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**Question 1**
**Report**

A particle of mass 3kg moving along a straight line under the action of a F N, covers a line distance, d, at time, t, such that d = t2 ${}^{\mathrm{}}$ + 3t. Find the magnitude of F at time t.

**Answer Details**

F = m * a

d = t2 ${}^{\mathrm{}}$ + 3t.

a = d2ddt2 $\frac{{}^{}}{}$

d[d]dt = 2t + 3

d2ddt2 = 2m/s2 ${}^{\mathrm{}}$

a = 2m/s2 ${}^{\mathrm{}}$

F = m * a

F = 3 × 2 = 6N

**Question 2**
**Report**

If 36, p,94 $\frac{\mathrm{}}{}$ and q are consecutive terms of an exponential sequence (G.P), find the sum of p and q.

**Answer Details**

GP : 36, P, q4 $\frac{\ufffd}{4}$, q, ... p + q = ?

Recall, | common | ratio, | r | = | TnTn-1 | = | T2T1 | = | T3T2 | = | T4T3 |

∴ | P36 | = | 94 | ÷ | p | ; | p2 ${}^{2}$ | = | 94 | x | 36 | ; | p2 ${}^{2}$ | = | 81 |

p | = | 9 | ∴ | r | = | T2T1 | = | 936 | = | 14 |

Also | r | = | T4T3 | = | q | ÷ | 94 |

∴ 14
$\frac{1}{4}$ = q ÷ 94
$\frac{9}{4}$ ;

94 $\frac{9}{4}$ = 4q

16q | = | 9 | , | q | = | 916 | ∴ | p | + | q | = | 9 | + | 916 | = | 9 | 916 |

**Question 3**
**Report**

If log10(3x+1)+log104=log10(9x+2)

, find the value of x

**Answer Details**

log10(3x+1)+log104=log10(9x+2) $$

log104(3x+1)=log10(9x+2) $$

4(3x+1) = 9x + 2

12x -4 = 9x + 2

12x - 9x = 2 + 4

3x = 6

x = 2

**Question 4**
**Report**

Evaluate 4p2+4C2−4p3

**Answer Details**

4p2+4C2−4p3 $\mathrm{}$

npr=n![n−r]!andnCr=n![n−r]!r! $$

= 4![4−2]!+4![4−2]!2!−4![4−3]!=4!2!+4!2!2!−4!1! $\frac{4!}{[4-2]!}+\frac{4!}{[4-2]!2!}-\frac{4!}{[4-3]!}=\frac{4!}{2!}+\frac{4!}{2!2!}-\frac{4!}{1!}$

= 4∗3∗2!2!+4∗3∗2!2!2!−4∗3∗2∗11! $\frac{4\ast 3\ast 2!}{2!}+\frac{4\ast 3\ast 2!}{2!2!}-\frac{4\ast 3\ast 2\ast 1}{1!}$

12 + 6 - 24 = -6

**Question 5**
**Report**

A body of mass 18kg moving with velocity 4ms-1 collides with another body of mass 6kg moving in the opposite direction with velocity 10ms-1. If they stick together after the collision, find their common velocity.

**Answer Details**

m1u1 + m2u2 = (m1 + m2)v

m1 = 18kg, m2 = 6kg, u1 = 4ms-1, u2 = -10m/s

18(4) + 6(-10) = (18+6)v

72 - 60 = 24v

12 = 24v

v = 12
$\frac{\mathrm{}}{}$ m/s

**Question 6**
**Report**

Given ∣∣∣21−34∣∣∣∣∣∣−6k∣∣∣∣∣∣3−26∣∣∣=15 Solve for k.

**Answer Details**

∣∣∣21−34∣∣∣∣∣∣−6k∣∣∣∣∣∣3−26∣∣∣=15 $|$

∣∣∣2[−6]1[−6]−3k+4k∣∣∣=∣∣∣3−26∣∣∣
$|\begin{array}{c}\mathrm{}\end{array}$

∣∣∣−12−6−3k+4k∣∣∣=∣∣∣3−26∣∣∣
$|\begin{array}{c}\end{array}$

-12 - 3k = 3

-3k = 3 + 12

k = 15−3 $\frac{\mathrm{}}{}$

k = -5

**Question 7**
**Report**

The probability that a student will graduate from college is 0.4. If 3 students are selected from the college, what is the probability that at least one student will graduate?

