1. QUESTION 9 The given inequality is 𝑦≤− 5 6 𝑥+ 2 3
The corresponding linear equation is 𝑦=− 5 6 𝑥+ 2 3
The intercepts are ( 4 5 ,0) , 0, 2 3
We plot the x-intercepts and draw a solid line through it.
We test the inequality 𝑦≤− 5 6 𝑥 with the origin (0,0).
This gives 0≤− 5 6 (0) ⇒0≤
This statement is true.
So we shade the lower half plane to obtain the graph in option B.
The correct answer is B.

2. QUESTION 10 The given system is : 𝑥≥0,𝑦≥𝑥,𝑦≤− 1 4 𝑥+5
For 𝑥≥0, we draw 𝑥=0 and shade the right half plane.
For 𝑦≥𝑥, we draw 𝑦=𝑥 and shade the upper half plane.
For 𝑦≤− 1 4 𝑥+5, we plot 𝑦=− 1 4 𝑥+5 and shade the lower half plane.
The intersection of all the regions is
The correct answer is A

3. QUESTION 11 The given system is −10𝑥−9𝑦+13𝑧=12 8𝑥−7𝑦−5𝑧=15
5𝑥+4𝑦−6𝑧=−12
The coefficient matrix is
−10 − −7 −5 5 4 −6
The constant matrix is
12 15 −12
The augmented matrix is
The correct answer is A
12 15 −12

4. The function given to us is 𝑦=2 𝑥+2 2 +3
QUESTION 12
The function given to us is
𝑦=2 𝑥
This function is written in the form 𝑦=𝑎 𝑥−ℎ 2 +𝑘
Where 𝑉(ℎ,𝑘) is the vertex of the function.
By comparing to the vertex form, 𝑎=2,ℎ=−2,𝑘=3
This implies that, 𝑉(−2,3) is the vertex.
Since 𝑎=2>0, the graph of the function must open upwards.
The graph that has its vertex at , 𝑉(−2,3) is the one in option B
The correct answer is B.

5. QUESTION 13 The pumpkins height is given by the equation;
𝑦=12+105𝑥−16 𝑥 2
⇒𝑦=−16 𝑥 𝑥+12
⇒𝑦=−16 (𝑥 2 − 𝑥)+12
⇒𝑦=−16 (𝑥 2 − 𝑥+ − )−−16 −
⇒𝑦=−16 𝑥− ) , ⇒𝑦=−16 𝑥− )
The vertex of this function is 𝑉(3.28,184.27). The y-value is the maximum height.
When the pumpkin hits the ground, its height is 0 feet.
This implies that; 0=−16 𝑥−
⇒−184.27=−16 𝑥−
⇒− −16 = 𝑥− , ⇒11.517= 𝑥− , ⇒ =𝑥−3.28,𝑥= =6.67
The pumpkins maximum height is and it hits the ground after 6.67 seconds.
The correct answer is C

6. Let the function be y=𝑎 𝑥 2 +𝑏𝑥+𝑐
QUESTION 14
Let the function be y=𝑎 𝑥 2 +𝑏𝑥+𝑐
When we substitute (−1,10), we get 10=𝑎−𝑏+𝑐…𝑒𝑞𝑛1
When we substitute 0,5 , we get 5=𝑐…𝑒𝑞𝑛2
When we substitute (3,13), we get 13=4𝑎+2𝑏+𝑐…𝑒𝑞𝑛3
We put equation (2) into equation (1) and equation (3) to get;
10=𝑎−𝑏+5,⇒𝑎−𝑏=5…𝑒𝑞𝑛4 and 13=4𝑎+2𝑏+5,⇒2𝑎+𝑏=4…𝑒𝑞𝑛5
We add equation (4) and (5) to get,3𝑎=9⇒𝑎=3
We put 𝑎=3 into equation (4) to get,3−𝑏=5⇒𝑏=−2.
We substitute all this value into the above function to get;
𝑦=3 𝑥 2 −2𝑥+5
The correct answer is B.

