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For ax² + bx + c = 0, the sum of the roots equals −b/a. Here a = 2 and b = −5, so the sum is −(−5)/2 = 5/2. The product of the roots would be c/a = 3/2, which is a different quantity.
The difference between the 13th and 7th terms spans 6 steps, so 6d = 64 − 34 = 30, giving d = 5. Substituting back, the first term satisfies a + 6(5) = 34, so a = 4.
This is a difference of cubes: a³ − b³ = (a − b)(a² + ab + b²) with a = x and b = 2. That yields (x − 2)(x² + 2x + 4). The quadratic factor has discriminant 4 − 16 < 0, so it cannot be factored further over the reals.
Equal roots require the discriminant b² − 4ac to vanish. Here that gives k² − 36 = 0, so k = ±6. Among the options only 6 satisfies this condition.
The first term is 3 and the common ratio is 2, so the nth term is 3·2^(n−1). For n = 10 this is 3·2⁹ = 3·512 = 1536.
The general term is C(5,r)(2x)^r·3^(5−r). For the x³ term, r = 3, giving C(5,3)·2³·3² = 10·8·9 = 720.
The (2,1) entry of AB is the dot product of row 2 of A with column 1 of B: 3·1 + 4·2 = 3 + 8 = 11. Each entry of the product combines one row of A with one column of B in this way.
For a 2×2 matrix [[a, b], [c, d]] the determinant is ad − bc. Here that is 3·4 − 5·2 = 12 − 10 = 2.
The sum of the roots is 6 and the product is (3)² − (√2)² = 9 − 2 = 7. The equation is x² − (sum)x + (product) = 0, that is x² − 6x + 7 = 0.
We have α + β = 4/3 and αβ = 1/3. Using α² + β² = (α + β)² − 2αβ gives 16/9 − 2/3 = 16/9 − 6/9 = 10/9.
The compound amount is P(1 + r)^n = 12000(1.1)² = 12000 × 1.21 = ₹14,520. The second year's interest is charged on the increased balance, which is why the result exceeds simple interest.
The nth term equals Sₙ − Sₙ₋₁. So the 8th term is S₈ − S₇ = (3·64 + 40) − (3·49 + 35) = 232 − 182 = 50. Equivalently, the progression starts at 8 with common difference 6, giving 8 + 7·6 = 50.
The general term is C(6,r)x^(6−r)(1/x)^r = C(6,r)x^(6−2r). For a constant term the exponent must vanish, so 6 − 2r = 0 and r = 3. The term is C(6,3) = 20.
The determinant is 1·4 − 2·3 = −2. The inverse is (1/det)[[d, −b], [−c, a]] = (−1/2)[[4, −2], [−3, 1]] = [[−2, 1], [3/2, −1/2]]. Multiplying by A must give the identity matrix.
A unique solution exists when the coefficient determinant is nonzero. The determinant is k·k − 2·4 = k² − 8, which vanishes only when k = ±2√2. Therefore any k other than these two values gives a unique solution.
Let the roots be a − d, a, a + d. Their sum equals 3a = 6 (the coefficient of x² with sign changed), so a = 2. Indeed the roots are 1, 2, 3, confirming the middle root is 2.
Using the laws of logarithms, log₂48 − log₂3 = log₂(48/3) = log₂16 = 4. Adding log₂2 = 1 gives 5. All terms share base 2, so the quotient and product rules apply directly.
For |r| < 1 the sum is a/(1 − r). Setting 2/(1 − r) = 8 gives 1 − r = 1/4, hence r = 3/4. This value lies within the convergence range, so the series does converge.
The discriminant is b² − 4ac = 16 − 40 = −24. A negative discriminant means the roots are non-real complex conjugates. Since the coefficients are real, the conjugates pair up automatically.
The x² coefficient collects terms where the exponents from the three factors sum to 2: three ways to choose x² once and 1 twice (3·1), and three ways to choose x from two factors (C(3,2) = 3). The total is 3 + 3 = 6.
From A² = A we get A(A − I) = 0. If A were invertible, multiplying by A⁻¹ would force A = I, contradicting the hypothesis. Hence A is singular and det A = 0. Idempotent matrices other than I are always singular.
The second equation is exactly twice the first, so it adds no new constraint. Two independent equations in three unknowns leave one free parameter, giving infinitely many solutions along a line. A linear system never has exactly two solutions.
Since ω³ = 1 and 1 + ω + ω² = 0, the sum becomes (1 + ω + ω²) + ω³(1 + ω) = 0 + 1·(1 + ω) = 1 + ω. But 1 + ω = −ω², and since ω² = −1 − ω, this equals ω after simplification using ω² + ω + 1 = 0.
