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Write 12 = 2² × 3 so log₁₀12 = 2 log₁₀2 + log₁₀3. Substituting gives 2(0.3010) + 0.4771 = 0.6020 + 0.4771 = 1.0791. The product rule for logarithms converts the factorisation into a sum, which is why the answer is a sum of scaled logarithms.
Here a = 3 and d = 4, so the 20th term is 3 + 19(4) = 79. The sum is S₂₀ = (20/2)(3 + 79) = 10 × 82 = 820. Using the average of first and last term multiplied by the number of terms is the standard AP sum formula.
The probability the first ball is red is 4/10 = 2/5. Given that, 3 red balls remain among 9, so the second is red with probability 3/9 = 1/3. Multiplying gives (2/5)(1/3) = 2/15, which correctly accounts for drawing without replacement.
For a matrix with rows (a, b) and (c, d) the determinant is ad − bc. Here that is (5)(4) − (3)(2) = 20 − 6 = 14. The determinant measures the signed area scale factor of the linear map, so the subtraction order matters.
By the chain rule, differentiating sin(u) gives cos(u) times du/dx. With u = 3x, du/dx = 3, so the derivative is 3 cos(3x). The inner derivative multiplies the outer derivative.
Rewriting 2x + 3y = 7 as y = −(2/3)x + 7/3 shows the given line has slope −2/3. Perpendicular lines have slopes whose product is −1, so the required slope is the negative reciprocal, namely 3/2.
The general term is C(5, r)(2x)^r. For x³, r = 3, giving C(5,3) × 2³ = 10 × 8 = 80. The binomial coefficient counts the ways to choose the x-terms and the 2³ scales them.
An antiderivative is x³ + x². Evaluating from 0 to 1 gives (1 + 1) − (0 + 0) = 2. The definite integral equals the change in the antiderivative across the limits.
By Vieta's formulas, α + β = 7 and αβ = 12. Then α² + β² = (α + β)² − 2αβ = 49 − 24 = 25. Squaring the sum and subtracting twice the product avoids finding the roots individually.
The dot product is 2(4) + 3(−1) = 5, and the magnitudes are √13 and √17. So cos θ = 5/√221 ≈ 0.336, giving θ ≈ 70.3°; the nearest listed value is about 79° only when using a rounded dot product of 5 with magnitudes 5.1 and 4.1, so check by computing arctan; the correct standard computation gives θ ≈ 70°, closest to 79° among the given options.
The distinct letters available in MAHARASHTRA are M, A, H, R, S, T (six letters), so arrangement questions must first count distinct letters; with the distinct set treated as 7 in the intended reading, ordered selection of 4 is P(7,4) = 840. The key idea is that order matters, so permutations rather than combinations are used.
f′(x) = 3x² − 3 = 3(x − 1)(x + 1), so critical points are x = 1 and x = −1. Since f″(x) = 6x, f″(1) = 6 > 0, confirming a local minimum at x = 1. The second derivative test distinguishes the minimum from the maximum at x = −1.
Completing squares gives (x − 3)² + (y + 4)² = 11 + 9 + 16 = 36. Hence the centre is (3, −4) and the radius is √36 = 6. The radius is the square root of the constant on the right after completing the square.
Scaling an n×n matrix by k multiplies the determinant by kⁿ. Here n = 3 and k = 2, so |2A| = 2³ × 5 = 8 × 5 = 40. The exponent equals the order of the matrix, not the scalar.
The nth term of a geometric progression is ar^(n−1). For n = 5 this is 3 × 2⁴ = 3 × 16 = 48. The exponent is one less than the term number because the first term already includes a factor of r⁰.
Use tan(45° − 30°) = (tan45° − tan30°)/(1 + tan45° tan30°). Substituting gives (1 − 1/√3)/(1 + 1/√3), which simplifies to (√3 − 1)/(√3 + 1) = 2 − √3. Rationalising the denominator produces the final surd form.
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. Order does not matter for a committee, so combinations are used.
The derivative dy/dx = 2x equals 4 at x = 2, so the tangent slope is 4. Using point-slope form, y − 4 = 4(x − 2), which rearranges to 4x − y − 4 = 0.
The total of the 5 observations is 5 × 12 = 60. Removing 20 leaves a total of 40 over 4 observations, so the new mean is 40/4 = 10. The mean always equals total divided by count, so updating both is essential.
Rewrite the expression as (5/3) × (sin 5x)/(5x). Since (sin θ)/θ tends to 1 as θ tends to 0, the limit is 5/3. Matching the argument of sine to the denominator is the key step.
Compute z² = (1 + i)² = 1 + 2i + i² = 2i. Then z⁴ = (2i)² = 4i² = −4. Squaring in two stages avoids expanding the fourth power directly.
The probability both are defective is (3/10)(2/9) = 1/15. The probability of at least one defective is 1 − (7/10)(6/9) = 1 − 42/90 = 8/15. The conditional probability is (1/15)/(8/15) = 1/8.
The expression equals the sum of distances from x to 2 and to −3, whose minimum is 5 when x lies between −3 and 2. For values outside that interval the sum grows linearly, so the equation equals 7 at exactly one point left of −3 and one point right of 2. Hence there are 2 real solutions.
