0
Precalculus B
5
QUIZ
, The A+Math Chicken
Precalculus B
Due date N/A
Time taken 3 hours
Grade out of 100.00
QUESTION 1
A farmer plans to grow rice and vegetables on a 100-hectare
farm. Each hectare of rice requires 3 tons of fertilizer and earns
R15,000. Each hectare of vegetables requires 5 tons of fertilizer
and earns R25,000. The farmer has only 360 tons of fertilizer
available.
Question: Determine the vertices/corner points of the feasible
region?
Choose the correct answer:
o a. (0; 72) , (70; 30) , (120; 0) , and (0; 0)
o b. None of the options
o c. (0; 100) , (0; 0) , and (100; 0)
o d. (30; 70) , (100; 0) , and (120; 0)
o e. (72; 0) , (0; 100) , (0; 0) , and (120; 0)
Let: • 𝑥 = Rice 3𝑥 + 5𝑦 ≤ 360
• 𝑦 = Vegetables 𝑥 + 𝑦 ≤ 100
𝑥≥0
𝑦≥0
Objective function: 𝑃 = 15 000𝑥 + 25 000𝑦
Page 1 of 23
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3
− 𝑥 + 72 = −𝑥 + 100
5
3
− 𝑥 + 𝑥 = 100 − 72
5
2
𝑥 = 28
5
2
𝑥 = 28 ÷
5
5
𝑥 = 28 ×
2
𝑥 = 70
𝑦 = −(70) + 100 = 30
∴ The vertices of the feasible region are at (0; 0) , (0; 72) , (70; 30) ,and (100; 0)
Page 2 of 23
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QUESTION 2
Water is flowing downhill at 𝟏𝟎 𝒎/𝒔 through a pipe that is at angle of
𝟑𝟎° with the horizontal. What are the components of its velocity in
vector form?
Choose the correct answer:
o a. 𝒗 = 5𝑖 − 5√3𝑗
o b. 𝒗 = 5(√3𝑖 − 𝑗)
o c. None of the options
o d. 𝒗 = −5√3𝑖 − 𝑗
o e. 𝒗 = 𝑖 + √3𝑗
❑ Vertical component (Vy ):
Because the water flows downhill, it moves downward making the
vertical component negative.
𝑉𝑦 = −𝑎2 ∙ sin 𝜃
∴ 𝒗 = [10cos 30°]𝑖 + [−10sin 30°]𝑗
√3 1
= (10 × ) 𝑖 + (−10 × ) 𝑗
2 2
= 5√3𝑖 − 5𝑗
= 5(√3𝑖 − 𝑗)
QUESTION 3
Consider the following statements:
A.
𝑛(𝑛+1)
1 + 2 + 3 + ⋯+ 𝑛 = ; for all natural
2
numbers 𝑛.
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B.
3 3 3 3 𝑛2 (𝑛+1)
1 + 2 +3 + ⋯+𝑛 = ; for all natural
3
numbers 𝑛.
Which of the statements are true?
Choose the correct option for the statements:
o a. Only A
o b. Only B
o c. Both A and B
o d. Neither A nor B
o e. None of the options
A:
𝑛(𝑛+1)
Let 𝑃(𝑛) denote the statement 1 + 2 + 3 + ⋯ + 𝑛 =
2
1. 𝑷(𝟏)
(1)[(1) + 1]
(1) =
2
1(2)
1=
2
1=1
∴ 𝑃(1) is true.
2. INDUCTION HYPOTHESIS
𝑷(𝒌)
𝑘(𝑘 + 1)
1 + 2 + 3 + ⋯ + (𝑘) =
2
𝑷(𝒌 + 𝟏)
(𝑘 + 1)[(𝑘 + 1) + 1]
1 + 2 + 3 + ⋯ + 𝑘 + (𝑘 + 1) =
2
𝑘(𝑘 + 1) (𝑘 + 1)(𝑘 + 2)
[ ] + (𝑘 + 1) =
2 2
2 (𝑘 2
𝑘 +𝑘 + 1)2 𝑘 + 2𝑘 + 𝑘 + 2
+ =
2 (1)2 2
2 2
𝑘 + 𝑘 + 2𝑘 + 2 𝑘 + 3𝑘 + 2
=
2 2
2 2
𝑘 + 3𝑘 + 2 𝑘 + 3𝑘 + 2
=
2 2
𝐿𝐻𝑆 = 𝑅𝐻𝑆
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∴ 𝑃(𝑘) and 𝑃(𝑘 + 1) is true.
