25 Math Problems for 9th Graders with Answers and Explanations
Welcome to a collection of 25 challenging math problems specifically designed for 9th graders. These problems cover a wide range of mathematical concepts, including algebra, geometry, trigonometry, and more. Each question is labeled with its corresponding math subcategory and difficulty level, ranging from easy to hard. To help you deepen your understanding, we have provided detailed step-by-step explanations for each problem. Let's embark on this mathematical journey and expand your horizons!
Problem 1: Solving Quadratic Equations (Algebra) - Easy
Solve the equation: x² + 5x - 6 = 0.
Solution:
Step 1: Factor the quadratic equation: (x + 6)(x - 1) = 0.
Step 2: Set each factor equal to zero and solve for x:
x + 6 = 0 or x - 1 = 0.
x = -6 or x = 1.
Answer: The solutions to the equation are x = -6 and x = 1.
Problem 2: Area of a Triangle (Geometry) - Easy
Find the area of a triangle with base length 10 cm and height 8 cm.
Solution:
Step 1: Use the formula for the area of a triangle: Area = (1/2) × base × height.
Step 2: Substitute the given values: Area = (1/2) × 10 cm × 8 cm.
Step 3: Calculate: Area = 40 cm².
Answer: The area of the triangle is 40 cm².
Problem 3: Trigonometric Ratios (Trigonometry) - Easy
Given that sinθ = 3/5, find the value of cosθ.
Solution:
Step 1: Use the Pythagorean identity: sin²θ + cos²θ = 1.
Step 2: Substitute the given value: (3/5)² + cos²θ = 1.
Step 3: Simplify: 9/25 + cos²θ = 1.
Step 4: Subtract 9/25 from both sides: cos²θ = 1 - 9/25.
Step 5: Simplify: cos²θ = 16/25.
Step 6: Take the square root of both sides: cosθ = ±4/5.
Answer: The value of cosθ is ±4/5.
Problem 4: Systems of Equations (Algebra) - Easy
Solve the system of equations:
2x + 3y = 8
x - 2y = 4
Solution:
Step 1: Use the method of substitution or elimination to solve the system of equations.
Step 2: Solving by substitution:
From the second equation, x = 2y + 4.
Substitute this value into the first equation: 2(2y + 4) + 3y = 8.
Simplify: 4y + 8 + 3y = 8.
Combine like terms: 7y + 8 = 8.
Subtract 8 from both sides: 7y = 0.
Divide by 7: y = 0.
Substitute y = 0 into the second equation: x - 2(0) = 4.
Simplify: x = 4.
Answer: The solution to the system of equations is x = 4 and y = 0.
Problem 5: Simplifying Radicals (Algebra) - Easy
Simplify √72.
Solution:
Step 1: Break down 72 into its prime factors: 72 = 2² × 3².
Step 2: Rewrite the square root using the prime factors: √72 = √(2² × 3²).
Step 3: Apply the property of square roots: √(a × b) = √a × √b.
√72 = √(2² × 3²) = 2√3.
Answer: √72 simplifies to 2√3.
Problem 6: Volume of a Cylinder (Geometry) - Medium
Find the volume of a cylinder with a radius of 5 cm and a height of 10 cm. (Use π ≈ 3.14)
Solution:
Step 1: Use the formula for the volume of a cylinder: Volume = πr²h.
Step 2: Substitute the given values: Volume = 3.14 × (5 cm)² × 10 cm.
Step 3: Calculate: Volume = 3.14 × 25 cm² × 10 cm.
Step 4: Multiply: Volume = 785 cm³.
Answer: The volume of the cylinder is 785 cm³.
Problem 7: Matrix Operations (Algebra) - Medium
Perform the matrix multiplication:
A = [2 3] B = [1 4]
[5 6] [2 3]
Solution:
Step 1: Multiply the corresponding elements and sum the products for each row and column:
AB = [2(1) + 3(2) 2(4) + 3(3)] = [8 17]
[5(1) + 6(2) 5(4) + 6(3)] [17 32]
Answer: The product of matrices A and B is:
AB = [8 17]
[17 32]
Problem 8: Pythagorean Theorem (Geometry) - Medium
In a right triangle, the length of one leg is 7 cm, and the length of the other leg is 9 cm. What is the length of the hypotenuse?
