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Question 1 Report
The shaded portion in the venn diagram above represents?
Question 2 Report
The number line represented by the inequality
x ≥ 2 is represented by an arrow with an filled-in dot above the 2.
The arrow points in the direction of all the numbers that are greater than 2.
Question 3 Report
Mr Adu spends his annual salary on food(f), rent(r), car maintenance, gifts(g), savings(s) and some miscellaneous (m) as indicate in the table below:
| F | R | C | G | S | M |
| 28% | 15% | 20% | 14% | 10% | 13% |
If the above information is represented on a pie chart. What angle represents his spending on food?
Question 4 Report
Zoology has 7 letters in total, with O repeated thrice
\(\frac{7!}{3!}\) → \(\frac{7*6*5*4*3*2*1}{3*2*1}\)
= 840ways
Question 5 Report
Question 6 Report
In the diagram above angle LNM and angle YNZ are represented by g and h respectively. Find ∠MNY
Question 7 Report
Question 8 Report
Using pythagoras formula:
Hyp2 = adj2 + opp2
52 = opp2 + 32
52 - 32 = adj2
4 = adj
length of the chord = 2 * 4 = 8cm
Question 9 Report
If \( \sin \theta = -\frac{3}{5} \) and \( \theta \) lies in the third quadrant, find \( \cos \theta \)
Where sin θ = opphyp →
opp = -3, hyp = 5
using pythagoras formula
hyp2 = adj2 + opp2
adj2 = hyp2 - opp2
adj2 = 52 - 32 → 25 - 9
adj2 = 16
adj = 4
cos θ = adjhyp →
In third quadrant: cos θ is negative → - 45
Question 10 Report
Question 11 Report
Integrate \( (2x+1)^3 \)
Recall chain rule:
u = 2x +1; du = 2dx → dx = du2
u3
= ∫ u3
du2
→ 12
∫ u3
= 1∗u42∗4
= u48 → 2x+148 + C
Question 12 Report
Given that \( r = \sqrt{\frac{3v}{\pi h}} \) make v the subject of the formula
square both sides to remove the big square root
→ r2 = 3vπh
cross multiply
3v = r2 * πh
v = πr2h3
Question 13 Report
If \( \sec^2 \theta + \tan^2 \theta = 3 \), then the angle \( \theta \) is equal to?
Question 14 Report
In the diagram above, XY = 8cm and OX = 5cm. Find Oz
hyp = 5cm, adj = 8cm2
Pythagoras theorem:
hyp22
= opp2
+ adj2
52 = x2 + 42
x
= 25 - 16
x = √9
x = 3cm
Question 15 Report
Question 16 Report
Given that \(S = 3t^2 + 5t - 10\) is displacement of a particle in metres, calculate it initial velocity.
To calculate the initial velocity of the particle, we need to find the first derivative of the displacement equation. The first derivative of the equation 3t^2 + 5t - 10 would give us the velocity equation.
The derivative of 3t^2 is 6t, the derivative of 5t is 5, and the derivative of a constant like -10 is 0.
So, the velocity equation would be 6t + 5. This is the initial velocity of the particle, which is 5 m/s.