7. QUESTION 15 The given quadratic equation is − 𝑥 2 +7𝑥=8
We rewrite the standard quadratic form to get;
𝑥 2 −7𝑥+8=0
Where 𝑎=1,𝑏=−7,𝑐=8
The quadratic formula is given by
𝑥= −𝑏± 𝑏 2 −4𝑎𝑐 2𝑎
This implies that;
𝑥= −−7± (−7) 2 −4(1)(8) 2(1)
𝑥= 7±
𝑥= 7 2 ±
The correct answer is B

8. QUESTION 16 The given expression is (−3𝑖)(−4𝑖)
We multiply the two complex numbers to get;
(−3𝑖)(−4𝑖) =12 𝑖 2
Recall that 𝑖 2 =−1
(−3𝑖)(−4𝑖) =12 −1
(−3𝑖)(−4𝑖) =−12
The correct answer is B

9. We make 𝑦 the subject in equation 1 to obtain; 𝑦=5−𝑥 …𝑒𝑞𝑛3
QUESTION 17
⇒2 𝑥 2 −10𝑥=0
𝑥 2 −5𝑥=0
𝑥 𝑥−5 =0
𝑥=0,5
𝑦=5−0=5 when 𝑥=0
𝑦=5−5=0 when 𝑥=5
The solution set is 0,5 ,(5,0)
The correct answer is D
The given system is
𝑥+𝑦=5….eqn1
𝑥 2 + 𝑦 2 =25…𝑒𝑞𝑛2
We make 𝑦 the subject in equation 1 to obtain;
𝑦=5−𝑥 …𝑒𝑞𝑛3
Put equation (3) into equation (2) to obtain;
𝑥 −𝑥 2 =25
𝑥 2 +25−10𝑥+ 𝑥 2 =25

10. QUESTION 18 The given function is 𝑦=(𝑥+3)(𝑥+2)(𝑥−2)
To find the zeros of the given function means we are looking for the x-values where the graph touches the x-axis.
At these x-intercepts 𝑦=0
This implies that;
𝑥+3 𝑥+2 𝑥−2 =0
By the zero product property,
𝑥+3=0 or 𝑥+2=0 or 𝑥−2=0
𝑥=−3 or 𝑥=−2 or 𝑥=2
The correct answer is A

11. QUESTION 19
A cubic function could have only one x-intercept, this means the other roots are complex.
This could also happen when the cubic equation has a root with a multiplicity of 3.
The same way, it could have two intercepts and still be a cubic equation.
A cubic equation could also have at most three x-intercepts, in this case
all the three roots are real.
Also taking into consideration the end behaviors,
The possible correct options are A and D.

12. The rational roots of 𝑥 4 +5 𝑥 3 +7 𝑥 2 −3𝑥−10=0 are −2,1
QUESTION 20
The given equation is
𝑥 4 +5 𝑥 3 +7 𝑥 2 −3𝑥−10=0.
According to the rational roots theorem ,the possible rational roots are ; ±1,±2, ±5,±10
Let 𝑝 𝑥 =𝑥 4 +5 𝑥 3 +7 𝑥 2 −3𝑥−10
According to the remainder theorem, if 𝑝 𝑎 =0, then 𝑥=𝑎 is a root of 𝑝(𝑥).
𝑝 1 =0,𝑝 −1 =−4,𝑝 2 =68,𝑝 −2 =0,𝑝 2 =68,𝑝 3 =260,𝑝 −3 =8,𝑝 −5 =180,𝑝 5 =1400,𝑝 −10 =5720,𝑝 10 =15660
Since 𝑝 1 =0 and 𝑝 −2 =0
The rational roots of 𝑥 4 +5 𝑥 3 +7 𝑥 2 −3𝑥−10=0 are −2,1
The correct answer is A