Solution:
Step 1: Apply the Pythagorean theorem: a² + b² = c².
Step 2: Substitute the given values: 7² + 9² = c².
Step 3: Simplify: 49 + 81 = c².
Step 4: Calculate: 130 = c².
Step 5: Take the square root of both sides: c = √130 ≈ 11.4.
Answer: The length of the hypotenuse is approximately 11.4 cm.
Problem 9: Arithmetic Sequences (Algebra) - Medium
Find the sum of the first 10 terms of an arithmetic sequence where the first term is 2 and the common difference is 3.
Solution:
Step 1: Use the formula for the sum of an arithmetic sequence: Sn = (n/2)(2a + (n-1)d).
Step 2: Substitute the given values: Sn = (10/2)(2(2) + (10-1)(3)).
Step 3: Simplify: Sn = (5)(4 + 9(3)).
Step 4: Calculate: Sn = 5(4 + 27).
Step 5: Add: Sn = 5(31).
Step 6: Multiply: Sn = 155.
Answer: The sum of the first 10 terms of the arithmetic sequence is 155.
Problem 10: Rationalizing Denominators (Algebra) - Medium
Simplify (3/√5) * (√5/√5).
Solution:
Step 1: Multiply the numerators: (3)(√5) = 3√5.
Step 2: Multiply the denominators: (√5)(√5) = 5.
Step 3: Simplify: 3√5/5.
Answer: The simplified expression is 3√5/5.
Problem 11: Exponent Rules (Algebra) - Hard
Simplify (x⁴y²) / (x²y³).
Solution:
Step 1: Apply the division rule for exponents: x⁴/x² = x^(4-2) = x².
Step 2: Apply the division rule for exponents: y²/y³ = y^(2-3) = y⁻¹ = 1/y.
Step 3: Combine the simplified terms: (x² * 1)/(1 * y) = x²/y.
Answer: The simplified expression is x²/y.
Problem 12: Laws of Logarithms (Algebra) - Hard
Simplify log₅(x⁵) + log₅(y²) - log₅(z).
Solution:
Step 1: Apply the product rule for logarithms: log₅(x⁵) + log₅(y²) = log₅(x⁵ * y²).
Step 2: Apply the quotient rule for logarithms: log₅(x⁵ * y²) - log₅(z) = log₅((x⁵ * y²)/z).
Answer: The simplified expression is log₅((x⁵ * y²)/z).
Problem 13: Trigonometric Identities (Trigonometry) - Hard
Prove the trigonometric identity: (1 + sin²θ)/(1 - cos²θ) = csc²θ.
Solution:
Step 1: Start with the left side of the equation: (1 + sin²θ)/(1 - cos²θ).
Step 2: Apply the Pythagorean identity: sin²θ + cos²θ = 1.
Step 3: Substitute the identity into the left side of the equation: (1 + sin²θ)/(1 - (1 - sin²θ)).
Step 4: Simplify: (1 + sin²θ)/(sin²θ).
Step 5: Take the reciprocal: 1/(sin²θ) = csc²θ.
Answer: The left side of the equation is equal to csc²θ.
Problem 14: Surface Area of a Sphere (Geometry) - Hard
Find the surface area of a sphere with a radius of 6 cm. (Use π ≈ 3.14)
Solution:
Step 1: Use the formula for the surface area of a sphere: Surface Area = 4πr².
Step 2: Substitute the given value: Surface Area = 4(3.14)(6 cm)².
Step 3: Calculate: Surface Area = 4(3.14)(36 cm²).
Step 4: Multiply: Surface Area = 452.16 cm².
Answer: The surface area of the sphere is 452.16 cm².
Problem 15: Logarithmic Equations (Algebra) - Hard
Solve the equation: log₃(x - 1) + log₃(x + 2) = 2.
Solution:
Step 1: Combine the logarithms using the product rule: log₃((x - 1)(x + 2)) = 2.
Step 2: Rewrite the equation in exponential form: ₃² = (x - 1)(x + 2).
Step 3: Simplify: 9 = (x² + x - 2).
Step 4: Rearrange the equation: x² + x - 11 = 0.
Step 5: Factor or use the quadratic formula to solve for x: (x - 3)(x + 4) = 0.
x - 3 = 0 or x + 4 = 0.
x = 3 or x = -4.
Answer: The solutions to the equation are x = 3 and x = -4.