CONCLUSION: Having proven Step 1 and Step 2, we can conclude by the
Principle of Mathematical Induction that 𝑃(𝑛) is true for all natural numbers
𝑛.
B:
𝑛2 (𝑛+1)
Let 𝑃(𝑛) denote the statement 13 + 23 + 33 + ⋯ + 𝑛3 =
3
1. 𝑷(𝟏)
(1)2 [(1) + 1]
(1)3≠
3
1(2)
1≠
3
2
1≠
3
∴ 𝑃(1) is not true.
CONCLUSION:
1 is a natural number, since 𝑃(1) is not true, 𝑃(𝑘) and 𝑃(𝑘 + 1) will
also not be true.
Thus 𝑃(𝑛) is not true.
QUESTION 4
3 2 𝑤 𝑥 1 0
If [ ][ ] = [ ] , then the sum of 𝑥 and 𝑦 is
5 3 𝑦 𝑧 0 1
Choose the correct answer:
o a. −8
o b. 0
o c. 7
o d. 4
o e. None of the options
3 2 𝑤 𝑥 1 0
[ ] × [𝑦 𝑧 ] = [ ]
5 3 0 1
3(𝑤) + 2(𝑦) 3(𝑥) + 2(𝑧) 1 0
[ ]=[ ]
5(𝑤) + 3(𝑦) 5(𝑥) + 3(𝑧) 0 1
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3𝑤 + 2𝑦 = 1 3𝑥 + 2𝑧 = 0
2𝑦 = 1 − 3𝑤 2𝑧 = −3𝑥
1 3 3
𝑦= − 𝑤 𝑧=− 𝑥
2 2 2
5𝑤 + 3𝑦 = 0 5𝑥 + 3𝑧 = 1
1 3 3
5𝑤 + 3 ( − 𝑤) = 0 5𝑥 + 3 (− 𝑥) = 1
2 2 2
3 9 9
5𝑤 + − 𝑤 = 0 5𝑥 − 𝑥 = 1
2 2 2
1 3 1
𝑤=− 𝑥=1
2 2 2
3 1 1
𝑤=− ÷ 𝑥 =1÷
2 2 2
𝑤 = −3 𝑥=2
1 3 3
𝑦= − (−3) = 5 𝑧 = − (2) = −3
2 2 2
∴𝑥+𝑦 = 2+5
=7
QUESTION 5
Simplify the following complex number and leave your answer in polar
form.
2
3𝜋 𝜋
(5 ∠ ) (2 ∠ 10)
5
4𝜋
10 ∠
5
Choose the correct answer:
11𝜋
o a. −4 ∠
6
o b. None of the options
𝜋
o c. 5∠
3
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𝜋
o d. 5∠
2
𝜋
o e. 4∠
6
3𝜋 𝜋
(52 ∠ 2 × ) × (2∠ 10)
= 5
4𝜋
10∠
5
6𝜋 𝜋
(25∠ ) × (2∠ 10)
= 5
4𝜋
10∠
5
6𝜋 𝜋
(25 × 2)∠ ( + )
= 5 10
4𝜋
10∠
5
13𝜋
50∠
= 10
4𝜋
10∠
5
13𝜋 4𝜋
= (50 ÷ 10)∠ ( − )
10 5
𝜋
= 5∠
2
QUESTION 6
An electronics factory makes smart speakers and wireless
headphones. It takes 3 hours of labour and 2 electronic kits to
make each smart speaker. It takes 2 hours of labour and 3
electronic kits to make each wireless headphone. The factory has
48 hours of labour and 42 electronic kits available per day. The
profit is R60 a smart speaker, R50 a wireless headphone.
