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Ajụjụ 1 Ripọtì
Given that 1/2 log10 P = 1, find the value Of P
Ajụjụ 3 Ripọtì
A chord of a circle of a diameter 42cm subtends an angle of 60o at the centre of the circle. Find the length of the mirror arc
Akọwa Nkọwa
Ajụjụ 4 Ripọtì
Differentiate 6x3−5x2+13x2 with respect to x
Akọwa Nkọwa
6x3−5x2+13x2
let y = 3x2
y = 6x33x2
- 6x23x2
+ 13x2
Y = 2x - 53
+ 13x2
dydx
= 2 + 13
(-2)x-3
= 2 - 23x3
Ajụjụ 5 Ripọtì
In a basket of fruits, there are 6 grapes, 11 bananas and 13 oranges. If one fruit is chosen at random, what is the probability that the fruit is either a grape or a banana?
Akọwa Nkọwa
There are a total of 6 + 11 + 13 = 30 fruits in the basket. The probability of selecting a grape is 6/30 and the probability of selecting a banana is 11/30. To find the probability of selecting either a grape or a banana, we add these probabilities: 6/30 + 11/30 = 17/30 So the probability of selecting either a grape or a banana is 17/30. Therefore, the answer is: - 17/30
Ajụjụ 6 Ripọtì
If U = (s, p, i, e, n, d, o, u, r), X = (s, p, e, n, d) Y = (s, e, n, o, u), Z = (p, n, o, u, r) find X ∩ Y ∪ Z
Ajụjụ 7 Ripọtì
Age202530354045Number of people351123
Find the median age of the frequency distribution in the table above.
Akọwa Nkọwa
To find the median age, we need to arrange the ages in ascending order and then find the middle value. However, since the table only gives us frequency counts for each age, we first need to construct a cumulative frequency distribution. Age | Number of People | Cumulative Frequency --- | --- | --- 20 | 3 | 3 25 | 5 | 8 30 | 1 | 9 35 | 1 | 10 40 | 2 | 12 45 | 3 | 15 The total number of people is 15, which is an odd number, so the median is simply the value that corresponds to the middle person. Since the cumulative frequency for 8 people is at age 25 and the cumulative frequency for 9 people is at age 30, the median age is 25. Therefore, the answer is: - 25
Ajụjụ 8 Ripọtì
Find the difference between the range and the variance of the following set of numbers 4, 9, 6, 3, 2, 8, 10, 5, 6, 7 where ∑d2 = 60
Ajụjụ 9 Ripọtì
The angle between the positive horizontal axis and a given line is 135o. Find the equation of the line if it passes through the point (2,3)
Akọwa Nkọwa
The problem requires finding the equation of a line that passes through a given point (2,3) and makes an angle of 135 degrees with the positive x-axis. To solve this problem, we need to first find the slope of the line. Since the given angle is measured from the positive x-axis and is 135 degrees, we know that the angle made with the negative x-axis is 45 degrees. Therefore, the slope of the line is the tangent of 45 degrees, which is 1. Now that we have the slope, we can use the point-slope form of the equation of a line to find the equation of the line. The point-slope form is: y - y1 = m(x - x1) where m is the slope and (x1, y1) is a point on the line. Plugging in the values we have, we get: y - 3 = 1(x - 2) Simplifying this equation gives us: y - x + 3 = 0 This equation is in the form of y = mx + b, where m is the slope and b is the y-intercept. We can see that the slope is 1, which we found earlier, and the y-intercept is 3. Therefore, the equation of the line that passes through the point (2,3) and makes an angle of 135 degrees with the positive x-axis is: y - x + 3 = 0 which is equivalent to: y = x - 3 So, the answer is neither (a) nor (b), but is (c) x + y = 5.
Ajụjụ 10 Ripọtì
Determine x + y if (2?3?14) (xy) = (?18)
Ajụjụ 11 Ripọtì
What value of g will make the expression 4x2 - 18xy + g a perfect square?
Akọwa Nkọwa
4x2 - 18xy + g = g →
(18y4
)2
= 18y24
Ajụjụ 12 Ripọtì
Find the range of values of x for which 3x - 7 ≤ 0 and x + 5 > 0
Akọwa Nkọwa
The first inequality 3x - 7 ≤ 0 can be solved as follows: 3x - 7 ≤ 0 3x ≤ 7 x ≤ 7/3 The second inequality x + 5 > 0 can be solved as follows: x + 5 > 0 x > -5 So the range of values of x that satisfies both inequalities is -5 < x ≤ 7/3. The correct answer is -5 < x ≤ 7/3.
Ajụjụ 13 Ripọtì
A survey of 100 students in an institution shows that 80 students speak Hausa and 20 students speak Igbo, while only 9 students speak both language. How many students speak neither Hausa nor Igbo?