Problem 16: Probability of Dependent Events (Statistics) - Hard
In a deck of cards, there are 26 red cards and 26 black cards. If two cards are drawn without replacement, what is the probability of drawing a red card followed by another red card?
Solution:
Step 1: Determine the total number of cards in the deck: 26 red cards + 26 black cards = 52 cards.
Step 2: Calculate the probability of drawing a red card first: 26/52 = 1/2.
Step 3: After the first card is drawn, there are 51 cards left in the deck, with 25 red cards remaining.
Step 4: Calculate the probability of drawing another red card: 25/51.
Step 5: Multiply the probabilities: (1/2) * (25/51) = 25/102.
Answer: The probability of drawing a red card followed by another red card is 25/102.
Problem 17: Volume of a Cone (Geometry) - Hard
Find the volume of a cone with a radius of 8 cm and a height of 12 cm. (Use π ≈ 3.14)
Solution:
Step 1: Use the formula for the volume of a cone: Volume = (1/3)πr²h.
Step 2: Substitute the given values: Volume = (1/3)(3.14)(8 cm)²(12 cm).
Step 3: Calculate: Volume = (1/3)(3.14)(64 cm²)(12 cm).
Step 4: Multiply: Volume = 803.84 cm³.
Answer: The volume of the cone is 803.84 cm³.
Problem 18: Arithmetic and Geometric Sequences (Algebra) - Hard
Given an arithmetic sequence with a first term of 3 and a common difference of 2, and a geometric sequence with a first term of 2 and a common ratio of 3, find the value of the expression (5th term of arithmetic sequence) × (4th term of geometric sequence).
Solution:
Step 1: Determine the 5th term of the arithmetic sequence: a₅ = a₁ + (n-1)d = 3 + (5-1)(2) = 11.
Step 2: Determine the 4th term of the geometric sequence: a₄ = a₁ * r^(n-1) = 2 * 3^(4-1) = 2 * 27 = 54.
Step 3: Calculate the expression: (5th term of arithmetic sequence) × (4th term of geometric sequence) = 11 × 54 = 594.
Answer: The value of the expression is 594.
Problem 19: Matrix Inverses (Algebra) - Hard
Given matrix A:
[3 4]
[2 5]
Find the inverse of matrix A, if it exists.
Solution:
Step 1: Calculate the determinant of matrix A: det(A) = (3)(5) - (4)(2) = 15 - 8 = 7.
Step 2: Check if the determinant is nonzero (i.e., det(A) ≠ 0).
Since det(A) = 7 ≠ 0, the inverse of matrix A exists.
Step 3: Find the inverse of matrix A using the formula: A⁻¹ = (1/det(A)) * adj(A).
adj(A) = [5 -4]
[-2 3]
A⁻¹ = (1/7) * [5 -4]
[-2 3]
Answer: The inverse of matrix A is:
[5/7 -4/7]
[-2/7 3/7]
Problem 20: Trigonometric Equations (Trigonometry) - Hard
Solve the equation: sin(2θ) = cos(θ).
Solution:
Step 1: Apply the double-angle formula for sine: 2sinθcosθ = cosθ.
Step 2: Rearrange the equation: 2sinθcosθ - cosθ = 0.
Step 3: Factor out cosθ: cosθ(2sinθ - 1) = 0.
Step 4: Set each factor equal to zero and solve for θ:
cosθ = 0 or 2sinθ - 1 = 0.
θ = π/2 + kπ or θ = π/6 + kπ, where k is an integer.
Answer: The solutions to the equation are θ = π/2 + kπ or θ = π/6 + kπ, where k is an integer.
Problem 21: Sequences and Series (Algebra) - Hard
Find the sum of the infinite geometric series: 2 + 1 + 1/2 + 1/4 + ...
Solution:
Step 1: Identify the common ratio: r = 1/2.
Step 2: Use the formula for the sum of an infinite geometric series: S = a / (1 - r), where a is the first term.
Step 3: Substitute the given value: S = 2 / (1 - 1/2) = 2 / (1/2) = 4.
Answer: The sum of the infinite geometric series is 4.
Problem 22: Logarithmic Properties (Algebra) - Hard
Simplify log₄(x³) - log₄(y²) + 2log₄(z).
Solution:
Step 1: Apply the quotient and power rules for logarithms:
log₄(x³) - log₄(y²) + 2log₄(z) = log₄(x³/y²) + log₄(z²).