Question: Determine the maximum profit?
Choose the correct answer:
o a. R1200
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o b. R1260
o c. R1020
o d. R960
o e. None of the options
Let: • 𝑥 = Smart speakers 3𝑥 + 2𝑦 ≤ 48
• 𝑦 = Wireless headphones 2𝑥 + 3𝑦 ≤ 42
𝑥≥0
𝑦≥0
Objective function: 𝑃 = 60𝑥 + 50𝑦
2 3
− 𝑥 + 14 = − 𝑥 + 24
3 2
3 2
𝑥 − 𝑥 = 24 − 14
2 3
5
𝑥 = 10
6
5
𝑥 = 10 ÷
6
6
𝑥 = 10 ×
5
60
𝑥=
5
𝑥 = 12
3
𝑦 = − (12) + 24 = 6
2
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The vertices of the feasible region are @ (0; 0) , (0; 14) , (12; 6) ,and (16; 0)
:
➢ For (12; 6) 𝑃 = 60(12) + 50(6) = R1020
➢ For (16; 0) 𝑃 = 60(16) + 50(0) = R960
➢ For (0; 14) 𝑃 = 60(0) + 50(14) = R700
➢ For (0; 0) 𝑃 = 60(0) + 50(0) = R0
∴The maximum profit is R1020
QUESTION 7
𝟑+𝟑𝒊
Simplify and express the final answer in polar form.
𝟏−𝒊
Choose the correct answer:
𝜋 𝜋
o a. 3 (cos 2 + 𝑖 sin 2 )
3𝜋 3𝜋
o b. 3 (cos + 𝑖 sin )
2 2
o c. None of the options
𝜋 𝜋
o d. 3 (cos 6 + 𝑖 sin 6 )
Page 9 of 23
Precalculus B
5
QUIZ
, The A+Math Chicken
Precalculus B
Due date N/A
Time taken 3 hours
Grade out of 100.00
QUESTION 1
A farmer plans to grow rice and vegetables on a 100-hectare
farm. Each hectare of rice requires 3 tons of fertilizer and earns
R15,000. Each hectare of vegetables requires 5 tons of fertilizer
and earns R25,000. The farmer has only 360 tons of fertilizer
available.
Question: Determine the vertices/corner points of the feasible
region?
Choose the correct answer:
o a. (0; 72) , (70; 30) , (120; 0) , and (0; 0)
o b. None of the options
o c. (0; 100) , (0; 0) , and (100; 0)
o d. (30; 70) , (100; 0) , and (120; 0)
o e. (72; 0) , (0; 100) , (0; 0) , and (120; 0)
Let: • 𝑥 = Rice 3𝑥 + 5𝑦 ≤ 360
• 𝑦 = Vegetables 𝑥 + 𝑦 ≤ 100
𝑥≥0
𝑦≥0
Objective function: 𝑃 = 15 000𝑥 + 25 000𝑦
Page 1 of 23
, The A+Math Chicken
3
− 𝑥 + 72 = −𝑥 + 100
5
3
− 𝑥 + 𝑥 = 100 − 72
5
2
𝑥 = 28
5
2
𝑥 = 28 ÷
5
5
𝑥 = 28 ×
2
𝑥 = 70
𝑦 = −(70) + 100 = 30
∴ The vertices of the feasible region are at (0; 0) , (0; 72) , (70; 30) ,and (100; 0)
Page 2 of 23
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QUESTION 2
Water is flowing downhill at 𝟏𝟎 𝒎/𝒔 through a pipe that is at angle of
𝟑𝟎° with the horizontal. What are the components of its velocity in
vector form?
Choose the correct answer:
o a. 𝒗 = 5𝑖 − 5√3𝑗
o b. 𝒗 = 5(√3𝑖 − 𝑗)
o c. None of the options
o d. 𝒗 = −5√3𝑖 − 𝑗
o e. 𝒗 = 𝑖 + √3𝑗
❑ Vertical component (Vy ):
Because the water flows downhill, it moves downward making the
vertical component negative.