Question 17 Report
Question 18 Report
Firstly; solving for x
6 = 2+5+x+1+x+2+7+96
cross multiply to have:
6 * 6 = 2 + 5 + x+1 + x+2 + 7 + 9
36 = 2x + 26
36 - 26 = 2x
10 = 2x
x = 5
Median = 7+62
→ 6.5
Question 19 Report
Question 20 Report
Question 21 Report
Given Data: x1 = -3, x2 = 2, y1 = 5, y2 = 10
coordinates of the mid-point of the line = (x1+x22 , y1+y22 )
(−3+22 , 5+102 )
= −12 ) , 152 )
Question 22 Report
Question 23 Report
Solve for k in the equation \( \left(\frac{1}{8}\right)^{k+2} = 1 \)
(8?1)k+2 = (80)
base 8 cancel out on both sides
-1(k+2) = 0
-k -2 = 0
: k = -2
Question 24 Report
Question 25 Report
Question 26 Report
Evaluate \(\mathrm{Log}_{2}\,8\mathbf{\sqrt{2}}\)
where Log2 8√2 → Log √128
→ Log22
12812
=
* (Log2
128) →
* (Log
2
)
= 7 * 12 * (Log
2)
where (Log 2) = 1
→ 7 *
* 1
=
or 3.5
Question 27 Report
Question 28 Report
If y varies inversely as x and x = 3 when y =4. Find the value of x when y = 12
Question 29 Report
Evaluate \(^{n^2+1}C_{n+5}\) if n = 3
32+1 C3+5
9+1 C3+5
10
C
= 10!8!2!
10∗9∗8!8!2! = 10∗92
= 45
Question 30 Report
Simplify \( \frac{1}{3-\sqrt{2}} \) in the form of p + q√2
Rationalization with conjugate 3+√2
→ 1∗[3+√2][3−√2][3+√2]
= 3+√29−3√2+3√2+√4
= 3+√29−2
→
=
+ √27
Question 31 Report
The common difference is 6.
Just add 6 to get the next term
8, 14, 20 and 26
Question 32 Report
What will be the result obtained when the numerator of \( \frac{96}{50} \) is decreased by 37.5% and its denominator decreased by 20%.
Numerator: 96 → 37.5% of 96 = 36
Decreased by 36 → 96 - 36
New numerator = 60
Denominator: 50 → 20% of 50 = 10
Decreased by 10 → 50 - 10 = 40
New Denominator = 40
New fraction = 6040 or 1.5
Question 33 Report
Question 34 Report
Evaluate \( (101_{two})^3 \)
5\(^3\) = 125
And 125 = \(1111101_{two}\)
Question 35 Report
Find the determinant of the matrix \( A = \begin{pmatrix} 2 & 3 \\ 1 & 3 \end{pmatrix} \)
|A| = (2*3) - (1*3)
→ 6 - 3
= 3
Question 36 Report
Question 37 Report
Let (*) be a binary operation on a natural number defined by \(a * b = a - b + (ab)^2\), then find \(3 * 5\)
3 * 5 = 3 - 5 + (3 x 5)2
-2 + (15)2
-2 + 225
= 223
Question 38 Report
find the limit of y = \( \frac{x^3 + 6x - 7}{x - 1} \) as x tends to 1
x3+6x−7x−1
When numerator is differentiated → 3x2 + 6
When denominator is differentiated → 1
: 3x2+61
substitute x for 1
3∗12+61 = 3+61
= 91
= 9
Question 39 Report
If \(A = \begin{pmatrix} 2 & 1 \\ 2 & 3 \\ 1 & 2 \end{pmatrix}\) and \(B = \begin{pmatrix} 3 & 2 \\ 4 & 2 \end{pmatrix}\). Find \(AB\)
Given A = ⎛⎝⎜221132⎞⎠⎟
and B = (3422)
We can multiply these matrices since the number of colums in A = number of rows in B
AB = ⎛⎝⎜(2∗3)+(1∗4)(2∗3)+(3∗4)(1∗3)+(2∗4)(2∗2)+(1∗2)(2∗2)+(3∗2)(1∗2)+(2∗2)⎞⎠⎟
AB = ⎛⎝⎜(6+4)(6+12)(3+8)(4+2)(4+6)(2+4)⎞⎠⎟
= ⎛⎝⎜1018116106⎞⎠⎟
Question 40 Report
A rectangular pyramid has an area \(24\text{cm}^2\) and height \(7.5\text{cm}\). Find its volume?
Volume of a rectangular pyramid = length∗width∗height3or area∗height3
= 24∗7.53 → 1803
Volume of the rectangular pyramid = 60cm3
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