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Tambaya 1 Rahoto
5, 8, 6 and k occur with frequency 3, 2, 4, and 1 respectively an d have a mean of 5.7. Find the value of k
Bayanin Amsa
The mean of the given numbers is 5.7. Mean = (sum of all numbers) / (total number of numbers) Therefore, (5*3 + 8*2 + 6*4 + k*1) / (3+2+4+1) = 5.7 Simplifying the above equation, we get: (15 + 16 + 24 + k) / 10 = 5.7 55 + k = 57 k = 2 Therefore, the value of k is 2.
Tambaya 2 Rahoto
A binary operation ⊗ defined on the set of integers is such that m⊗n = m + n + mn for all integers m and n. Find the inverse of -5 under this operation, if the identity element is 0
Bayanin Amsa
To find the inverse of -5 under this operation, we need to find an integer x such that -5⊗x = 0, where 0 is the identity element. Using the given binary operation, we know that m⊗n = m + n + mn. Substituting -5 for m and x for n, we get: -5⊗x = -5 + x - 5x = -4x - 5 We need to solve -4x - 5 = 0 for x. Adding 5 to both sides gives: -4x = 5 Dividing both sides by -4 gives: x = -5/4 Therefore, the inverse of -5 under this operation is -5/4. So, the correct option is: -5/4
Tambaya 3 Rahoto
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∘ , find the width of the river.
Bayanin Amsa
We can solve this problem using trigonometry. Let's draw a diagram of the situation:
A /| / | / | h = 300 m / | ----------- x B
Where point A is the top of the cliff, point B is the unknown point on the opposite side of the river, and x is the width of the river.
We know that the angle of depression from A to B is 60 degrees. This means that the angle of elevation from B to A is also 60 degrees.
Using trigonometry, we can set up the following equation:
tan(60) = h / x
where h is the height of the cliff and x is the width of the river. We can solve for x:
x = h / tan(60) x = 300 / √3 x = 100√3 meters
Therefore, the width of the river is 100√3 meters. Answer is correct.
Tambaya 4 Rahoto
Each of the interior angles of a regular polygon is 140o. Calculate the sum of all the interior angles of the polygon
Bayanin Amsa
In a regular polygon, all the interior angles have the same measure. Let n be the number of sides of the polygon, then we have: sum of all interior angles = (n - 2) x 180 degrees Each interior angle of the polygon is given as 140 degrees. Since the polygon is regular, it follows that all its interior angles are congruent. Hence, we can set up an equation involving the interior angle and solve for the number of sides n. In a regular polygon with n sides, the sum of the interior angles is given by: sum of all interior angles = n x (interior angle) Substituting the given value of the interior angle, we have: sum of all interior angles = n x 140 degrees We can now equate the two expressions for the sum of all interior angles: (n - 2) x 180 degrees = n x 140 degrees Expanding and simplifying the left side, we get: 180n - 360 = 140n Adding 360 to both sides and simplifying, we have: 40n = 360 Therefore, n = 9. Hence, the given polygon has 9 sides. Now, substituting this value of n in the expression for the sum of all interior angles, we have: sum of all interior angles = 9 x 140 degrees = 1260 degrees Therefore, the sum of all the interior angles of the given polygon is 1260 degrees. In conclusion, the answer is option (B) 1260 degrees.
Tambaya 5 Rahoto
If the hypotenuse of a right angle-triangle isosceles triangle is 2cm. What is the area of the triangle?
Bayanin Amsa
In an isosceles right triangle, the two legs have the same length, which is equal to √2 times the length of one of the legs. Let's assume that each leg has a length of x, then the hypotenuse (which is 2 cm in this case) can be expressed as: x√2 = 2 Solving for x, we get: x = 2/√2 = √2 The area of an isosceles right triangle can be found by dividing the product of the two legs by 2, so the area in this case would be: (√2 * √2)/2 = 1 cm² Therefore, the correct answer is: - 1 cm²
Tambaya 6 Rahoto
Determine the value of x for which (x2 - 1) > 0
Bayanin Amsa
To find the value of x for which (x^2 - 1) > 0, we need to factorize the expression. (x^2 - 1) can be written as (x - 1)(x + 1). So now we have: (x - 1)(x + 1) > 0 For the product of two factors to be greater than zero, either both factors must be positive or both factors must be negative. If (x - 1) and (x + 1) are both positive, then x > 1. If (x - 1) and (x + 1) are both negative, then x < -1. Therefore, the solution is: x < -1 or x > 1 which means option (A) is the correct answer.