Akọwa Nkọwa
In the survey, 80 students speak Hausa, 20 students speak Igbo and 9 students speak both Hausa and Igbo. If we add the number of students speaking only Hausa and the number of students speaking only Igbo, we get 80 + 20 - 9 = 91. Therefore, there are 100 - 91 = 9 students who speak neither Hausa nor Igbo. So, the answer is 9.
Ajụjụ 14 Ripọtì
Simplify 2?3+3?53?5?2?3
Akọwa Nkọwa
(2?3+3?5)(3?5+2?3)(3?5?2?3)(3?5?2?3)
= 5+12?1533
=19+4?1511
Ajụjụ 16 Ripọtì
Find the standard derivation of the following data -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5
Akọwa Nkọwa
x = ∑xN
= 011
= 0
x(x−x)(x−x)2−5−525−4−416−3−39−2−24−1−1100011122433944165525110
S.D = √∑(x−x)2∑f
= √11011
= √10
Ajụjụ 18 Ripọtì
Solve the simultaneous equations 2x−2x = 2, 4x+3y = 10
Akọwa Nkọwa
2x−2x
= 2.....(1)
4x+3y
= 10
6x
= 12 →
x = 612
x = 12
put x = 12
in equation (i)
= 4 - 3y
= 2
= 4 - 2
= 3y
therefore y = 32
Ajụjụ 20 Ripọtì
The nth term of a sequence is given 31 - n , find the sum of the first terms of the sequence.
Akọwa Nkọwa
Tn = 31 - n
S3 = 31 - 1 + 31 - 2 + 31 - 3
= 1 + 13
+ 19
= 139
Ajụjụ 21 Ripọtì
A cone with the sector angle of 45o is cut out of a circle of radius of the cone.
Akọwa Nkọwa
Ajụjụ 22 Ripọtì
The table above shows that the scores of a group of students in a test. If the average score is 3.5, find the value of x
Akọwa Nkọwa
| mean = | 60 + 5x |
| 18 + x | |
| 3.5 = | 60 + 5x |
| 18 + x | |
| 7 = | 60 + 5x |
| 2 | 18 + x |
7(18+x) = 2(60+5x)
126 + 7x = 120 + 10x
10x - 7x = 126 - 120
3x = 6
x = 2
Ajụjụ 23 Ripọtì
Find the non-zero positive value of x which satisfies the equation ∣∣ ∣∣x101xx01x∣∣ ∣∣ = 0
Akọwa Nkọwa
x(x2 - 1) - x = 0
= x3 - 2x = 0
x(x2 - 2) = 0
x = 2
Ajụjụ 24 Ripọtì
Evaluate 64.7642 - 35.2362 correct to 3 significant figures
Akọwa Nkọwa
To evaluate the expression 64.7642 - 35.2362, we simply subtract the second number from the first number. 64.7642 - 35.2362 = 29.528 To round this answer to three significant figures, we count three digits starting from the left-most nonzero digit. In this case, the left-most nonzero digit is 2, so we round to three digits after the 2. 29.5 Therefore, the answer is 29.5 to 3 significant figures.
Ajụjụ 25 Ripọtì
A number is selected at random between 20 and 30, both numbers inclusive. Find the probability that the number is a prime
Akọwa Nkọwa
Possible outcomes are 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30. Prime numbers has only two factors
itself and 1
The prime numbers among the group are 23, 29. Probability of choosing a prime number
= Number of primeNo. of total Possible Outcomes
= 211
Ajụjụ 26 Ripọtì
Find the distance between the point Q (4,3) and the point common to the lines 2x - y = 4 and x + y = 2
Akọwa Nkọwa
To find the distance between point Q and the point common to the lines 2x - y = 4 and x + y = 2, we need to first find the coordinates of the point of intersection of the two lines. We can solve the system of equations: 2x - y = 4 x + y = 2 by either substitution or elimination to get the coordinates of the point of intersection, which is (1, 1). Then, we can use the distance formula to find the distance between point Q (4, 3) and (1, 1): d = √[(4 - 1)² + (3 - 1)²] = √[9 + 4] = √13. Therefore, the answer is √13.
Ajụjụ 27 Ripọtì
Two binary operations ∗ and ⊕ are defines as m ∗ n = mn - n - 1 and m ⊕ n = mn + n - 2 for all real numbers m, n.
Find the value of 3 ⊕ (4 ∗ 50)
Akọwa Nkọwa
m ∗
n = mn - n - 1, m ⊕
n = mn + n - 2
3 ⊕
(4 ∗
5) = 3 ⊕
(4 x 5 - 5 - 1) = 3 ⊕
14
3 ⊕
14 = 3 x 14 + 14 - 2
= 54
Ajụjụ 28 Ripọtì
Integrate 1x + cos x with respect to x
Ajụjụ 30 Ripọtì
If the function f(fx) = x3 + 2x2 + qx - 6 is divisible by x + 1, find q
Akọwa Nkọwa
To find the value of q, we can use the fact that the given function f(fx) is divisible by x+1. If a polynomial f(x) is divisible by x-a, then f(a) = 0. Using this fact, we can substitute x = -1 in the given function f(fx) = x^3 + 2x^2 + qx - 6, since x+1 is a factor of f(fx): f(f-1) = (-1)^3 + 2(-1)^2 + q(-1) - 6 f(f-1) = -1 + 2 - q - 6 f(f-1) = -5 - q Since f(f-1) is divisible by x+1, we have: f(f-1) = -5 - q = 0 Therefore, q = -5. Hence, the value of q is -5 when f(fx) = x^3 + 2x^2 + qx - 6 is divisible by x+1.