**Answer Details**

To find the probability that at least one student will graduate, we can calculate the probability that no student will graduate and then subtract that from 1. The probability that a student will not graduate from college is 1 - 0.4 = 0.6. So, the probability that none of the three selected students will graduate is: 0.6 * 0.6 * 0.6 = 0.216 Therefore, the probability that at least one student will graduate is: 1 - 0.216 = 0.784 So, the answer is 0.78. In summary, the probability that at least one student will graduate from college given that 3 students are selected is 0.784 or 0.78, which is obtained by subtracting the probability that none of the three students will graduate from 1.

**Question 8**
**Report**

The table shows the distribution of the distance (in km) covered by 40 hunters while hunting.

What is the mode of the distribution?

Distance(km) | 3 | 4 | 5 | 6 | 7 | 8 |

Frequency | 5 | 4 | x | 9 | 2x | 1 |

**Answer Details**

To find the mode of the distribution, we need to identify the value with the highest frequency. In the table, the frequency is given for each distance (in km) covered by the hunters. Let's first find the frequency of distance 5 km covered by the hunters. From the table, we know that the sum of all frequencies should be equal to 40, the total number of hunters. Sum of frequencies = 5 + 4 + x + 9 + 2x + 1 = 40 Simplifying the equation, we get: 3x + 19 = 40 3x = 21 x = 7 Now we know that the frequency of distance 5 km is x = 7. The frequencies for the other distances are: - Frequency of distance 3 km = 5 - Frequency of distance 4 km = 4 - Frequency of distance 6 km = 9 - Frequency of distance 7 km = 2x = 14 - Frequency of distance 8 km = 1 The highest frequency is 14, which corresponds to the distance of 7 km. Therefore, the mode of the distribution is 7 km. Hence, the answer is option (C) 7.

**Question 9**
**Report**

Solve: 4sin2 ${}^{2}$θ + 1 = 2, where 0º < θ < 180º

**Answer Details**

4sin2

2
${}^{}$θ + 1 = 2

4sin2 ${}^{\mathrm{}}$θ = 2 - 1

4sin2 ${}^{2}$θ = 1

s√in2θ

= 14−−√

sinθ = 12 $\frac{\mathrm{}}{}$

θ = sin−112 $$

θ = 30º 0r 150º

**Question 10**
**Report**

Given that P = (-4, -5) and Q = (2,3), express →PQ in the form (k,θ). where k is the magnitude and θ the bearing.

**Answer Details**

To find →PQ, we need to subtract the coordinates of point P from those of point Q, giving us: →PQ = Q - P = (2-(-4), 3-(-5)) = (6, 8) The magnitude, or length, of →PQ is found using the distance formula: k = √(6² + 8²) = √100 = 10 To find the bearing θ, we use trigonometry. The tangent of θ is the ratio of the opposite side (the change in y-coordinates) to the adjacent side (the change in x-coordinates): tan θ = 8/6 = 4/3 Using a calculator, we can find that θ is approximately 53.13º. However, we need to adjust this value depending on which quadrant →PQ lies in. Since both x and y are positive, →PQ lies in the first quadrant, so we don't need to make any adjustments. Therefore: θ = 53.13º So the vector →PQ can be expressed in the form (k,θ) as: (10 units, 53.13º) Therefore, the correct answer is (a) (10 units, 053º).

**Question 11**
**Report**

(3√6√5+√543√5 $\frac{3\surd 6}{\surd 5}+\frac{\surd 54}{3\surd 5}$)−1

**Answer Details**

(3√6√5+√543√5 $\frac{3\surd 6}{\surd 5}+\frac{\surd 54}{3\surd 5}$)−1 ${}^{-1}$

= √5(3√5)3√6+3√6 $\frac{\surd 5(3\surd 5)}{3\surd 6+3\surd 6}$

= 3∗56√6=52√6 $\frac{3\ast 5}{6\surd 6}=\frac{5}{2\surd 6}$

= 5∗2√62√6+2√6=10√64∗6 $\frac{5\ast 2\surd 6}{2\surd 6+2\surd 6}=\frac{10\surd 6}{4\ast 6}$

= 5√612

**Question 12**
**Report**

A particle moving with a velocity of 5m/s accelerates at 2m/s2 ${}^{2}$. Find the distance it covers in 4 seconds.

**Answer Details**

from the equation of motion

u = 5m/s, a = 2m/s2
${}^{\mathrm{}}$, t = 4s

s = ut + 12at2 $\frac{\mathrm{}}{}$

s = 5*4 + 122∗42 $\frac{\mathrm{}}{}$

s = 20 + 16

s = 36m

**Question 13**
**Report**

In how many ways can six persons be paired?