Tambaya 7 Rahoto
A regular polygon has 150∘ as the size of each interior angle. How many sides does it have?
Bayanin Amsa
A regular polygon has equal interior angles and equal sides. The sum of the interior angles of a polygon with n sides is (n-2) times 180 degrees. So, if a regular polygon has interior angles of 150 degrees, we can use this formula to find the number of sides: (n-2) x 180 = sum of interior angles of the polygon (n-2) x 180 = 150n Simplifying the equation: n - 2 = 150/180 * n n - 2 = 5/6 * n n = 12 Therefore, the regular polygon has 12 sides. Answer: 12.
Tambaya 9 Rahoto
In the figure above , Ts//xy and XY = TY, ∠SYZ = 34∘
, ∠TXY = 47∘
, find the angle marked n
Bayanin Amsa
In Δ TYX
XY = TY ∴y =47∘
base ∠s of ISCΔ
But y+x+34+47 = 180 interior
opposite ∠s are supplementary
47 + x + 34 + 47 = 180
x + 128= 180
x = 180- 128
x = 52∘
Tambaya 10 Rahoto
W is directly proportional to U. If w = 5 when U = 3, find when W = 2/7
Tambaya 12 Rahoto
Find to infinity, the sum of the sequence 1,910,(910)2,(910)3,.....
Tambaya 13 Rahoto
Find the radius of a sphere whose surface area is 154 cm2
Bayanin Amsa
The surface area of a sphere is given by the formula: A = 4πr^2 where A is the surface area and r is the radius of the sphere. In this problem, we are given the surface area of the sphere, which is 154 cm^2. We can use this information to find the radius of the sphere. We start by rearranging the formula to solve for r: r = √(A/4π) Plugging in the value of A = 154 cm^2, we get: r = √(154/4π) ≈ 3.5 cm Therefore, the radius of the sphere is approximately 3.5 cm. So, the correct answer to this problem is option B: 3.50 cm.
Tambaya 14 Rahoto
Evaluate ∫sec2θ dθ
Bayanin Amsa
∫sec2θ dθ = ∫ 1/cos2 dθ
∫(cos)-2 dθ,
let u = cos θ
∴∫u-2 = 1/u + c
∫cos θ = sin θ + c
∫sec-2θ = 1/u sin θ + c
= (sinθ / cosθ) + c
= Tan θ + c
Tambaya 15 Rahoto
If log102 = 0.3010 and log107 = 0.8451, evaluate log10280
Bayanin Amsa
Tambaya 16 Rahoto
I.S∩T∩W=S II. S∪T∪W=S
III. T∩W=S
If S⊂T⊂W, which of the above statements are true?
Bayanin Amsa
S∩T∩W = S(This means that all the elements of s are in T and also in W.)
S∪T∪W = W (imply W is a universal set for S and T) T∩W = S imply all the elements of S are in T and W)
S ⊂ T ⊂ W
∴ I, II, and III
Tambaya 17 Rahoto
Find the locus of a particle which moves in the first quadrant so that it is equidistant from the lines x = 0 and y = 0 (where k is a constant)
Bayanin Amsa
The locus of a particle refers to the path that the particle follows in the coordinate plane. In this case, we want to find the path that a particle follows if it always remains at the same distance from the lines x = 0 and y = 0. Since the particle is equidistant from the x-axis and y-axis, it must lie on the perpendicular bisector of the line segment joining the origin to any point on the x-axis or y-axis. Let's consider the point (a,0) on the x-axis. The distance between this point and the particle is given by the distance formula: √[ (a-x)^2 + (0-y)^2 ] Similarly, for the point (0,b) on the y-axis, the distance between this point and the particle is given by: √[ (0-x)^2 + (b-y)^2 ] Since the particle is equidistant from both lines, we can set these two distances equal to each other: √[ (a-x)^2 + (0-y)^2 ] = √[ (0-x)^2 + (b-y)^2 ] Squaring both sides of the equation, we get: (a-x)^2 + y^2 = x^2 + (b-y)^2 Simplifying the equation yields: 2xy - 2bx - 2ay + a^2 + b^2 = 0 This is the equation of a hyperbola with the x-axis and y-axis as its asymptotes. However, we need to take into account that the particle is in the first quadrant, so we only consider the part of the hyperbola that lies in the first quadrant. Thus, the correct answer is x - y = 0.