Step 2: Combine the logarithms using the product rule:
log₄(x³/y²) + log₄(z²) = log₄((x³/y²)z²).
Step 3: Simplify the expression inside the logarithm: (x³/y²)z².
Step 4: Rewrite the expression using exponents: (x³z²) / y².
Answer: The simplified expression is log₄((x³z²) / y²).
Problem 23: Trigonometric Equations with Multiple Angles (Trigonometry) - Hard
Solve the equation: cos(2θ) + sin(θ) = 0.
Solution:
Step 1: Apply the double-angle formula for cosine: 1 - 2sin²θ + sinθ = 0.
Step 2: Rearrange the equation: -2sin²θ + sinθ + 1 = 0.
Step 3: Factor the quadratic equation: (sinθ - 1)(2sinθ + 1) = 0.
Step 4: Set each factor equal to zero and solve for θ:
sinθ - 1 = 0 or 2sinθ + 1 = 0.
θ = π/2 + 2kπ or θ = 7π/6 + 2kπ, where k is an integer.
Answer: The solutions to the equation are θ = π/2 + 2kπ or θ = 7π/6 + 2kπ, where k is an integer.
Problem 24: Vectors (Geometry) - Hard
Given vectors u = 3i + 2j and v = -i + 4j, find the magnitude of the vector u - v.
Solution:
Step 1: Subtract the components of the vectors: u - v = (3 - (-1))i + (2 - 4)j = 4i - 2j.
Step 2: Calculate the magnitude using the Pythagorean theorem: |u - v| = √((4)² + (-2)²) = √(16 + 4) = √20 = 2√5.
Answer: The magnitude of the vector u - v is 2√5.
Problem 25: Complex Numbers (Algebra) - Hard
Find the square root of -16.
Solution:
Step 1: Rewrite -16 as 16i².
Step 2: Take the square root: √(16i²) = 4i.
Answer: The square root of -16 is 4i.
Congratulations on completing the 25 challenging math problems for 9th graders! These problems have tested your understanding of various mathematical concepts, allowing you to expand your skills and knowledge. By following the step-by-step solutions, you have gained deeper insights into algebra, geometry, trigonometry, and more. Keep practicing and exploring new mathematical concepts to further enhance your abilities. Remember, math is a journey of continuous learning and growth. Well done!
Problem 1: Solving Quadratic Equations (Algebra) - Easy
Solve the equation: x² + 5x - 6 = 0.
Solution:
Step 1: Factor the quadratic equation: (x + 6)(x - 1) = 0.
Step 2: Set each factor equal to zero and solve for x:
x + 6 = 0 or x - 1 = 0.
x = -6 or x = 1.
Answer: The solutions to the equation are x = -6 and x = 1.
Problem 2: Area of a Triangle (Geometry) - Easy
Find the area of a triangle with base length 10 cm and height 8 cm.
Solution:
Step 1: Use the formula for the area of a triangle: Area = (1/2) × base × height.
Step 2: Substitute the given values: Area = (1/2) × 10 cm × 8 cm.
Step 3: Calculate: Area = 40 cm².
Answer: The area of the triangle is 40 cm².
Problem 3: Trigonometric Ratios (Trigonometry) - Easy
Given that sinθ = 3/5, find the value of cosθ.
Solution:
Step 1: Use the Pythagorean identity: sin²θ + cos²θ = 1.
Step 2: Substitute the given value: (3/5)² + cos²θ = 1.
Step 3: Simplify: 9/25 + cos²θ = 1.
Step 4: Subtract 9/25 from both sides: cos²θ = 1 - 9/25.
Step 5: Simplify: cos²θ = 16/25.
Step 6: Take the square root of both sides: cosθ = ±4/5.
Answer: The value of cosθ is ±4/5.
Problem 4: Systems of Equations (Algebra) - Easy
Solve the system of equations:
2x + 3y = 8
x - 2y = 4
Solution:
Step 1: Use the method of substitution or elimination to solve the system of equations.
Step 2: Solving by substitution:
From the second equation, x = 2y + 4.
Substitute this value into the first equation: 2(2y + 4) + 3y = 8.
Simplify: 4y + 8 + 3y = 8.
Combine like terms: 7y + 8 = 8.
