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For any n × n matrix A and scalar k, the determinant satisfies det(kA) = k^n det(A). Here n = 3 and k = 2, so det(2A) = 2^3 × 5 = 8 × 5 = 40. The scalar multiplies every one of the three rows, which is why the factor is cubed rather than doubled.
Powers of i repeat with period 4: i^1 = i, i^2 = -1, i^3 = -i, i^4 = 1. Since 2023 = 4 × 505 + 3, we get i^2023 = (i^4)^505 × i^3 = 1 × (-i) = -i.
Write sin(5x)/(3x) as (5/3) · sin(5x)/(5x). As x → 0, the angle 5x → 0 and sin(5x)/(5x) → 1, so the limit equals 5/3. Both numerator and denominator tend to 0, so the standard limit sin θ/θ → 1 resolves the indeterminacy.
Split the sum: Σk^2 + Σk = 10·11·21/6 + 10·11/2 = 385 + 55 = 440. Equivalently, use the closed form n(n + 1)(n + 2)/3 with n = 10, giving 10·11·12/3 = 440.
Taking logarithms gives ln y = x ln x, then differentiating implicitly yields (1/y)(dy/dx) = ln x + 1. Multiplying by y = x^x gives dy/dx = x^x(1 + ln x). The power rule does not apply because the exponent is itself a function of x.
Use the identity sin²x = (1 - cos 2x)/2, so the integral becomes [x/2 - sin 2x/4] evaluated from 0 to π/2. At the upper limit this is π/4 - 0, and at the lower limit it is 0, so the value is π/4.
Use the identity |a × b|² = |a|²|b|² - (a·b)², which gives 9 × 16 - 36 = 144 - 36 = 108. Hence |a × b| = √108 = 6√3. The dot product fixes the cosine of the angle, and the cross product magnitude follows from the sine.
The favourable outcomes are (3,6), (4,5), (5,4) and (6,3), giving 4 favourable cases out of 36 equally likely outcomes. Therefore the probability is 4/36 = 1/9.
Dividing by 144 gives x²/16 - y²/9 = 1, so a² = 16 and b² = 9. For a hyperbola, e = √(1 + b²/a²) = √(1 + 9/16) = √(25/16) = 5/4.
The integrating factor is e^(∫tan x dx) = sec x, so the equation becomes d/dx(y sec x) = sec²x. Integrating gives y sec x = tan x + C, and y(0) = 0 forces C = 0, so y = tan x · cos x = sin x.
For a skew-symmetric matrix A we have Aᵀ = −A. Taking determinants, det(Aᵀ) = det(−A), so det(A) = (−1)³det(A) = −det(A), which forces det(A) = 0. The odd order 3 is what makes the sign factor −1 rather than +1.
The point (−1, √3) lies in the second quadrant, so its argument is π minus the reference angle. Since tan α = √3/1 gives α = π/3, the principal argument is π − π/3 = 2π/3.
The word has 6 letters in which the letter E occurs twice and all other letters are distinct. The number of distinct permutations is therefore 6!/2! = 720/2 = 360.
Continuity at x = 2 requires f(2) to equal the limit of f(x) as x → 2. Since x² − 4 = (x − 2)(x + 2), the quotient simplifies to x + 2 for x ≠ 2, whose limit at x = 2 is 4, so k must be 4.
Put x = tan θ, so that 2x/(1 + x²) = sin 2θ and θ = tan⁻¹x. Then y = 2θ = 2tan⁻¹x, and differentiating gives dy/dx = 2/(1 + x²).
Substituting u = x² gives du = 2x dx, so the integral becomes (1/2)∫ from 0 to 1 of e^u du. This equals (1/2)(e − 1), since e⁰ = 1.
Expanding the squared magnitude gives |a + b|² = |a|² + |b|² + 2(a·b) = 1 + 1 + 0 = 2. Hence |a + b| = √2, which reflects the fact that a and b are perpendicular unit vectors.
The probability that the first ball is red is 3/8, and given that, the probability that the second is also red is 2/7. Multiplying these dependent probabilities gives (3/8)(2/7) = 6/56 = 3/28.
The general term is C(9, r)·x^(9 − r)·(2/x²)^r = C(9, r)·2^r·x^(9 − 3r). For the term independent of x we need 9 − 3r = 0, so r = 3, giving C(9, 3)·2³ = 84·8 = 672.
The order is the order of the highest derivative present, which is the second derivative, so the order is 2. The degree is the exponent of that highest derivative when the equation is written as a polynomial in derivatives, which here is 3.
For a quadratic with roots α and β, α + β = 5 and αβ = 6. Using α² + β² = (α + β)² − 2αβ gives 25 − 12 = 13.
Rewriting 3x + 4y = 7 gives slope −3/4. A perpendicular line has slope equal to the negative reciprocal, so its slope is 4/3.
A set with 5 elements has 2⁵ = 32 subsets in total. Excluding the set A itself leaves 32 − 1 = 31 subsets.
