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Rewrite the ratio as (sin 5x / x) ÷ (sin 2x / x). Since sin(ax)/x tends to a as x tends to 0, the numerator approaches 5 and the denominator approaches 2, giving 5/2. Direct substitution produces the indeterminate form 0/0, so the standard limit (or L'Hôpital's rule) must be applied.
For a square matrix of order n, det(kA) = k^n · det(A), because the scalar k multiplies each of the n rows. Here n = 3 and k = 2, so det(2A) = 2³ × 5 = 8 × 5 = 40.
The derivative of tan⁻¹x is exactly 1/(1+x²), so tan⁻¹x is the antiderivative. The arbitrary constant C must be attached because all antiderivatives of a function differ only by a constant.
The dot product of two vectors in component form is the sum of the products of corresponding components: (2)(1) + (3)(−2) + (−1)(2). This gives 2 − 6 − 2 = −6, a negative value because the vectors point partly in opposite directions.
Applying the product rule gives dy/dx = 2x sin x + x² cos x. At x = 0 both terms vanish because each contains a factor of x, so the derivative equals 0.
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-order derivative when the equation is polynomial in the derivatives, and here d²y/dx² is raised to the power 3.
There are 6 × 6 = 36 equally likely ordered outcomes. The favourable outcomes summing to 9 are (3,6), (4,5), (5,4) and (6,3), giving 4 outcomes, so the probability is 4/36 = 1/9.
Using the addition formula, tan(tan⁻¹1 + tan⁻¹2) = (1+2)/(1−1·2) = −3, and since that sum lies between π/2 and π it equals π − tan⁻¹3. Adding tan⁻¹3 then gives π exactly.
The curves meet where x² = x, that is at x = 0 and x = 1, and the line lies above the parabola on that interval. The area is ∫₀¹ (x − x²) dx = 1/2 − 1/3 = 1/6.
Integrating by parts with u = x and dv = sin x dx gives ∫ x sin x dx = −x cos x + sin x. Evaluating from 0 to π/2 yields (−0 + 1) − (0 + 0) = 1, so the definite integral equals 1.
For a complex number z = a + bi, the modulus is |z| = √(a² + b²). With a = 3 and b = −4, |z| = √(9 + 16) = √25 = 5. The sign of the imaginary part does not affect the modulus.
This is an infinite geometric series with first term a = 1 and common ratio r = 1/3. Because |r| < 1, the sum is a/(1 − r) = 1/(1 − 1/3) = 1/(2/3) = 3/2.
Rewriting 3x − 4y + 7 = 0 as y = (3/4)x + 7/4 shows that its slope is 3/4. Perpendicular lines have slopes whose product is −1, so the required slope is −1 ÷ (3/4) = −4/3.
For A = [[a, b], [c, d]], A⁻¹ = (1/(ad − bc))[[d, −b], [−c, a]]. Here ad − bc = 1(4) − 2(3) = −2, so A⁻¹ = (−1/2)[[4, −2], [−3, 1]] = [[−2, 1], [3/2, −1/2]].
The general term is C(6,r) x^(6−r)(2/x²)^r = C(6,r)2^r x^(6−3r). The term independent of x requires 6 − 3r = 0, so r = 2. Its coefficient is C(6,2)·2² = 15·4 = 60.
The standard limit lim(x→0)(e^(ax) − 1)/x = a follows by substituting u = 3x or by L'Hôpital's rule. Here a = 3, so the limit is 3.
The mean is (2 + 4 + 6 + 8 + 10)/5 = 30/5 = 6. The squared deviations are 16, 4, 0, 4, 16, whose sum is 40. Therefore variance = 40/5 = 8.
Let u = x², so du = 2x dx and x dx = du/2. When x = 0, u = 0, and when x = 1, u = 1. Thus the integral is (1/2)∫₀¹ e^u du = (e − 1)/2.
Complete the square: x² + 4x + 13 = (x + 2)² + 9 = (x + 2)² + 3². Using ∫ du/(u² + a²) = (1/a)tan⁻¹(u/a) + C with u = x + 2 and a = 3 gives (1/3)tan⁻¹((x + 2)/3) + C.
By the conditional probability formula, P(first red | second red) = P(both red)/P(second red). P(both red) = (3/7)(2/6) = 6/42, while P(second red) = (3/7)(2/6) + (4/7)(3/6) = 6/42 + 12/42 = 18/42. Hence the required probability is 6/18 = 1/3.
For z = a + bi, the product z z̄ equals a² + b², because (a + bi)(a − bi) = a² + b². Here a = 2 and b = 3, so z z̄ = 4 + 9 = 13.
Using sin²θ + cos²θ = 1, we get cos²θ = 1 − 9/25 = 16/25. Since θ is in the first quadrant, cos θ is positive, so cos θ = 4/5.
Choosing a committee ignores order, so the number of selections is C(7,3) = 7!/(3!4!) = 35. This is a combination problem, not a permutation problem.
