For the complex number z = 3 − 4i, what is the modulus |z|?
- 7
- √7
- 5 Correct answer.
- 25
Explanation
The modulus of a complex number a + bi is √(a² + b²). Here a = 3 and b = −4, so |z| = √(9 + 16) = √25 = 5.
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The modulus of a complex number a + bi is √(a² + b²). Here a = 3 and b = −4, so |z| = √(9 + 16) = √25 = 5.
The word NEPAL has 5 distinct letters, so the number of arrangements is 5! = 5 × 4 × 3 × 2 × 1. This equals 120.
For AB to be defined, the number of columns of A must equal the number of rows of B, which holds here. The product then has the number of rows of A and the number of columns of B, so the order is 2 × 4.
The general term is C(8,r) x^(8−r) 2^r. For x^5 we need 8 − r = 5, so r = 3. The coefficient is C(8,3)·2^3 = 56·8 = 448.
The point (−1, 1) lies in the second quadrant, so its principal argument is between π/2 and π. The reference angle is π/4, giving arg z = π − π/4 = 3π/4.
Choose 2 women from 4 in C(4,2) = 6 ways, and 1 man from 5 in C(5,1) = 5 ways. By the multiplication principle, the total is 6 × 5 = 30.
Multiplying a 3 × 3 matrix by 2 multiplies each of its three rows by 2, so the determinant is multiplied by 2^3 = 8. Therefore det(2A) = 8 · det(A) = 8 · 5 = 40.
Evaluate Z at every corner: (0,0) gives 0, (4,0) gives 20, (2,3) gives 10 + 12 = 22, and (0,2) gives 8. The largest value is 22.
For infinitely many solutions the two equations must be dependent, so the coefficient determinant must vanish: 2k − 12 = 0, giving k = 6. With k = 6 the second equation is exactly twice the first, so the system is consistent.
The general term is C(10,r) x^(10−r)(1/x)^r = C(10,r) x^(10−2r). For the term independent of x, set 10 − 2r = 0, so r = 5. Its coefficient is C(10,5) = 252.
For any complex number z, the product z times its conjugate equals the square of its modulus, so z z̄ = 2^2 + 3^2 = 4 + 9 = 13. The imaginary parts cancel when the conjugate is multiplied.
The hundreds digit can be chosen in 4 ways, the tens digit in 3 remaining ways, and the units digit in 2 remaining ways. Thus the number of 3-digit numbers is 4 × 3 × 2 = 24.
The trace of a square matrix is the sum of the entries on its main diagonal. For this matrix the diagonal entries are 1 and 4, so trace(A) = 1 + 4 = 5.
For (x + 2/x)^6 there are seven terms, so the middle term is the fourth term. Using the general term, T4 = C(6,3) x^3 (2/x)^3 = 20 × 8 = 160.
Multiply numerator and denominator by the conjugate 1 + i. The denominator becomes (1 - i)(1 + i) = 2, while the numerator becomes (1 + i)^2 = 2i, so the quotient is i.
An octagon has 8 vertices, and joining any two vertices gives C(8,2) = 28 segments. The 8 sides are not diagonals, so the number of diagonals is 28 - 8 = 20.
For an n × n matrix A, the determinant of its adjoint satisfies |adj A| = |A|^(n-1). Here n = 2 and |A| = 3, so |adj A| = 3^(2-1) = 3.
The feasible region has corner points (0,0), (4,0), and (0,4). At these points Z = 3x + 2y equals 0, 12, and 8 respectively, so the maximum value is 12.
Since A^2 = A, also A^3 = A. Expanding gives (I + A)^3 = I + 3A + 3A^2 + A^3 = I + 7A, so subtracting 7A leaves I.
For a non-real cube root of unity, 1 + ω + ω^2 = 0, so 1 + ω = -ω^2. Then 1 + ω - ω^2 = -2ω^2, and cubing gives (-2ω^2)^3 = -8ω^6 = -8.
Since z² = (1 + 2i)² = 1 + 4i + 4i², and i² = −1, this simplifies to −3 + 4i. The real part is therefore −3; the imaginary part is 4.
A 2×2 matrix is singular exactly when its determinant is zero. Here det M = x·x − 2·8 = x² − 16, so x² = 16 and therefore x = ±4.
Treat the two particular books as one block, giving 3! orders of the block and the other two books. The two books inside the block can be arranged in 2! ways, so the total is 3! × 2! = 12.
Putting x = 1 in (1 + x)^n gives (1 + 1)^n = 2^n, which is the sum of all binomial coefficients. This substitution adds every coefficient exactly once.
Write z in polar form: modulus 2 and argument 60°. Cubing gives modulus 2³ = 8 and argument 180°, so z³ = 8(cos 180° + i sin 180°) = −8.
For any square matrix, A(adj A) = |A|I. Here |A| = −2, so A adj A = −2I, and the determinant of a 3×3 scalar multiple is (−2)³ = −8.
The corner points are (0,0), (2,0), (2,4), and (0,6). Evaluating Z = 2x + 3y gives 0, 4, 16, and 18 respectively, so the maximum value is 18.