13. We can rewrite as a quadratic equation in 𝑥 2 to obtain;
QUESTION 21
𝑥 2 = −3± (3) 2 −4(1)(−4) 2(1)
𝑥 2 = −3±
𝑥 2 = −3±5 2
𝑥 2 =1, 𝑥 2 =−4
𝑥=±1,𝑥=±2𝑖
∴𝑥=1,−1,2𝑖,−2𝑖
The correct answer is D
The given equation is
3 𝑥 2 −4=− 𝑥 4
We rearrange to get;
𝑥 4 +3 𝑥 2 −4=0
We can rewrite as a quadratic equation in 𝑥 2 to obtain;
𝑥 𝑥 2 −4=0
Where 𝑎=1,𝑏=3,𝑐=−4
Using the quadratic formula, 𝑥 2 = −𝑏± 𝑏 2 −4𝑎𝑐 2𝑎 , we have

14. QUESTION 22
The binomial expression is 𝑠+3𝑣 5
The binomial theorem is given by
𝑎+𝑏 𝑛 = 𝑟=0 𝑛 𝑛𝐶 𝑟 𝑎 𝑛−𝑟 𝑏 𝑟
By comparing, 𝑎=𝑠,𝑏=3𝑣,𝑛=5
𝑠+3𝑣 5 = 5𝐶 0 𝑠 5−0 3𝑣 𝐶 1 𝑠 5−1 3𝑣 𝐶 2 𝑠 5−2 3𝑣 𝐶 3 𝑠 5−3 3𝑣 𝐶 4 𝑠 5−4 3𝑣 𝐶 5 𝑠 5−5 3𝑣 5
𝑠+3𝑣 5 = 5𝐶 0 𝑠 5 +5 𝐶 1 𝑠 4 (3𝑣)+5 𝐶 2 𝑠 3 (9 𝑣 2 )+5 𝐶 3 𝑠 2 (27 𝑣 3 )+5 𝐶 4 𝑠 1 (81 𝑣 4 )+5 𝐶 5 𝑠 0 (243 𝑣 5 )
𝑠+3𝑣 5 = 𝑠 𝑠 4 𝑣+90 𝑠 3 𝑣 𝑠 2 𝑣 𝑠 𝑣 𝑣 5
The correct answer is B

15. QUESTION 23
The line of best fit has equation 𝑦=𝑚𝑥+𝑐
Where 𝑐= 𝑦 −𝑚 𝑥
𝑐=10− 20 7 ×4=− 10 7
𝑚= ∑𝑥𝑦− ∑𝑥∑𝑦 𝑛 ∑ 𝑥 2 − ∑𝑥 2 𝑛
𝑚= 360− 28× −
𝑚= 20 7
This implies that;
𝑦= 20 7 𝑥− 10 7
𝑦=2.86𝑥−1.43
Therefore the best equation that represents the regression line is
𝑦=3𝑥−2
The correct answer is B. 𝑥 𝑦 𝑥𝑦
𝒙 𝟐 1 4 2 3 5 15 9 10 40 16 80 25 6 19
114 36 7
105 49
∑𝑥=28
∑𝑦=70
∑𝑥𝑦=360
∑ 𝑥 2 =140

16. The initial population is 295. This is the same as 𝑃 0 =295
QUESTION 24
The initial population is This is the same as 𝑃 0 =295
The rate of increment is 7%=0.07=r%
The exponential model is given by the formula;
𝑓 𝑥 = 𝑃 0 (1+𝑟%) 𝑥
This implies that
𝑓 𝑥 = 𝑥
𝑓 𝑥 = 𝑥
After 2 years, the population will be;
𝑓 2 =
𝑓 2 =338
The correct answer is C

17. The half-life formula is 𝑦= 𝑦 0 1 2 𝑥 ℎ
QUESTION 25
The half-life formula is
𝑦= 𝑦 𝑥 ℎ
Where 𝑦 0 =477𝑘𝑔 is the initial amount and ℎ=37𝑑𝑎𝑦𝑠 is the half-life.
We substitute the values into the formula to obtain;
𝑦= 𝑥 37
When x=6, 𝑦= = 𝑘𝑔
The correct answer is B