A skew-symmetric matrix satisfies Aᵀ = −A. For any diagonal entry aᵢᵢ, this condition gives aᵢᵢ = −aᵢᵢ, so aᵢᵢ = 0.
Rewrite the second line as 3x + 4y = 11/2 so that the coefficients match. The distance is |5 − 11/2| / √(3² + 4²) = (1/2)/5 = 1/10.
Let the first term be a and the common difference be d. Then a + 2d = 7 and a + 6d = 19, giving 4d = 12, so d = 3 and a = 1. Hence a₁₀ = a + 9d = 1 + 27 = 28.
Since d/dx[tan(3x)] = 3 sec²(3x), the integral must be divided by 3. Therefore ∫ sec²(3x) dx = tan(3x)/3 + C.
The area is 1/2 |a × b|. Here (i + j) × (j + k) = i − j + k, whose magnitude is √3. Thus the area is √3/2.
The first head on toss 2k has probability (1/2)^(2k). Summing over k ≥ 1 gives 1/4 + (1/4)² + ⋯ = (1/4)/(1 − 1/4) = 1/3.
Separating variables gives dy/y = dx/x, so ln|y| = ln|x| + C and y = Kx. The condition y(1) = 2 gives K = 2, hence y = 2x.
The number −1 + i√3 lies in the second quadrant, with reference angle arctan(√3/1) = π/3. The principal argument is therefore π − π/3 = 2π/3, which lies in (−π, π].
For a square matrix A of order n, det(kA) = k^n det(A). Here n = 3 and k = 2, so det(2A) = 2^3 × 5 = 40.
The complement of 'at least one head' is 'no heads', which occurs only for TT. Since P(TT) = 1/4, the required probability is 1 − 1/4 = 3/4.
The radius is the distance from the centre (2, −3) to the point (5, 1): sqrt((5−2)^2 + (1+3)^2) = sqrt(9 + 16) = 5. Hence the circle is (x − 2)^2 + (y + 3)^2 = 25.
As x→0, sin(5x) is asymptotic to 5x and tan(3x) is asymptotic to 3x. Therefore the quotient tends to 5x/(3x) = 5/3.
For dy/dx + P(x)y = Q(x), the integrating factor is exp(∫P(x)dx). Here P(x) = 2/x, so the factor is exp(2 ln x) = x^2 for x > 0.
With n = 5 and p = 1/2, the binomial probability is C(5,2)(1/2)^2(1/2)^3. This equals 10/32 = 5/16.
The area of the parallelogram is |a × b|. Computing a × b = −7i + 5j − k gives magnitude sqrt(49 + 25 + 1) = sqrt(75) = 5√3.
Let I be the integral. Using ∫₀^a f(x)dx = ∫₀^a f(a−x)dx with a = π/2 gives I = ∫₀^(π/2) cos x/(cos x+sin x) dx. Adding the two forms gives 2I = ∫₀^(π/2) 1 dx = π/2, so I = π/4.
Using 1 + z = 2cos(θ/2)e^(iθ/2) and 1 − z = −2i sin(θ/2)e^(iθ/2), their quotient is (2cos)/(−2i sin) = i cot(θ/2). Thus the quotient simplifies to i cot(θ/2).
Since i^4 = 1, reduce each exponent modulo 4: 2024 ≡ 0 mod 4, so i^{2024} = 1, and 2025 ≡ 1 mod 4, so i^{2025} = i. Therefore the sum is 1 + i.
Use the compound-angle identity sin A cos B + cos A sin B = sin(A + B) with A = π/3 and B = π/6. Then sin(π/3 + π/6) = sin(π/2) = 1.
The total number of balls is 3 + 5 = 8, and the number of favourable red balls is 3. Hence the probability of drawing a red ball is 3/8.
For any square matrix A, A adj(A) = |A| I. For a 2 × 2 matrix, taking determinants gives |A adj(A)| = |A|^2 |I| = |A|^2. With |A| = 3, this is 9.
The scalar triple product is the determinant of the component matrix with rows (1,1,0), (0,1,1), and (1,0,1). Expanding gives 1(1·1 − 1·0) − 1(0·1 − 1·1) + 0 = 1 + 1 = 2.
The order is the highest derivative present, which is the second derivative, so the order is 2. The degree is the exponent of the highest-order derivative after the equation is polynomial in derivatives, giving degree 3.
Use the identity 1 − cos 2x = 2 sin² x, so the expression becomes 2(sin x/x)². As x → 0, sin x/x → 1, so the limit is 2.
Using integration by parts with u = x and dv = e^x dx gives ∫ x e^x dx = x e^x − ∫ e^x dx. Therefore the result is (x − 1)e^x + C.
For cube roots of unity, 1 + ω + ω² = 0, so 1 + ω = −ω². Then 1 + ω − ω² = −ω² − ω² = −2ω², and cubing gives (−2ω²)^3 = −8ω^6 = −8.
The sum of the coefficients is obtained by setting x = 1, so 2^n = 64 and hence n = 6. The coefficient of x² is C(6,2) = 15.