𝑉𝑦 = −𝑎2 ∙ sin 𝜃
∴ 𝒗 = [10cos 30°]𝑖 + [−10sin 30°]𝑗
√3 1
= (10 × ) 𝑖 + (−10 × ) 𝑗
2 2
= 5√3𝑖 − 5𝑗
= 5(√3𝑖 − 𝑗)
QUESTION 3
Consider the following statements:
A.
𝑛(𝑛+1)
1 + 2 + 3 + ⋯+ 𝑛 = ; for all natural
2
numbers 𝑛.
Page 3 of 23
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B.
3 3 3 3 𝑛2 (𝑛+1)
1 + 2 +3 + ⋯+𝑛 = ; for all natural
3
numbers 𝑛.
Which of the statements are true?
Choose the correct option for the statements:
o a. Only A
o b. Only B
o c. Both A and B
o d. Neither A nor B
o e. None of the options
A:
𝑛(𝑛+1)
Let 𝑃(𝑛) denote the statement 1 + 2 + 3 + ⋯ + 𝑛 =
2
1. 𝑷(𝟏)
(1)[(1) + 1]
(1) =
2
1(2)
1=
2
1=1
∴ 𝑃(1) is true.
2. INDUCTION HYPOTHESIS
𝑷(𝒌)
𝑘(𝑘 + 1)
1 + 2 + 3 + ⋯ + (𝑘) =
2
𝑷(𝒌 + 𝟏)
(𝑘 + 1)[(𝑘 + 1) + 1]
1 + 2 + 3 + ⋯ + 𝑘 + (𝑘 + 1) =
2
𝑘(𝑘 + 1) (𝑘 + 1)(𝑘 + 2)
[ ] + (𝑘 + 1) =
2 2
2 (𝑘 2
𝑘 +𝑘 + 1)2 𝑘 + 2𝑘 + 𝑘 + 2
+ =
2 (1)2 2
2 2
𝑘 + 𝑘 + 2𝑘 + 2 𝑘 + 3𝑘 + 2
=
2 2
2 2
𝑘 + 3𝑘 + 2 𝑘 + 3𝑘 + 2
=
2 2
𝐿𝐻𝑆 = 𝑅𝐻𝑆
Page 4 of 23
, The A+Math Chicken
∴ 𝑃(𝑘) and 𝑃(𝑘 + 1) is true.
CONCLUSION: Having proven Step 1 and Step 2, we can conclude by the
Principle of Mathematical Induction that 𝑃(𝑛) is true for all natural numbers
𝑛.
B:
𝑛2 (𝑛+1)
Let 𝑃(𝑛) denote the statement 13 + 23 + 33 + ⋯ + 𝑛3 =
3
1. 𝑷(𝟏)
(1)2 [(1) + 1]
(1)3≠
3
1(2)
1≠
3
2
1≠
3
∴ 𝑃(1) is not true.
CONCLUSION:
1 is a natural number, since 𝑃(1) is not true, 𝑃(𝑘) and 𝑃(𝑘 + 1) will
also not be true.
Thus 𝑃(𝑛) is not true.