Subtract 8 from both sides: 7y = 0.
Divide by 7: y = 0.
Substitute y = 0 into the second equation: x - 2(0) = 4.
Simplify: x = 4.
Answer: The solution to the system of equations is x = 4 and y = 0.
Problem 5: Simplifying Radicals (Algebra) - Easy
Simplify √72.
Solution:
Step 1: Break down 72 into its prime factors: 72 = 2² × 3².
Step 2: Rewrite the square root using the prime factors: √72 = √(2² × 3²).
Step 3: Apply the property of square roots: √(a × b) = √a × √b.
√72 = √(2² × 3²) = 2√3.
Answer: √72 simplifies to 2√3.
Problem 6: Volume of a Cylinder (Geometry) - Medium
Find the volume of a cylinder with a radius of 5 cm and a height of 10 cm. (Use π ≈ 3.14)
Solution:
Step 1: Use the formula for the volume of a cylinder: Volume = πr²h.
Step 2: Substitute the given values: Volume = 3.14 × (5 cm)² × 10 cm.
Step 3: Calculate: Volume = 3.14 × 25 cm² × 10 cm.
Step 4: Multiply: Volume = 785 cm³.
Answer: The volume of the cylinder is 785 cm³.
Problem 7: Matrix Operations (Algebra) - Medium
Perform the matrix multiplication:
A = [2 3] B = [1 4]
[5 6] [2 3]
Solution:
Step 1: Multiply the corresponding elements and sum the products for each row and column:
AB = [2(1) + 3(2) 2(4) + 3(3)] = [8 17]
[5(1) + 6(2) 5(4) + 6(3)] [17 32]
Answer: The product of matrices A and B is:
AB = [8 17]
[17 32]
Problem 8: Pythagorean Theorem (Geometry) - Medium
In a right triangle, the length of one leg is 7 cm, and the length of the other leg is 9 cm. What is the length of the hypotenuse?
Solution:
Step 1: Apply the Pythagorean theorem: a² + b² = c².
Step 2: Substitute the given values: 7² + 9² = c².
Step 3: Simplify: 49 + 81 = c².
Step 4: Calculate: 130 = c².
Step 5: Take the square root of both sides: c = √130 ≈ 11.4.
Answer: The length of the hypotenuse is approximately 11.4 cm.
Problem 9: Arithmetic Sequences (Algebra) - Medium
Find the sum of the first 10 terms of an arithmetic sequence where the first term is 2 and the common difference is 3.
Solution:
Step 1: Use the formula for the sum of an arithmetic sequence: Sn = (n/2)(2a + (n-1)d).
Step 2: Substitute the given values: Sn = (10/2)(2(2) + (10-1)(3)).
Step 3: Simplify: Sn = (5)(4 + 9(3)).
Step 4: Calculate: Sn = 5(4 + 27).
Step 5: Add: Sn = 5(31).
Step 6: Multiply: Sn = 155.
Answer: The sum of the first 10 terms of the arithmetic sequence is 155.
Problem 10: Rationalizing Denominators (Algebra) - Medium
Simplify (3/√5) * (√5/√5).
Solution:
Step 1: Multiply the numerators: (3)(√5) = 3√5.
Step 2: Multiply the denominators: (√5)(√5) = 5.
Step 3: Simplify: 3√5/5.
Answer: The simplified expression is 3√5/5.
Problem 11: Exponent Rules (Algebra) - Hard
Simplify (x⁴y²) / (x²y³).
Solution:
Step 1: Apply the division rule for exponents: x⁴/x² = x^(4-2) = x².
Step 2: Apply the division rule for exponents: y²/y³ = y^(2-3) = y⁻¹ = 1/y.
Step 3: Combine the simplified terms: (x² * 1)/(1 * y) = x²/y.
Answer: The simplified expression is x²/y.
Problem 12: Laws of Logarithms (Algebra) - Hard
Simplify log₅(x⁵) + log₅(y²) - log₅(z).
Solution:
Step 1: Apply the product rule for logarithms: log₅(x⁵) + log₅(y²) = log₅(x⁵ * y²).
Step 2: Apply the quotient rule for logarithms: log₅(x⁵ * y²) - log₅(z) = log₅((x⁵ * y²)/z).
Answer: The simplified expression is log₅((x⁵ * y²)/z).