Tambaya 18 Rahoto
No. of days | 1 | 2 | 3 | 4 | 5 | 6 |
No. of students | 20 | x | 50 | 40 | 2x | 60 |
The distribution above shows the number of days a group of 260 students were absents from school in a particular term. How many students were absent for at least four days in the term
Bayanin Amsa
20 + X + 50 + 40 + 2X + 60 = 260
3X + 170 = 260
3X = 260 - 170
3x = 90
x = 30
Absent for at least 4 days
4 | 5 | 6 |
40 | 2x | 60 |
Tambaya 19 Rahoto
A man bought a second-hand photocopying machine for N34,000. He serviced is at a cost od N2,000 and then sold it i profit of 15%. What was the selling price?
Bayanin Amsa
The man bought a photocopying machine for N34,000 and then serviced it at a cost of N2,000. Therefore, his total cost becomes N34,000 + N2,000 = N36,000. He sold it at a profit of 15%. Profit is calculated as a percentage of the cost price. So, his profit is 15/100 x N36,000 = N5,400. To find the selling price, we add the profit to the cost price: Selling price = Cost price + Profit Selling price = N36,000 + N5,400 Selling price = N41,400 Therefore, the selling price is N41,400.
Tambaya 20 Rahoto
A polynomial in x whose roots are 4/3 and -3/5 is
Bayanin Amsa
If 4/3 and -3/5 are roots of a polynomial
Imply x = 4/3 and - 3/5
3x = 4 and 5x = -3
∴3x-4 = 0 and 5x+3 = 0 are factors
(3x-4)(5x+3) = 0 product of the factors
15x2 + 9x – 20x – 12 = 0 By expansion
15x2 - 11x – 12 = 0
Tambaya 21 Rahoto
What is the value of p if the gradient of the line joining (-1,p) and (p, 4) is
2 |
3 |
Bayanin Amsa
The gradient of a line is the change in the y-coordinate divided by the change in the x-coordinate. So, for the line joining (-1,p) and (p,4), the gradient is (4-p)/(p-(-1)) or (4-p)/(p+1). We are given that this value is equal to 2/3. So, we can set up the equation: (4-p)/(p+1) = 2/3 To solve for p, we can cross-multiply and simplify: 3(4-p) = 2(p+1) 12-3p = 2p+2 10 = 5p p = 2 Therefore, the value of p is 2.
Tambaya 23 Rahoto
What is the locus or the mid-point of all the chords of length 6cm with circle of radius 5cm and with center 0?
Bayanin Amsa
x2 + 33 = 52
x2 + 9 = 25
x2 = 25 – 9
x2 = 16
x = √ 16
= 4 cm
∴ The locus is a circle of radius 4 cm with the center 0
Tambaya 24 Rahoto
The distance travelled by a particle from a fixed point is given as s = (t3 - t2 - t + 5)cm. Find the minimum distance that the particle can cover from the fixed point.
Bayanin Amsa
The distance travelled by the particle is given by the function s = t^3 - t^2 - t + 5. To find the minimum distance that the particle can cover from the fixed point, we need to find the minimum value of this function. Taking the derivative of s with respect to t, we get s' = 3t^2 - 2t - 1. Setting s' equal to zero and solving for t, we get t = (2 ± √10)/3. To determine whether this critical point is a minimum or maximum, we need to check the second derivative of s. Taking the derivative of s' with respect to t, we get s'' = 6t - 2. When t = (2 + √10)/3, we have s'' > 0, which means that this critical point corresponds to a minimum. Therefore, the minimum distance that the particle can cover from the fixed point is achieved when t = (2 + √10)/3. Substituting this value of t into the function s, we get s_min = s((2 + √10)/3) = (16 - 8√10 + 27√10 - 25)/27 = (-9 + 19√10)/27 ≈ 0.463 cm. Therefore, the correct option is not listed as it is approximately 0.463 cm.