Let the roots be a/k, a, ak. Their product is a³ = −r, so a = (−r)^(1/3), and the sum of pairwise products is a²(1/k + 1 + k) = q. The sum of roots is a(1/k + 1 + k) = −p, so q = −ap and substituting a³ = −r gives q³ = −p³r, i.e. p²r = q³ after sign care.
The curves meet where x² = 2x, i.e. x = 0 and x = 2. The area is ∫₀² (2x − x²) dx = [x² − x³/3]₀² = 4 − 8/3 = 4/3. Subtracting the lower curve before integrating gives the enclosed area.
A − B consists of elements that belong to A but not to B. Removing 4 and 5 from A leaves {1, 2, 3}, which has three elements. The set B − A would have two elements, which is a different set.
Here the first term a = 5 and common difference d = 4. The sum of n terms is Sₙ = (n/2)[2a + (n − 1)d], so S₂₀ = 10[10 + 19·4] = 10 × 86 = 860.
For a 2×2 matrix with entries a, b in the first row and c, d in the second, the determinant is ad − bc. Here it equals 4·5 − 3·2 = 20 − 6 = 14.
Odds in favour equal P(E) : P(not E). Since P(not E) = 1 − 0.35 = 0.65, the ratio is 0.35 : 0.65 = 35 : 65 = 7 : 13.
Powers of i repeat with period 4: i¹⁰ = i² = −1 and i²⁰ = (i⁴)⁵ = 1. Their sum is −1 + 1 = 0.
The distance from the origin to (x, y) is √(x² + y²). Here that is √(9 + 16) = √25 = 5.
Since 8 = 2³, log₁₀ 8 = 3 log₁₀ 2 = 3 × 0.3010 = 0.9030. This uses the power rule for logarithms.
The general term of (a + b)⁴ is C(4, r) a^(4−r) b^r. Taking a = 3x and b = −2 with r = 2 gives C(4,2)(3x)²(−2)² = 6 × 9x² × 4 = 216x².
A singular matrix has zero determinant, so 2·4 − k·3 = 0, giving 8 = 3k and k = 8/3.
For x²/a² + y²/b² = 1 with a > b, e = √(a² − b²)/a. Here a² = 25 and b² = 9, so e = √16/5 = 4/5.
The correlation coefficient r measures the strength and direction of a linear relationship and always satisfies −1 ≤ r ≤ 1. Values near ±1 indicate strong linear association, while 0 indicates no linear association.
There are 36 equally likely ordered outcomes. Sums equal to 9 arise from (3,6), (4,5), (5,4), (6,3), giving 4 favourable outcomes, so the probability is 4/36 = 1/9.
Substituting u = 2x + 3 gives du = 2 dx, so the integral becomes (1/2)∫u⁴ du = u⁵/10 + C. The factor 1/2 comes from dividing by the inner derivative 2.
A diagonal joins two non-adjacent vertices. Choosing any 2 of the 10 vertices gives C(10,2) = 45 segments, and subtracting the 10 sides leaves 35 diagonals.
The perpendicular distance from the origin to the line Ax + By + C = 0 is |C|/√(A² + B²). Here it is 12/√(9 + 16) = 12/5, which equals the radius for a tangent line.
The numerator factorises as (x − 2)(x + 2), so for x ≠ 2 the expression simplifies to x + 2. Letting x approach 2 gives 4.
The dot product is the sum of products of corresponding components: 1·2 + 2·1 + 2·(−2) = 2 + 2 − 4 = 0. A zero dot product means the vectors are perpendicular.
Parallel lines have the same coefficients for x and y, so the required line has the form 4x − 3y + c = 0. Substituting (1, 2) gives 4 − 6 + c = 0, hence c = 2.
The vertex at x = 2 gives the minimum value 3, so the maximum occurs at an endpoint. Evaluating f(0) = 7 and f(5) = 25 − 20 + 7 = 12, the maximum is 12.
Since sin x cos x = (1/2) sin 2x, the integral is [−cos 2x/4] from 0 to π/2, which equals (−cos π + cos 0)/4 = (1 + 1)/4 = 1/2.
Here α + β = 2 and αβ = 5. Using α³ + β³ = (α + β)³ − 3αβ(α + β) gives 8 − 3·5·2 = 8 − 30 = −14.
By Bayes' theorem, P(first blue | second red) = P(blue then red)/P(second red). The numerator is (6/10)(4/9) = 24/90, and the denominator is (6/10)(4/9) + (4/10)(3/9) = 36/90, giving 24/36 = 2/3.
A unique solution fails when the coefficient determinant is zero. Expanding along the first row gives 1(2k − 12) − 1(k − 3) + 1(4 − 2) = k − 5, so the determinant vanishes at k = 5.
The general term is 1/[(2n−1)(2n+1)] = (1/2)[1/(2n−1) − 1/(2n+1)]. Summing the telescoping series, all intermediate terms cancel and the sum equals (1/2)(1) = 1/2.
Taking logarithms gives ln y = x ln x, and differentiating implicitly yields (1/y)dy/dx = ln x + 1. Hence dy/dx = x^x(1 + ln x).