Problem 13: Trigonometric Identities (Trigonometry) - Hard
Prove the trigonometric identity: (1 + sin²θ)/(1 - cos²θ) = csc²θ.
Solution:
Step 1: Start with the left side of the equation: (1 + sin²θ)/(1 - cos²θ).
Step 2: Apply the Pythagorean identity: sin²θ + cos²θ = 1.
Step 3: Substitute the identity into the left side of the equation: (1 + sin²θ)/(1 - (1 - sin²θ)).
Step 4: Simplify: (1 + sin²θ)/(sin²θ).
Step 5: Take the reciprocal: 1/(sin²θ) = csc²θ.
Answer: The left side of the equation is equal to csc²θ.
Problem 14: Surface Area of a Sphere (Geometry) - Hard
Find the surface area of a sphere with a radius of 6 cm. (Use π ≈ 3.14)
Solution:
Step 1: Use the formula for the surface area of a sphere: Surface Area = 4πr².
Step 2: Substitute the given value: Surface Area = 4(3.14)(6 cm)².
Step 3: Calculate: Surface Area = 4(3.14)(36 cm²).
Step 4: Multiply: Surface Area = 452.16 cm².
Answer: The surface area of the sphere is 452.16 cm².
Problem 15: Logarithmic Equations (Algebra) - Hard
Solve the equation: log₃(x - 1) + log₃(x + 2) = 2.
Solution:
Step 1: Combine the logarithms using the product rule: log₃((x - 1)(x + 2)) = 2.
Step 2: Rewrite the equation in exponential form: ₃² = (x - 1)(x + 2).
Step 3: Simplify: 9 = (x² + x - 2).
Step 4: Rearrange the equation: x² + x - 11 = 0.
Step 5: Factor or use the quadratic formula to solve for x: (x - 3)(x + 4) = 0.
x - 3 = 0 or x + 4 = 0.
x = 3 or x = -4.
Answer: The solutions to the equation are x = 3 and x = -4.
Problem 16: Probability of Dependent Events (Statistics) - Hard
In a deck of cards, there are 26 red cards and 26 black cards. If two cards are drawn without replacement, what is the probability of drawing a red card followed by another red card?
Solution:
Step 1: Determine the total number of cards in the deck: 26 red cards + 26 black cards = 52 cards.
Step 2: Calculate the probability of drawing a red card first: 26/52 = 1/2.
Step 3: After the first card is drawn, there are 51 cards left in the deck, with 25 red cards remaining.
Step 4: Calculate the probability of drawing another red card: 25/51.
Step 5: Multiply the probabilities: (1/2) * (25/51) = 25/102.
Answer: The probability of drawing a red card followed by another red card is 25/102.
Problem 17: Volume of a Cone (Geometry) - Hard
Find the volume of a cone with a radius of 8 cm and a height of 12 cm. (Use π ≈ 3.14)
Solution:
Step 1: Use the formula for the volume of a cone: Volume = (1/3)πr²h.
Step 2: Substitute the given values: Volume = (1/3)(3.14)(8 cm)²(12 cm).
Step 3: Calculate: Volume = (1/3)(3.14)(64 cm²)(12 cm).
Step 4: Multiply: Volume = 803.84 cm³.
Answer: The volume of the cone is 803.84 cm³.
Problem 18: Arithmetic and Geometric Sequences (Algebra) - Hard
Given an arithmetic sequence with a first term of 3 and a common difference of 2, and a geometric sequence with a first term of 2 and a common ratio of 3, find the value of the expression (5th term of arithmetic sequence) × (4th term of geometric sequence).
Solution:
Step 1: Determine the 5th term of the arithmetic sequence: a₅ = a₁ + (n-1)d = 3 + (5-1)(2) = 11.
Step 2: Determine the 4th term of the geometric sequence: a₄ = a₁ * r^(n-1) = 2 * 3^(4-1) = 2 * 27 = 54.
Step 3: Calculate the expression: (5th term of arithmetic sequence) × (4th term of geometric sequence) = 11 × 54 = 594.
Answer: The value of the expression is 594.
Problem 19: Matrix Inverses (Algebra) - Hard
Given matrix A:
[3 4]
[2 5]
Find the inverse of matrix A, if it exists.
Solution:
Step 1: Calculate the determinant of matrix A: det(A) = (3)(5) - (4)(2) = 15 - 8 = 7.