Ajụjụ 32 Ripọtì
Find the value of (0.006)3 + (0.004)3 in standard form
Akọwa Nkọwa
To solve this problem, we simply need to evaluate the expressions inside the parentheses first, then add the two results. (0.006)3 = 0.000000216, which can be written in standard form as 2.16 x 10-7. (0.004)3 = 0.000000064, which can be written in standard form as 6.4 x 10-8. Adding these two values gives: 2.16 x 10-7 + 6.4 x 10-8 = 2.76 x 10-7 Therefore, the answer is 2.8 x 10-7.
Ajụjụ 33 Ripọtì
The figure shows circles of radii 3cm and 2cm with centres at X and Y respectively. The circles have a transverse common tangent of length 25cm. Calculate XY.
Akọwa Nkọwa
Ajụjụ 34 Ripọtì
In the figure, XYZ is a triangle with XY = 5cm, XZ = 2cm and XZ is produced to E making the angle < YZE = 150∘ . If the angle XYZ = θ , calculate the value of sinθ
Ajụjụ 35 Ripọtì
If X ∗ Y = X + Y - XY, find x when (x ∗ 2) + (x ∗ 3) = 68
Akọwa Nkọwa
x ∗
y = x + y - xy
(x ∗
2) + (x ∗
3) = 68
= x + 2 - 2x + x + 3 - 3x
= 86
3x = 63
x = -21
Ajụjụ 37 Ripọtì
Find the volume of the prism.
Akọwa Nkọwa
To find the volume of a prism, we need to multiply the area of the base by the height of the prism. In this case, the base is a triangle with base 11cm and height 15cm, so its area is 1/2 * base * height = 1/2 * 11cm * 15cm = 82.5cm^2. The height of the prism is given as 6cm. Therefore, the volume of the prism is 82.5cm^2 * 6cm = 495cm^3. Hence, the answer is 495cm^3.
Ajụjụ 38 Ripọtì
An arc of a circle subtends an angle 70o at the centre. If the radius of the circle is 6cm, calculate the area of the sector subtended by the given angle.(π = 227 )
Akọwa Nkọwa
Ajụjụ 39 Ripọtì
The angle of elevation of a building from a measuring instrument placed on the ground is 30o. If the building is 40m high, how far is the instrument from the foot of the building?
Akọwa Nkọwa
40x
= tan 30o
x = 40tan36
= 401√3
= 40√3m
Ajụjụ 40 Ripọtì
Given that loga2 = 0.693 and loga3 = 1.097, find loga 13.5
Akọwa Nkọwa
loga 13.5 = loga 272
= 3loga 3 - log2a
= 3 x 1.097 - 0.693
= 2.598
Ajụjụ 41 Ripọtì
A point P moves so that is equidistant from point L and M. If LM is 6cm, find the distance of P from LM when P is 10cm from L
Akọwa Nkọwa
p from LM = √102
- 82
= √36
= 6cm
Ajụjụ 44 Ripọtì
Sn is the sum of the first n terms of a series given by Sn = n2n - 1. Find the nth term
Akọwa Nkọwa
The given series is: 1 + 3 + 5 + 7 + ... + (2n - 1) Notice that each term in the series is an odd number, and the difference between consecutive terms is 2. To find the nth term, we can use the formula for the nth term of an arithmetic sequence: an = a1 + (n - 1)d where a1 is the first term, d is the common difference, and n is the term number. In this case, a1 = 1 and d = 2, so we have: an = 1 + (n - 1)2 an = 2n - 1 Therefore, the answer is (D) 2n - 1.
Ajụjụ 45 Ripọtì
Find the simple interest rate percent per annum at which ₦1,000 accumulates to ₦1,240 in 3 years
Akọwa Nkọwa
The formula for simple interest is: I = P * r * t Where: I = Interest P = Principal (initial amount borrowed or invested) r = Interest rate per annum (as a decimal) t = Time in years From the given information: P = ₦1,000 I = ₦1,240 - ₦1,000 = ₦240 t = 3 years Substituting these values into the formula, we get: 240 = 1000 * r * 3 Simplifying the equation, we get: r = 240 / (1000 * 3) = 0.08 Converting to a percentage, we get: r = 0.08 * 100% = 8% Therefore, the simple interest rate percent per annum is 8%. Answer: 8%
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