The coefficient of x⁴ comes from terms with j + 2k = 4 in i + j + k = 5. The possibilities (j,k) = (4,0), (2,1), and (0,2) give multinomial counts 5, 30, and 10; their sum is 45.
If the last digit is 0, the first three places can be filled in 5P3 = 60 ways. If the last digit is 2 or 4, the first digit has 4 choices, then 4 and 3 choices follow, giving 48 per last digit; total 60 + 2(48) = 156.
Total committees of 5 from 10 people: C(10,5) = 252. The only ones with no girl are the all-boy committees, C(6,5) = 6, so the number containing at least one girl is 252 − 6 = 246.
By Vieta's formulas, α + β = 5 and αβ = 6. The identity α² + β² = (α + β)² − 2αβ then gives 25 − 12 = 13, which also matches the direct computation 2² + 3² using the roots 2 and 3.
For n distinct beads on a necklace, fixing one bead removes rotational symmetry and leaves (n − 1)! orders; dividing by 2 then identifies each arrangement with its mirror image, since a necklace can be flipped over. Thus the count is 5!/2 = 120/2 = 60.
A skew-symmetric matrix satisfies Aᵀ = −A, so det(A) = det(Aᵀ) = det(−A) = (−1)³ det(A) = −det(A). Therefore 2 det(A) = 0, forcing det(A) = 0 for every skew-symmetric matrix of odd order.
The general term is C(7, r)(2x)^(7−r)(−x⁻²)^r = C(7, r)·2^(7−r)·(−1)^r·x^(7−3r). Setting 7 − 3r = −2 gives r = 3, so the coefficient is C(7, 3)·2⁴·(−1)³ = 35·16·(−1) = −560.
To convert degrees to radians, multiply by π/180. Thus 150° = 150π/180 = 5π/6. Equivalently, 150° is 30° short of 180°, so the radian measure is π − π/6 = 5π/6.
For a function cos(kx), the period is 2π/|k| because the argument kx must increase by 2π. Here k = 3, so the period is 2π/3. The coefficient 3 compresses the usual 2π period by a factor of three.
The cosine function equals zero at the odd multiples of π/2, namely π/2, 3π/2, 5π/2, and their negatives. These are exactly the values generated by (2n+1)π/2 for integer n. The expression nπ gives the zeros of sine, not cosine.
The principal value of sin⁻¹ lies in [−π/2, π/2]. Since sin(−π/6) = −1/2 and −π/6 lies in that interval, the principal value is −π/6. Angles such as 7π/6 and 5π/6 lie outside the principal range.
Rearrange to sin²θ = 1/2, so sinθ = ±1/√2. The solutions combine to θ = nπ ± π/4, where the ± sign accounts for both positive and negative values of sinθ. The term nπ supplies the period π of the squared equation.
Use the double-angle identities 1 − cos 2θ = 2 sin²θ and sin 2θ = 2 sinθ cosθ. Substituting gives (2 sin²θ)/(2 sinθ cosθ) = sinθ/cosθ = tanθ. The quotient is therefore the tangent of θ, not its reciprocal.
For constants a and b, the maximum of a sinθ + b cosθ is √(a² + b²). Here a = 3 and b = 4, so the maximum is √(9 + 16) = 5. This value is attained when the sine and cosine terms align with the coefficient vector.
By the cosine rule, cos A = (b² + c² − a²)/(2bc), where a is the side opposite angle A. Substituting 13, 14 and 15 gives (196 + 225 − 169)/(2·14·15) = 252/420 = 3/5. Hence angle A is acute and cos A = 3/5.
Rewrite cotθ as tan(π/2 − θ), which is valid where both functions are defined. Then tan2θ = tan(π/2 − θ) gives 2θ = nπ + π/2 − θ, so 3θ = nπ + π/2 and θ = (2n+1)π/6. The integer n records the periodicity of the tangent equation.
Use the double-angle formula tan(2 tan⁻¹x) = 2x/(1 − x²) when |x| < 1. With x = 1/3 this becomes (2/3)/(1 − 1/9) = (2/3)/(8/9) = 3/4. Because 2 tan⁻¹(1/3) lies in (0, π/2), it equals tan⁻¹(3/4).
For a circle, arc length is s = rθ when θ is measured in radians. Substituting r = 9 cm and θ = π/4 gives s = 9π/4 cm, so the arc length is 9π/4 cm.
The angle 7π/6 lies in the third quadrant, where sine is negative. Its reference angle is π/6, and sin(π/6) = 1/2, so sin(7π/6) = −1/2.
The area of a triangle with two sides and the included angle is (1/2)bc sin A. Substituting b = 8, c = 6 and A = 30° gives (1/2)(8)(6)(1/2) = 12 cm².
The principal range of cos⁻¹ is [0, π], so the answer must lie in that interval. The angle in [0, π] whose cosine is −1/√2 is 3π/4.
Since tan(π/4) = 1, and tangent has period π, all solutions are given by θ = nπ + π/4. The nπ term accounts for every full period of the tangent function.
Use the identities 1 + tan²x = sec²x and 1 + cot²x = csc²x. Then the quotient is sec²x / csc²x = (1/cos²x)/(1/sin²x) = sin²x/cos²x = tan²x.