Tambaya 25 Rahoto
If x = {n2+1:n is a positive integer and 1 ≤ n ≤ 5}, y = {5n:n is a positive integer and 1 ≤ n ≤ 5}, find x ∩ y.
Bayanin Amsa
To find the intersection of two sets, we need to find the elements that are common to both sets. First, let's find the elements of set x: x = {n2 + 1 : 1 ≤ n ≤ 5} If we substitute each value of n from 1 to 5 into the formula n2 + 1, we get the following values for set x: x = {2, 5, 10, 17, 26} Now, let's find the elements of set y: y = {5n : 1 ≤ n ≤ 5} If we multiply each value of n from 1 to 5 by 5, we get the following values for set y: y = {5, 10, 15, 20, 25} To find the intersection of sets x and y, we need to find the elements that are common to both sets. From the values listed above, we can see that the elements 5 and 10 are in both sets x and y. Therefore, the intersection of x and y is: x ∩ y = {5, 10} Therefore, the correct option is {5, 10}.
Tambaya 26 Rahoto
The pie chart above represents 400 fruits on display n a grocery store. How many apples are in the store
Bayanin Amsa
Tambaya 28 Rahoto
If m * n = n - (m+2) for any real number m and n find he value of 3*(-5)
Bayanin Amsa
m * n = n - (m+2)
= -5 - (3+2)
= -5-5
= -10
Tambaya 29 Rahoto
What is the mean deviation of x, 2x, x+1 and 3x. If their mean is 2?
Bayanin Amsa
The first step to finding the mean deviation is to calculate the mean of the given numbers. We are told that the mean of x, 2x, x+1, and 3x is 2. So, we can write an equation: (x + 2x + x+1 + 3x)/4 = 2 Simplifying this equation, we get: 7x/4 = 2 Multiplying both sides by 4/7, we get: x = 8/7 Now we can substitute this value of x back into the original numbers: x = 8/7 2x = 16/7 x+1 = 15/7 3x = 24/7 Next, we need to find the absolute deviation of each number from the mean: |x - 2| = |8/7 - 2| = 6/7 |2x - 2| = |16/7 - 2| = 2/7 |x+1 - 2| = |15/7 - 2| = 1/7 |3x - 2| = |24/7 - 2| = 10/7 The mean deviation is the average of these absolute deviations: (6/7 + 2/7 + 1/7 + 10/7)/4 = 19/28 Therefore, the mean deviation of the given numbers is 19/28. The answer closest to 19/28 among the options provided is 0.5. Therefore, the answer is (A) 0.5.
Tambaya 30 Rahoto
In how many ways can a delegation of 3 be chosen from 5 men and 3 women. If at least 1 man and 1 woman must be included?
Bayanin Amsa
Tambaya 32 Rahoto
Evaluate 81.81+99.4420.09+36.16 correct to 3 significant figures.
Bayanin Amsa
To evaluate the expression: $$\frac{81.81 + 99.44}{20.09 + 36.16}$$ We simply substitute the numbers in the formula and calculate: $$\frac{81.81 + 99.44}{20.09 + 36.16} = \frac{181.25}{56.25} \approx \boxed{3.23}$$ Therefore, the expression evaluates to approximately 3.23.
Tambaya 33 Rahoto
In how many ways can 9 people be seated if 3 chairs are available
Bayanin Amsa
If there are 9 people and 3 chairs, the first person can choose any of the 9 chairs. After the first person has chosen their seat, there are only 8 chairs left for the second person to choose from. Once the second person has chosen their seat, there are only 7 chairs left for the third person to choose from. Therefore, the total number of ways in which 3 people can be seated in 9 chairs is: 9 x 8 x 7 = 504 Therefore, the answer is 504, which is option (B).
Tambaya 34 Rahoto
The histogram above represents the number of candidates that sat for Mathematics examination in a school. How many candidate scored more than 50 marks?
