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+ 2 3 with the origin (0,0). This gives 0≤− 5 6 (0)+ 2 3 ⇒0≤ 2 3 . This statement is true. So we shade the lower half plane to obtain the graph in option B. The correct answer is B.

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

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

The function given to us is 𝑦=2 𝑥+2 2 +3 QUESTION 12 The function given to us is 𝑦=2 𝑥+2 2 +3 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.

QUESTION 13 The pumpkins height is given by the equation; 𝑦=12+105𝑥−16 𝑥 2 ⇒𝑦=−16 𝑥 2 +105𝑥+12 ⇒𝑦=−16 (𝑥 2 − 105 16 𝑥)+12 ⇒𝑦=−16 (𝑥 2 − 105 16 𝑥+ − 105 32 2 )−−16 − 105 32 2 +12 ⇒𝑦=−16 𝑥− 105 32 2 )+184.27, ⇒𝑦=−16 𝑥−3.28 2 )+184.27 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−3.28 2 +184.27 ⇒−184.27=−16 𝑥−3.28 2 ⇒− 184.27 −16 = 𝑥−3.28 2 , ⇒11.517= 𝑥−3.28 2 , ⇒ 11.517 =𝑥−3.28,𝑥=3.394+3.28=6.67 The pumpkins maximum height is 184.27 and it hits the ground after 6.67 seconds. The correct answer is C

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.

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± 17 2 𝑥= 7 2 ± 17 2 The correct answer is B

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

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 2 + 5−𝑥 2 =25 𝑥 2 +25−10𝑥+ 𝑥 2 =25

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

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.

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

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± 25 2 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 2 2 +3 𝑥 2 −4=0 Where 𝑎=1,𝑏=3,𝑐=−4 Using the quadratic formula, 𝑥 2 = −𝑏± 𝑏 2 −4𝑎𝑐 2𝑎 , we have

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 0 +5 𝐶 1 𝑠 5−1 3𝑣 1 +5 𝐶 2 𝑠 5−2 3𝑣 2 +5 𝐶 3 𝑠 5−3 3𝑣 3 +5 𝐶 4 𝑠 5−4 3𝑣 4 +5 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 5 +15 𝑠 4 𝑣+90 𝑠 3 𝑣 2 +270 𝑠 2 𝑣 3 +405𝑠 𝑣 4 +243 𝑣 5 The correct answer is B

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− 28×70 7 140− 28 2 7 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

The initial population is 295. This is the same as 𝑃 0 =295 QUESTION 24 The initial population is 295. 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 =295 1+0.07 𝑥 𝑓 𝑥 =295 1.07 𝑥 After 2 years, the population will be; 𝑓 2 =295 1.07 2 𝑓 2 =338 The correct answer is C

The half-life formula is 𝑦= 𝑦 0 1 2 𝑥 ℎ QUESTION 25 The half-life formula is 𝑦= 𝑦 0 1 2 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=477 1 2 𝑥 37 When x=6, 𝑦=477 1 2 6 37 =426.288𝑘𝑔 The correct answer is B

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

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 900 3ln 5 𝑥=4.2265655051188≈4.23 The correct answer is C

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 0.6157546749109≈0.62. The correct answer is D

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

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 .

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)

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

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

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

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) 14 400 +22 500 −900=15700 B(0,100) 14 0 +22 100 −900=1300 C(0,700) 14 0 +22 700 −900=14500 b. Producing 400 units of x, and 500 units of y. The maximum profit is 15700

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

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.