**Answer Details**

6C2=6![6−2]![2!] ${}_{2}$

6∗5∗4!4!∗2! $\frac{\mathrm{}}{}$

= 6∗52 $\frac{\mathrm{}}{}$

= 15

**Question 14**
**Report**

If x2+y2+−2x−6y+5=0 ${}^{}$, evaluate dy/dx when x=3 and y=2.

**Answer Details**

x2+y2+−2x−6y+5=0 ${}^{}$

When differentiated:

x2+y2+−2x−6y+5=0 ${\mathrm{\to \; 2x\; +\; 2y\; -\; 2\; -\; 6\; =\; 0}}^{}$

where x=3 and y=2

2[3] + 2[2] - 8 = 0

6 + 4 - 8 = 2

**Question 15**
**Report**

Given that 8x+mx2−3x−4≡5x+1+3x−4

**Answer Details**

8x+mx2−3x−4≡5x+1+3x−4 $\frac{\mathrm{}}{}$

8x+mx2−3x−4 $\frac{\mathrm{}}{}$ ≡ 5(x−1)+3(x+4)x2−3x−4 $\frac{\mathrm{}}{}$

multiplying both sides by x2-3x-4

8x+m ≡ 5(x-4)+3(x+1)

8x + m ≡ 5x - 20 + 3x + 3

8x - 5x - 3x + m = -20 + 3

m = -17

**Question 16**
**Report**

Simplify 9∗3n+1−3n+23n+1−3n

**Answer Details**

9∗3n+1−3n+23n+1−3n $\frac{\mathrm{}}{}$

= 3n∗3n∗31∗−32∗323n∗31−3n $\frac{{\mathrm{}}^{}}{}$

= 3n(32∗31)3n(31−1) $\frac{{\mathrm{}}^{}}{}$

= 27−93−1 $\frac{27-9}{3-1}$

= 182 $\frac{18}{2}$

= 9

**Question 17**
**Report**

A binary operation ∆ is defined on the set of real numbers R, by x∆y = x+y−xy4−−−−−−−−−√

, where x, yER. Find the value of 4∆3

**Answer Details**

x∆y = x+y−xy4−−−−−−−−−√

4∆3 = 4+3−4∗34−−−−−−−−−√ $\sqrt{4+3-\frac{4\ast 3}{4}}$

= 4+3−3−−−−−−−−√ $\sqrt{4+3-3}$

= 4–√ $\sqrt{4}$

= 2.

**Question 18**
**Report**

If Un = kn2 ${}^{2}$ + pn, U1 ${}_{1}$ = -1, U5 ${}_{5}$ = 15, find the values of k and p.

**Answer Details**

Un = kn2 ${}^{2}$ + pn,

U1 ${}_{1}$ = -1,

U5 ${}_{5}$ = 15,

when n = 1

U1 = k(1)2
${}^{2}$+ p(1) = -1

k + p = -1 --------eqn1

when n = 5

U5
${}_{5}$ = k(5)2
${}^{2}$+ p(5) = 15

25k + 5p = 15 --------eqn2

multiply eqn1 by 5 and eqn2 by 1

5k + 5p = -5 -------eqn3

25k + 5p = 15 -------eqn4

eqn4 - eqn3

20k = 20

k = 1

sub for k in eqn1

1 + p = -1

p = -1 -1 = -2

**Question 19**
**Report**

Which of the following is the semi-interquartile range of a distribution?

**Answer Details**

The semi-interquartile range is a measure of variability in a dataset. To calculate it, we first need to find the median of the dataset, which is the middle value when the data is arranged in order from lowest to highest. Then, we need to find the quartiles, which are the values that divide the dataset into four equal parts. The semi-interquartile range is half of the difference between the upper quartile and the lower quartile. The upper quartile is the value that separates the highest 25% of the data from the lowest 75%, while the lower quartile is the value that separates the lowest 25% of the data from the highest 75%. Looking at the given options, the formula that corresponds to the semi-interquartile range is: 1/2 (Upper Quartile - Lower Quartile) Therefore, the correct answer is option d) "1/2 (Upper Quartile - Lower Quartile)".

**Question 20**
**Report**

Evaluate1∫0x2(x3+2)3

**Answer Details**

1∫0x2(x3+2)3 ${\mathrm{}}_{}^{}{3}^{}$dx

let u=x3+2,du=3x2d