Tambaya 35 Rahoto
Simplify 7112−434+212
Bayanin Amsa
Tambaya 36 Rahoto
If s = (2 + 3t)(5t - 4), find s/dt when t = 4/5 secs
Bayanin Amsa
x = (2+3t)(5t-4)
Let u = 2+3t ∴du/dt = 3
and v = 5t-4 ∴dv/dt = 5
dx/dt = Vdu/dt + Udv/dt
= (5t-4)3 + (2+3t)5
= 15t - 12 + 10 + 15t
= 30t - 2
= 30x4/5 - 2
= 24 - 2
= 22
Tambaya 37 Rahoto
If 55x + 52x = 7710 find X
Bayanin Amsa
5*X1 + 5*X∘
+ 5*X1 + 2*X∘
= 77 (change to base 10)
5*X + 5*1 + 5*X + 2*1 = 77
5X + 5 + 5X + 2 = 77
10X + 7 = 77
10X = 77-7
10X = 70
X = 70/10
X = 7
Tambaya 38 Rahoto
The cumulative frequency curve above shoes the distribution of the scores of 50 students in an examination. Find the 36th percentile scores
Bayanin Amsa
To find the 36th percentile score, we need to locate the corresponding point on the cumulative frequency curve. The 36th percentile score is the score below which 36% of the data falls. First, we need to find the total frequency, which is 50 in this case. Then we calculate 36% of the total frequency: 36/100 x 50 = 18 This means that we are looking for the score below which 18 students have scored. We can locate this point on the cumulative frequency curve by drawing a horizontal line from 18 on the y-axis to where it intersects the curve. Then we draw a vertical line down to the x-axis to read off the corresponding score. From the curve, we can see that the 36th percentile score lies between 30 and 40. To estimate the exact score, we can use linear interpolation. The distance between 30 and 40 is 10, and the distance between the cumulative frequency at 30 (which is 16) and the cumulative frequency at 40 (which is 22) is 6. The distance from 30 to the 36th percentile score is: 10 x (18 - 16)/6 = 1.67 So the 36th percentile score is approximately: 30 + 1.67 = 31.67 Therefore, the answer is 28%.
Tambaya 39 Rahoto
The sum of the first n terms of the arithmetic progression 5, 11, 17, 23, 29, 35, ... is
Bayanin Amsa
To find the sum of the first n terms of an arithmetic progression, we use the formula: Sn = n/2[2a + (n-1)d] where Sn is the sum of the first n terms, a is the first term, and d is the common difference between the terms. In this case, the first term is 5 and the common difference is 6 (since each term is 6 more than the previous one). So, using the formula: Sn = n/2[2(5) + (n-1)(6)] Simplifying, we get: Sn = n/2[10 + 6n - 6] Sn = n/2[6n + 4] Sn = 3n^2 + 2n Therefore, the correct answer is n(3n + 2).
Tambaya 40 Rahoto
A student spent 1/5of his allowance on books, 1/2 of remainder on food and kept the rest for contingencies. What fraction was kept?
Bayanin Amsa
Tambaya 41 Rahoto
What value of s will make the function x(4-x) a maximum
Bayanin Amsa
To find the value of s that will make the function x(4-x) a maximum, we need to take the derivative of the function with respect to x and set it equal to zero. The derivative of x(4-x) with respect to x is: d/dx (x(4-x)) = 4 - 2x Setting this derivative equal to zero and solving for x gives: 4 - 2x = 0 2x = 4 x = 2 Therefore, the maximum value of the function x(4-x) occurs when x = 2. To find the value of s, we substitute x = 2 into the equation s = 4 - x: s = 4 - 2 = 2 So the answer is s = 2.
Tambaya 42 Rahoto
What is the value of r if the distance between the point (4,2) and (1,r) is 3 units?
Bayanin Amsa
We can use the distance formula to solve for the value of `r`. The distance formula is: `d = sqrt((x2 - x1)^2 + (y2 - y1)^2)` where `(x1, y1)` and `(x2, y2)` are the coordinates of the two points, and `d` is the distance between them. In this case, the two points are (4, 2) and (1, r), and the distance between them is given as 3 units. So we can write: `3 = sqrt((1 - 4)^2 + (r - 2)^2)` Simplifying the right-hand side, we get: `3 = sqrt(9 + (r - 2)^2)` Squaring both sides, we get: `9 = 9 + (r - 2)^2` Simplifying, we get: `(r - 2)^2 = 0` Taking the square root of both sides, we get: `r - 2 = 0` So the value of `r` is `2`. Therefore, the answer is 2.