QUESTION 4
3 2 𝑤 𝑥 1 0
If [ ][ ] = [ ] , then the sum of 𝑥 and 𝑦 is
5 3 𝑦 𝑧 0 1
Choose the correct answer:
o a. −8
o b. 0
o c. 7
o d. 4
o e. None of the options
3 2 𝑤 𝑥 1 0
[ ] × [𝑦 𝑧 ] = [ ]
5 3 0 1
3(𝑤) + 2(𝑦) 3(𝑥) + 2(𝑧) 1 0
[ ]=[ ]
5(𝑤) + 3(𝑦) 5(𝑥) + 3(𝑧) 0 1
Page 5 of 23
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3𝑤 + 2𝑦 = 1 3𝑥 + 2𝑧 = 0
2𝑦 = 1 − 3𝑤 2𝑧 = −3𝑥
1 3 3
𝑦= − 𝑤 𝑧=− 𝑥
2 2 2
5𝑤 + 3𝑦 = 0 5𝑥 + 3𝑧 = 1
1 3 3
5𝑤 + 3 ( − 𝑤) = 0 5𝑥 + 3 (− 𝑥) = 1
2 2 2
3 9 9
5𝑤 + − 𝑤 = 0 5𝑥 − 𝑥 = 1
2 2 2
1 3 1
𝑤=− 𝑥=1
2 2 2
3 1 1
𝑤=− ÷ 𝑥 =1÷
2 2 2
𝑤 = −3 𝑥=2
1 3 3
𝑦= − (−3) = 5 𝑧 = − (2) = −3
2 2 2
∴𝑥+𝑦 = 2+5
=7
QUESTION 5
Simplify the following complex number and leave your answer in polar
form.
2
3𝜋 𝜋
(5 ∠ ) (2 ∠ 10)
5
4𝜋
10 ∠
5
Choose the correct answer:
11𝜋
o a. −4 ∠
6
o b. None of the options
𝜋
o c. 5∠
3
Page 6 of 23
, The A+Math Chicken
𝜋
o d. 5∠
2
𝜋
o e. 4∠
6
3𝜋 𝜋
(52 ∠ 2 × ) × (2∠ 10)
= 5
4𝜋
10∠
5
6𝜋 𝜋
(25∠ ) × (2∠ 10)
= 5
4𝜋
10∠
5
6𝜋 𝜋
(25 × 2)∠ ( + )
= 5 10
4𝜋
10∠
5
13𝜋
50∠
= 10
4𝜋
10∠
5
13𝜋 4𝜋
= (50 ÷ 10)∠ ( − )
10 5
𝜋
= 5∠
2
QUESTION 6
An electronics factory makes smart speakers and wireless
headphones. It takes 3 hours of labour and 2 electronic kits to
make each smart speaker. It takes 2 hours of labour and 3
electronic kits to make each wireless headphone. The factory has
48 hours of labour and 42 electronic kits available per day. The
profit is R60 a smart speaker, R50 a wireless headphone.
Question: Determine the maximum profit?
Choose the correct answer:
o a. R1200
Page 7 of 23
, The A+Math Chicken
o b. R1260
o c. R1020
o d. R960
o e. None of the options
Let: • 𝑥 = Smart speakers 3𝑥 + 2𝑦 ≤ 48
• 𝑦 = Wireless headphones 2𝑥 + 3𝑦 ≤ 42
𝑥≥0
𝑦≥0
Objective function: 𝑃 = 60𝑥 + 50𝑦
2 3
− 𝑥 + 14 = − 𝑥 + 24
3 2
3 2
𝑥 − 𝑥 = 24 − 14
2 3
5
𝑥 = 10
6
5
𝑥 = 10 ÷
6
6
𝑥 = 10 ×
5
60
𝑥=
5
𝑥 = 12
3
𝑦 = − (12) + 24 = 6
2
Page 8 of 23
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The vertices of the feasible region are @ (0; 0) , (0; 14) , (12; 6) ,and (16; 0)
:
➢ For (12; 6) 𝑃 = 60(12) + 50(6) = R1020
➢ For (16; 0) 𝑃 = 60(16) + 50(0) = R960
➢ For (0; 14) 𝑃 = 60(0) + 50(14) = R700
➢ For (0; 0) 𝑃 = 60(0) + 50(0) = R0
∴The maximum profit is R1020
QUESTION 7
𝟑+𝟑𝒊
Simplify and express the final answer in polar form.
𝟏−𝒊
Choose the correct answer:
𝜋 𝜋
o a. 3 (cos 2 + 𝑖 sin 2 )
3𝜋 3𝜋
o b. 3 (cos + 𝑖 sin )
2 2
o c. None of the options
𝜋 𝜋
o d. 3 (cos 6 + 𝑖 sin 6 )
Page 9 of 23