combining like terms: 9 * x + 5 * N20.00 = N18,100.00 Subtracting 5 * N20.00 from both sides: 9 * x = N18,100.00 - 5 * N20.00 Calculating N18,100.00 - 5 * N20.00: 9 * x = N18,100.00 - 5 * N20.00 = N18,100.00 - N100.00 = N18,000.00 Dividing both sides by 9: x = N18,000.00 / 9 = N2,000.00 So each girl's share is N2,000.00, and each boy's share is N2,000.00 + N20.00 = N2,020.00.

**Question 39**
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Simplify the expression \(\frac{a^2 b^4 - b^2 a^4}{ab(a + b)}\)

**Answer Details**

The expression can be simplified as follows: First, we can factor out the common factor of \(ab\) from the numerator: \begin{align*} \frac{a^2 b^4 - b^2 a^4}{ab(a + b)} &= \frac{ab(a^2 b^3 - b^2 a^3)}{ab(a + b)} \\ &= \frac{ab(a^2 b^3 - b^2 a^3)}{ab(a + b)} \\ &= \frac{a^2 b^3 - b^2 a^3}{a + b}. \end{align*} Next, we can simplify the numerator: \begin{align*} a^2 b^3 - b^2 a^3 &= (a^2)(b^3) - (b^2)(a^3) \\ &= a^2b^2 (b - a) \\ &= ab^2 (a - b)(b + a) \\ &= ab^2 (a^2 - b^2). \end{align*} Therefore, the expression simplifies to: \[\frac{ab^2 (a^2 - b^2)}{a + b} = ab (a - b) = \boxed{ab^2 - a^2b}.\]

**Question 40**
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One factor of \(7x^2 + 33x - 10\) is

**Question 41**
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In the diagram, PR is a diameter of the circle RSP, RP is produced to T and TS is a tangent to the circle at S. If < PRS = 24\(^o\), calculate the value of < STR

**Question 42**
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Solve the equation: \(\frac{1}{5x} + \frac{1}{x}\)= 3

**Question 43**
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In the diagram, NQ//TS, <RTS = 50\(^o\) and <PRT = 100\(^o\). Find the value of <NPR

**Question 44**
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The average age of a group of 25 girls is 10year. If one girl, aged 12 years and 4 months joins the group, find the new average age of the group

**Answer Details**

To find the new average age of the group, we need to calculate the sum of the ages of all the girls in the group after the new girl joins, and then divide by the total number of girls in the group. The sum of the ages of the original 25 girls is 25 * 10 = 250 years. When the new girl joins the group, the total number of girls becomes 26, and the sum of their ages becomes 250 + 12.33 (12 years and 4 months converted to decimal form) = 262.33 years. To find the new average age, we divide the sum of the ages by the number of girls: New average age = 262.33 / 26 ≈ 10.1 years Therefore, the new average age of the group is 10.1 years.

**Question 45**
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If x = \(\frac{2}{3}\) and y = - 6, evaluate xy - \(\frac{y}{x}\)

**Answer Details**

We are asked to evaluate the expression xy - $\frac{y}{x}$, where x = $\frac{2}{3}$ and y = -6. First, we can find xy: xy = $\frac{2}{3}$ * -6 = -4 Next, we can find $\frac{y}{x}$: $\frac{y}{x}$ = -6 * $\frac{3}{2}$ = -9 Finally, we can subtract $\frac{y}{x}$ from xy: xy - $\frac{y}{x}$ = -4 - (-9) = -4 + 9 = 5 So the expression xy - $\frac{y}{x}$ = 5.

**Question 46**
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Given that cos 30\(^o\) = sin 60\(^o\) = \(\frac{3}{2}\) and sin 30\(^o\) = cos 60\(^o\) = \(\frac{1}{2}\), evaluate \(\frac{tan 60^o - q}{1 - tan 30^o}\)

**Question 47**
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The diagonal of a square is 60 cm. Calculate its peremeter