18. The given logarithmic expression is log 3 2187
QUESTION 26
The given logarithmic expression is
log
We need to express 2187 as an index number to a base of 3.
This will give us
log
We now apply the following property of logarithm.
log 𝑎 𝑎 𝑛 =𝑛 log 𝑎 𝑎 =𝑛
We apply this property to obtain;
log =7 log 3 3 =7
The correct answer is C

19. We cannot express 900 as an index number to a base of 5.
QUESTION 27
The given equation is
5 3𝑥 =900
We cannot express 900 as an index number to a base of 5.
So we take the logarithm of both sides to base 𝑒 to obtain;
ln 5 3𝑥 = ln 900
Recall that ln 𝑎 𝑛 =𝑛 ln 𝑎 . We apply this property to obtain;
3𝑥 ln 5 = ln 900
We divide through by 3ln 5 to get;
𝑥= 3ln ln 5
𝑥= ≈4.23
The correct answer is C

20. We want to solve the equation ln 4 + ln 3𝑥 =2
QUESTION 28
We want to solve the equation
ln 4 + ln 3𝑥 =2
We apply the product rule of logarithm to obtain;
ln 𝐴 + ln 𝐵 = ln 𝐴𝐵
We apply this property to obtain;
ln 4 3𝑥 =2
ln 12𝑥=2
We take the logarithm of both sides to base 𝑒 to get;
𝑒 ln 12𝑥 = 𝑒 2
⇒12𝑥= 𝑒 2
⇒𝑥= 𝑒 2 12
This evaluates to ≈0.62.
The correct answer is D

21. We want to make 𝑚 the subject.
QUESTION 29
The given equation is
𝐿=5𝑚 𝑛 2
We want to make 𝑚 the subject.
We divide both sides by 5 𝑛 2 to obtain;
𝐿 5 𝑛 2 = 5𝑚 𝑛 2 5 𝑛 2
We cancel the common factors to obtain;
𝐿 5 𝑛 2 =𝑚 Or
𝑚= 𝐿 5 𝑛 2
The correct answer is C

22. QUESTION 30 The equation of variation now becomes 𝑥=− 1 4 𝑦
When 𝑦=−164, we obtain, 𝑥=− 1 4 (−164)
This implies that; 𝑥=41
The correct answer is C.
If 𝑥 varies directly as 𝑦.
We write this as
𝑥∝𝑦
When we introduce the constant of variation.
𝑥=𝑘𝑦
When 𝑥=24, 𝑦=−96.
We substitute these values to determine the constant of variation.
24=−96𝑘
We divide both sides by −96 to obtain 24 −96 =𝑘
This implies that 𝑘=− 1 4 .

23. The system of equations is 2𝑥−3𝑦+𝑧=−19…𝑒𝑞𝑛1 5𝑥+𝑦−𝑧=−7…𝑒𝑞𝑛2
QUESTION 31
The system of equations is
2𝑥−3𝑦+𝑧=−19…𝑒𝑞𝑛1
5𝑥+𝑦−𝑧=−7…𝑒𝑞𝑛2
−𝑥+6𝑦−𝑧=35…𝑒𝑞𝑛3
We add equations (1) and (2) to get; 7𝑥−2𝑦=−26…𝑒𝑞𝑛4
We again add equation (1) and (3) to get; 𝑥+3𝑦=16…𝑒𝑞𝑛5
We make 𝑥 the subject in equation (5) to get; 𝑥=16−3𝑦…𝑒𝑞𝑛6
We substitute equation (6) into equation (4) to get;7 16−3𝑦 −2𝑦=−26
This implies that; 112−21𝑦−2𝑦=−26⇒−23𝑦=−138⇒𝑦=6
We put 𝑦=6 into equation (6) to get: 𝑥=16−3 6 =−2.
We put 𝑥=−2,𝑦=6 into equation (2) to get; −−2+6 6 −𝑧=35⇒𝑧=3
The correct answer is B. (−2,6,3)