Step 2: Check if the determinant is nonzero (i.e., det(A) ≠ 0).
Since det(A) = 7 ≠ 0, the inverse of matrix A exists.
Step 3: Find the inverse of matrix A using the formula: A⁻¹ = (1/det(A)) * adj(A).
adj(A) = [5 -4]
[-2 3]
A⁻¹ = (1/7) * [5 -4]
[-2 3]
Answer: The inverse of matrix A is:
[5/7 -4/7]
[-2/7 3/7]
Problem 20: Trigonometric Equations (Trigonometry) - Hard
Solve the equation: sin(2θ) = cos(θ).
Solution:
Step 1: Apply the double-angle formula for sine: 2sinθcosθ = cosθ.
Step 2: Rearrange the equation: 2sinθcosθ - cosθ = 0.
Step 3: Factor out cosθ: cosθ(2sinθ - 1) = 0.
Step 4: Set each factor equal to zero and solve for θ:
cosθ = 0 or 2sinθ - 1 = 0.
θ = π/2 + kπ or θ = π/6 + kπ, where k is an integer.
Answer: The solutions to the equation are θ = π/2 + kπ or θ = π/6 + kπ, where k is an integer.
Problem 21: Sequences and Series (Algebra) - Hard
Find the sum of the infinite geometric series: 2 + 1 + 1/2 + 1/4 + ...
Solution:
Step 1: Identify the common ratio: r = 1/2.
Step 2: Use the formula for the sum of an infinite geometric series: S = a / (1 - r), where a is the first term.
Step 3: Substitute the given value: S = 2 / (1 - 1/2) = 2 / (1/2) = 4.
Answer: The sum of the infinite geometric series is 4.
Problem 22: Logarithmic Properties (Algebra) - Hard
Simplify log₄(x³) - log₄(y²) + 2log₄(z).
Solution:
Step 1: Apply the quotient and power rules for logarithms:
log₄(x³) - log₄(y²) + 2log₄(z) = log₄(x³/y²) + log₄(z²).
Step 2: Combine the logarithms using the product rule:
log₄(x³/y²) + log₄(z²) = log₄((x³/y²)z²).
Step 3: Simplify the expression inside the logarithm: (x³/y²)z².
Step 4: Rewrite the expression using exponents: (x³z²) / y².
Answer: The simplified expression is log₄((x³z²) / y²).
Problem 23: Trigonometric Equations with Multiple Angles (Trigonometry) - Hard
Solve the equation: cos(2θ) + sin(θ) = 0.
Solution:
Step 1: Apply the double-angle formula for cosine: 1 - 2sin²θ + sinθ = 0.
Step 2: Rearrange the equation: -2sin²θ + sinθ + 1 = 0.
Step 3: Factor the quadratic equation: (sinθ - 1)(2sinθ + 1) = 0.
Step 4: Set each factor equal to zero and solve for θ:
sinθ - 1 = 0 or 2sinθ + 1 = 0.
θ = π/2 + 2kπ or θ = 7π/6 + 2kπ, where k is an integer.
Answer: The solutions to the equation are θ = π/2 + 2kπ or θ = 7π/6 + 2kπ, where k is an integer.
Problem 24: Vectors (Geometry) - Hard
Given vectors u = 3i + 2j and v = -i + 4j, find the magnitude of the vector u - v.
Solution:
Step 1: Subtract the components of the vectors: u - v = (3 - (-1))i + (2 - 4)j = 4i - 2j.
Step 2: Calculate the magnitude using the Pythagorean theorem: |u - v| = √((4)² + (-2)²) = √(16 + 4) = √20 = 2√5.
Answer: The magnitude of the vector u - v is 2√5.
Problem 25: Complex Numbers (Algebra) - Hard
Find the square root of -16.
Solution:
Step 1: Rewrite -16 as 16i².
Step 2: Take the square root: √(16i²) = 4i.
Answer: The square root of -16 is 4i.
Congratulations on completing the 25 challenging math problems for 9th graders! These problems have tested your understanding of various mathematical concepts, allowing you to expand your skills and knowledge. By following the step-by-step solutions, you have gained deeper insights into algebra, geometry, trigonometry, and more. Keep practicing and exploring new mathematical concepts to further enhance your abilities. Remember, math is a journey of continuous learning and growth. Well done!