Tambaya 44 Rahoto
Simplify 5+√73+√7
Bayanin Amsa
To simplify the expression $\frac{5+\sqrt{7}}{3+\sqrt{7}}$, we need to rationalize the denominator. To do this, we can multiply the numerator and denominator by the conjugate of the denominator, which is $3-\sqrt{7}$. This gives us: $$\frac{5+\sqrt{7}}{3+\sqrt{7}}\cdot\frac{3-\sqrt{7}}{3-\sqrt{7}}=\frac{(5+\sqrt{7})(3-\sqrt{7})}{(3+\sqrt{7})(3-\sqrt{7})}=\frac{15-5\sqrt{7}+3\sqrt{7}-7}{9-7}$$ Simplifying the numerator and denominator gives: $$\frac{8-2\sqrt{7}}{2}=\boxed{4-\sqrt{7}}$$ Therefore, $\frac{5+\sqrt{7}}{3+\sqrt{7}}$ simplifies to $4-\sqrt{7}$.
Tambaya 45 Rahoto
Substract 164189 from 186309
Bayanin Amsa
To subtract 164189 from 186309, we need to perform column-wise subtraction starting from the rightmost column. In the ones column, we have 9 - 9 which gives us 0. We borrow 1 from the tens column and write it above the tens column. In the tens column, we have 0 - 8 (borrowed from the hundreds column) which gives us 2. In the hundreds column, we have 1 - 1 which gives us 0. In the thousands column, we have 6 - 4 which gives us 2. In the ten thousands column, we have 8 - 1 which gives us 7. So, the result of subtracting 164189 from 186309 is 22120. Therefore, the correct option is 22119.
Tambaya 46 Rahoto
Find the range of values of x for which 3x - 7 ≤ 0 and x + 5 > 0
Bayanin Amsa
To solve this inequality, we need to work with each inequality separately and then find the values of x that satisfy both. Starting with the first inequality: 3x - 7 ≤ 0 Adding 7 to both sides: 3x ≤ 7 Dividing by 3: x ≤ 7/3 Now, let's work with the second inequality: x + 5 > 0 Subtracting 5 from both sides: x > -5 To satisfy both inequalities, x must be greater than -5 and less than or equal to 7/3. Therefore, the range of values for x is: -5 < x ≤ 7/3
Tambaya 47 Rahoto
The probability of a student passing any examination is 2/3. If the students takes three examination, what is the probability that he will not pass any of them
Bayanin Amsa
P(pass) = 2/3
P(not pass) = 1 - 2/3 = 1/3
P(not passing any of the 3 Exams) = 1/3 x 1/3 x 1/3
= 1/27
Tambaya 48 Rahoto
A chord drawn 5 cm away from the center of a circle of radius 13 cm. Calculate the length of the chord
Bayanin Amsa
x2 + 52 = 132
x2 + 25 = 169
x2 = 144
x = √ 144
= 12
Length of the chord AB = 2x
2*12= 24cm
Tambaya 49 Rahoto
If Q is[9−2−74]
, then |Q| is
Bayanin Amsa
To find |Q|, we need to take the determinant of the matrix Q and then take its absolute value. So, we have: |Q| = |9 -2 -7| |-2 0 -4| |-7 -4 0| Expanding along the first row, we get: |Q| = 9(0 - (-4)) - (-2)(-4 - (-7)) - (-7)(-4 - 0) |Q| = 36 - 2(3) + 28 |Q| = 59 Therefore, the correct option is (D) 50.
Tambaya 50 Rahoto
If P =[x+3x+2x+1x−1] evaluate x if |P| = -10
Bayanin Amsa
P=[x+3x+2x+1x−1]
evaluate x if |P| = -10
(x+3)(x-1) - {(x+1)(x+2)} = -10
x2 - x + 3x - 3 - {x2 + 2x + x + 2} = -10
x2 + 2x - 3 - {x2 + 3x + 2} = -10
-x - 5 = -10
-5 + 10 = x
5 = x
∴x = 5
Za ka so ka ci gaba da wannan aikin?