24. The given expression is 4 𝑥 2 +31𝑥+21 This is a quadratic trinomial.
QUESTION 32
The given expression is
4 𝑥 2 +31𝑥+21
This is a quadratic trinomial.
𝑎=4,𝑏=31,𝑐=21
𝑎𝑐=4×21=4×7×3=28×3
We split the middle term to get;
4 𝑥 2 +28𝑥+3𝑥+21
We factor to get
4𝑥 𝑥+7 +3(𝑥+7)
We factor further to obtain;
𝑥+7 (4𝑥+3)
The correct answer is B

25. We want to solve the equation 5 2𝑥 =15,625
QUESTION 33
We want to solve the equation
5 2𝑥 =15,625
The left hand side of the equation is in the index form so we need to rewrite the right hand side also in the index form to obtain.
5 2𝑥 = 5 6
Since the bases are the same, we equate the exponents to get
2𝑥=6
We divide through by 2, to obtain
𝑥=3
The correct answer is B

26. We make y the subject in equation (2) to get; 𝑦=3.25𝑥+14…𝑒𝑞𝑛3
QUESTION 34
The given system is
2.5𝑥−3𝑦=−13…𝑒𝑞𝑛1
3.25𝑥−𝑦=−14…𝑒𝑞𝑛2
We make y the subject in equation (2) to get; 𝑦=3.25𝑥+14…𝑒𝑞𝑛3
We substitute equation (3) into equation (1) to get; 2.5𝑥−3 3.25𝑥+14 =−13
This implies that;
2.5𝑥−9.75𝑥−42=−13⇒−7.25𝑥=29⇒𝑥=−4
We substitute 𝑥=4 into equation (3) to get 𝑦=3.25(−4)+14=1
Therefore the solution is
𝑥=−4, 𝑦=1

27. The objective function is 𝑃=14𝑥+22𝑦−900 The constraints are;
QUESTION 35
The objective function is
𝑃=14𝑥+22𝑦−900
The constraints are;
𝑦−𝑥≥100,𝑥+2𝑦≤1400
The non-negative constraints are:𝑥≥0,𝑦≥0
a. The feasible region has vertices
𝐴 400,500 ,𝐵 0,100 ,𝐶(0,700)
Vertex
𝑷=𝟏𝟒𝒙+𝟐𝟐𝒚−𝟗𝟎𝟎
A(400,500)
−900=15700
B(0,100)
−900=1300
C(0,700)
−900=14500
b. Producing 400 units of x, and 500 units of y.
The maximum profit is

28. The given quadratic equation is 4 𝑥 2 −9𝑥−9=0
QUESTION 36
The given quadratic equation is 4 𝑥 2 −9𝑥−9=0
We split the middle term to get; 4 𝑥 2 +3𝑥−12𝑥−9=0
We factor to obtain; 𝑥 4𝑥+3 −3 4𝑥+3 =0
This implies that; 4𝑥+3 𝑥−3 =0
4𝑥+3=0 or 𝑥−3=0
𝑥=− 3 4 or 𝑥=3

29. We apply long division to obtain;
QUESTION 37
We apply long division to obtain;
2 𝑥 2 −3𝑥+2
𝑥−3
2 𝑥 3 −9 𝑥 2 +11𝑥−6
− 2 𝑥 3 −6 𝑥 2
−3 𝑥 2 +11𝑥
−(−3 𝑥 2 +9𝑥)
2𝑥−6
−(2𝑥−3)
Therefore when 2 𝑥 3 −9 𝑥 2 +11𝑥−6 is divided by 𝑥−3 the quotient is 2 𝑥 2 −3𝑥+2 and the remainder is 0.