The term containing x⁴ is obtained by choosing x from 4 factors and 2 from the remaining 2 factors. Its coefficient is C(6,4)·2²·(−1)⁴ = 15·4 = 60.
For the ellipse x²/a² + y²/b² = 1, the latus rectum length is 2b²/a. Here a² = 25 and b² = 9, so the length is 2·9/5 = 18/5.
There are 12 face cards in a standard pack, and 4 of them are kings. Therefore the conditional probability is 4/12 = 1/3.
Differentiate e^(sin x) using the chain rule. The derivative is e^(sin x) multiplied by the derivative of sin x, which is cos x.
Use integration by parts with u = x and dv = cos x dx. Then du = dx and v = sin x, so the integral is x sin x − ∫ sin x dx = x sin x + cos x + C.
Let f(x) = x³ − 3x + 1. Its derivative is 3(x² − 1), so the critical points are x = −1 and x = 1. Since f(−1) = 3 and f(1) = −1, and the cubic has opposite end behaviour at ±∞, its graph crosses the x-axis once in each of three intervals, giving three distinct real roots.
The nth term is 1/((2n − 1)(2n + 1)) = 1/2[1/(2n − 1) − 1/(2n + 1)]. The partial sums telescope to 1/2[1 − 1/(2N + 1)], whose limit as N → ∞ is 1/2.
For an n × n matrix A and a scalar k, the rule is det(kA) = k^n det(A). Here n = 3 and k = 2, so det(2A) = 2^3 · 2 = 16.
Simplify by multiplying numerator and denominator by 1 + i: (1 + i)/(1 − i) = ((1 + i)^2)/(1 + 1) = i. The modulus of i is 1.
The number of circular arrangements of n distinct objects is (n − 1)!. With n = 5, this gives (5 − 1)! = 4! = 24.
Use 1 − cos 2x = 2 sin² x. Then the quotient becomes 2(sin x/x)², and since sin x/x → 1, the limit is 2.
For x > −1, tan⁻¹((1 − x)/(1 + x)) = π/4 − tan⁻¹ x. Differentiating gives dy/dx = −1/(1 + x²), so at x = 0 the value is −1.
Integrate by parts with u = ln x and dv = dx. This gives ∫ ln x dx = x ln x − ∫1 dx = x ln x − x + C.
By Bayes' theorem, P(A | white) = [(1/2)(2/5)] / [(1/2)(2/5) + (1/2)(3/5)] = (1/5)/(1/2) = 2/5.
Such a circle has equation x² + y² − 2ry = 0, where r is its radius. Differentiating gives x + y y′ = r y′, and eliminating r using r = (x² + y²)/(2y) yields (x² − y²)y′ = 2xy.
The transverse axis length is 2a = 6, so a = 3. The foci give c = 5, hence b² = c² − a² = 25 − 9 = 16. Therefore the hyperbola is x²/9 − y²/16 = 1.
The dot product of a = a1i + a2j + a3k and b = b1i + b2j + b3k is a1b1 + a2b2 + a3b3. Here it is (2)(1) + (3)(−1) + (1)(4) = 2 − 3 + 4 = 3.
Compute A² = [[7, 10], [15, 22]] and 5A = [[5, 10], [15, 20]]. Subtracting entrywise gives [[2, 0], [0, 2]], which is 2I.
Square both sides: sin²θ + 2sinθcosθ + cos²θ = 1/4. Since sin²θ + cos²θ = 1, this gives 1 + sin 2θ = 1/4, so sin 2θ = −3/4.
The general term is C(9, r)(x²)^(9−r)(1/x)^r = C(9, r)x^(18−3r). For the x-power to vanish, 18 − 3r = 0, so r = 6. The coefficient is C(9, 6) = 84.
The probability is (4/10)(3/9), because after one red is removed there are 3 red among 9 balls. Multiplying gives 12/90 = 2/15.
For parallel lines ax + by = c1 and ax + by = c2, the distance is |c2 − c1| / √(a² + b²). Here it is |12 − 7| / √(9 + 16) = 5/5 = 1.
The direction ratios are B − A = (3, 4, 0), whose length is √(9 + 16 + 0) = 5. Dividing each direction ratio by 5 gives direction cosines (3/5, 4/5, 0).
Write the equation as y′ + P(x)y = Q(x), so P(x) = 1/x. The integrating factor is e^(∫(1/x)dx) = e^(ln x) = x because x > 0.
f′(x) = 3x² − 6x + 6 = 3[(x − 1)² + 1] > 0 for all x, so f is strictly increasing on [0, 3]. Hence the maximum occurs at x = 3, and f(3) = 27 − 27 + 18 − 4 = 14.
The foci are at 1 and −1, so the focal distance is c = 1. The constant sum is 2a = 4, hence a = 2 and the eccentricity is e = c/a = 1/2.
By Vieta’s formulas, α + β = 5 and αβ = 6. Therefore α² + β² = (α + β)² − 2αβ = 25 − 12 = 13. The roots are 2 and 3, and 2² + 3² = 13 confirms the result.