Since α + β + γ = 0, the identity α³ + β³ + γ³ = 3αβγ applies. The product of the roots of x³ + px + q equals −q (constant term with sign changed). Hence the sum of cubes is 3(−q) = −3q.
Let the first term be a. The sum to infinity is a/(1 − r) = 4a, so 1/(1 − r) = 4 and 1 − r = 1/4, giving r = 3/4. Since 0 < r < 1, the series converges and all terms stay positive.
The general term is C(4,r)·3^(4−r)·(2x)^r. For x² we take r = 2, giving C(4,2)·3²·2² = 6·9·4 = 216.
Dividing ar⁷ by ar⁴ gives r³ = 384/48 = 8, so r = 2. Then ar⁴ = 48 gives a·16 = 48, hence a = 3.
The determinant is 4·2 − 3·1 = 5, and since det(A²) = (det A)², this gives 25. So the two required values are 5 and 25, matching the pair 5 and 25; the option listing 25 and 625 is incorrect because det A = 5, not 25.
For a cubic x³ + bx² + cx + d, the sum of pairwise products of roots equals c. Here b = −4, c = 1, d = 6, so αβ + βγ + γα = 1.
Irrational roots of a rational quadratic occur in conjugate pairs, so the other root is 2 + √3. The sum is 4 and the product is 4 − 3 = 1, giving x² − 4x + 1 = 0.
Using the quotient law, log₃ 54 − log₃ 2 = log₃(54/2) = log₃ 27 = 3.
We need (1 + r)² = 9261/8000 = 1.157625, so 1 + r = 1.07625, giving r = 0.07625 ≈ 7.5% per annum.
Infinitely many solutions require proportional coefficients: 2/(k+1) = 3/6 = 7/14. From 3/6 = 1/2 we need 2/(k+1) = 1/2, giving k + 1 = 4, so k = 3.
Since 1 + ω + ω² = 0, we have 1 + ω = −ω², so 1 + ω − ω² = −2ω². Cubing gives −8ω⁶ = −8(ω³)² = −8.
A² = [[7, 10], [15, 22]] and 5A = [[5, 10], [15, 20]]. Subtracting gives [[2, 0], [0, 2]] = 2I, so A² − 5A = 2I. The option −2I is wrong because the diagonal entries are positive 2.
The nth term is Sₙ − Sₙ₋₁. Here S₁₀ = 130 and S₉ = 108, so t₁₀ = 130 − 108 = 22. Equivalently, tₙ = 2n + 2 gives 22 when n = 10.
Let the roots be a and a + 3 with a(a + 3) = 10, so a² + 3a − 10 = 0 and a = 2 or a = −5. The roots are 2 and 5 (or −5 and −2), whose sum is 7 or −7, so the positive p is 7.
The expansion has 7 terms, so the middle term is the 4th, with r = 3. It equals C(6,3)·x³·2³ = 20·8·x³ = 160x³.
det A = 4 − 1 = 3, and the adjugate swaps the diagonal and negates the off-diagonal entries. Hence A⁻¹ = (1/3)[[2, −1], [−1, 2]], and checking A·A⁻¹ gives the identity.
The coefficients are n, n(n−1)/2, n(n−1)(n−2)/6. Setting twice the middle equal to the sum of the other two gives 2·n(n−1)/2 = n + n(n−1)(n−2)/6, which simplifies to n² − 9n + 14 = 0, so n = 7 or n = 2. Only n = 7 allows a 4th term.
sin 30° equals 1/2 and cos 60° also equals 1/2, since 30° and 60° are complementary angles. Adding the two gives 1/2 + 1/2 = 1.
tan 45° = 1 and cot 45° = 1, because the tangent and cotangent of 45° are reciprocal and both equal 1. Their product is therefore 1.
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.
To convert degrees to radians, multiply by π/180. Thus 150° × π/180 = 150π/180 = 5π/6.
For y = sin(kx), the period is 2π/|k|. Here k = 3, so the period is 2π/3.
Sine is an odd function, so sin(−x) = −sin x. Therefore sin(−30°) = −sin 30° = −1/2.
Sine vanishes at every integer multiple of π, since sin 0 = 0 and the sine curve repeats with period 2π while crossing zero again at π. Hence θ = nπ for all integers n.
Taking opposite = 3 and adjacent = 4 gives hypotenuse = 5 by the Pythagorean theorem. So sin θ = 3/5 and cos θ = 4/5, whose sum is 7/5.
Write 75° = 45° + 30° and apply cos(A+B) = cos A cos B − sin A sin B. This gives (√2/2)(√3/2) − (√2/2)(1/2) = (√6 − √2)/4.
From sin A = 5/13 with A acute, cos A = 12/13. Then sin 2A = 2 sin A cos A = 2 × (5/13) × (12/13) = 120/169.