**Answer Details**

To find the perimeter of a square, we need to know the length of one of its sides. Let's call the length of one side of the square "x". Since the square has four equal sides, we can say that the perimeter of the square is equal to 4 times x, or 4x. We are given that the diagonal of the square is 60 cm. We know that the diagonal of a square forms a right triangle with two sides that are equal to the length of one side of the square. We can use the Pythagorean theorem to find the length of one side of the square. The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse (the longest side) is equal to the sum of the squares of the lengths of the other two sides. In this case, the hypotenuse is the diagonal of the square, which we know is 60 cm. We can let one side of the square be "x", as we mentioned earlier. The other side of the square is also "x" because the square has four equal sides. So we have: 60^2 = x^2 + x^2 Simplifying this equation, we get: 3600 = 2x^2 Dividing both sides by 2, we get: 1800 = x^2 Taking the square root of both sides, we get: x = sqrt(1800) We can simplify this answer by factoring out the square of a perfect square: x = sqrt(900 * 2) x = sqrt(900) * sqrt(2) x = 30 * sqrt(2) Now that we know the length of one side of the square, we can find its perimeter by multiplying by 4: perimeter = 4x = 4(30 * sqrt(2)) = 120 * sqrt(2) Therefore, the answer is 120\(\sqrt{2}\).

**Question 49**
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Express 0.0000407, correct to 2 significant figures

**Answer Details**

When expressing a number to a certain number of significant figures, you need to look at the digits that are significant and those that are not significant. Significant digits are those that carry meaning, while non-significant digits are just placeholders. In this case, the number 0.0000407 has 4 significant digits. To express it to 2 significant figures, you need to round it to the nearest value with 2 significant digits. The third significant digit in 0.0000407 is 7. Since it is greater than 5, we round up the second significant digit (which is 4) to 5. Therefore, the final answer, rounded to 2 significant figures, is 0.000041.

**Question 50**
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(a) Using completing the square method, solve, correct to 2 decimal places, the equation \(3y^{2} - 5y + 2 = 0\).

(b) Given that \(M = \begin{pmatrix} 1 & 2 \\ 4 & 3 \end{pmatrix}, N = \begin{pmatrix} m & x \\ n & y \end{pmatrix}\) and \(MN = \begin{pmatrix} 2 & 1 \\ 3 & 4 \end{pmatrix}\), find the matrix N.

None

**Question 51**
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Out of 120 customers in a shop, 45 bought both bags and shoes. If all the customers bought either bags or shoes and 11 more customers bought shoes than bags:

(a) Illustrate the this information in a diagram;

(b) find the number of customers who bought shoes;

(c) calculate the probability that a customer selected at random bought bags.

None

**Question 52**
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(a) Copy and complete the table of values for the equation \(y = 2x^{2} - 7x - 9\) for \(-3 \leq x \leq 6\).

x | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 | 5 | 6 |

y | 13 | -9 | -14 | -12 | 6 |

(b) Using scales of 2cm to 1 unit on the x- axis and 2cm to 4 units on the y- axis, draw the graphs of \(y = 2x^{2} - 7x - 9\) for \(-3 \leq x \leq 6\).

(c) Use the graph to estimate the :

(i) roots of the equation \(2x^{2} - 7x = 26\);

(ii) coordinates of the minimum point of y;

(iii) range of values for which \(2x^{2} - 7x < 9\).

None

**Answer Details**

None

**Question 53**
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(a) The operation (*) is defined on the set of real numbers, R, by \(x * y = \frac{x + y}{2}, x, y \in R\).

(i) Evaluate \(3 * \frac{2}{5}\).

(ii) If \(8 * y = 8\frac{1}{4}\), find the value of y.

(b) In \(\Delta ABC, \overline{AB} = \begin{pmatrix} -4 \\ 6 \end{pmatrix}\) and \(\overline{AC} = \begin{pmatrix} 3 \\ -8 \end{pmatrix}\). If P is the midpoint of \(\overline{AB}\), express \(\overline{CP}\) as a column vector.

None

**Answer Details**

None

**Question 54**
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(a) The angle of depression of a point P on the ground from the top T of a building is 23.6°. If the distance from P to the foot of the building is 50m, calculate, correct to the nearest metre, the height if the building.

(b)

In the diagram, \(PT // SU, QS // TR, /SR/ = 6cm\) and \(/RU/ = 10 cm\). If the area of \(\Delta TRU = 45 cm^{2}\), calculate the area of the trapezium QTUS.

None