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Expanding along the first row gives 1·(1·0 − 4·6) − 2·(0·0 − 4·5) + 3·(0·6 − 1·5) = −24 + 40 − 15 = 1. Checking by the rule of Sarrus confirms the same value, so the determinant equals 1.
Let the roots be n and n + 1. Their sum is 2n + 1 = 7, giving n = 3, so the roots are 3 and 4. The product of the roots equals k, hence k = 3 × 4 = 12.
The word has 10 letters with S appearing 3 times, T appearing 3 times, I appearing 2 times, and A and C once each. Dividing the total permutations by the repeated-letter factorials gives 10!/(3!·3!·2!) = 3628800/72 = 50400.
The antiderivative of 1/(1 + x²) is arctan x. Evaluating from 0 to 1 gives arctan 1 − arctan 0 = π/4 − 0 = π/4. Hence the definite integral equals π/4.
Perpendicular vectors have zero dot product: (2)(1) + (−1)(2) + (1)(λ) = 2 − 2 + λ = λ. Setting this equal to zero gives λ = 0. So the required value is 0.
By the binomial theorem, the term containing x⁵ in (1 + x)¹⁰ is C(10, 5)·x⁵. Computing C(10, 5) = 10!/(5!·5!) = 252 gives the coefficient.
Here the first term a = 3 and the common difference d = 4. The sum formula Sₙ = (n/2)[2a + (n − 1)d] gives S₂₀ = 10·[6 + 19·4] = 10·82 = 820.
Rewrite the expression as 5·(sin 5x)/(5x). Since (sin θ)/θ tends to 1 as θ tends to 0, the limit equals 5·1 = 5.
The probability the first ball is red is 4/10 = 2/5. Given that, 3 red remain among 9 balls, so the second is red with probability 3/9 = 1/3. Multiplying gives (2/5)(1/3) = 2/15.
First square z: (1 + i)² = 1 + 2i + i² = 2i. Squaring again gives (2i)² = 4i² = −4. Hence z⁴ = −4.
For the circle x² + y² = r², the tangent at (x₁, y₁) is xx₁ + yy₁ = r². Substituting (3, 4) and r² = 25 gives 3x + 4y = 25.
Apply the product rule: d/dx(x·ln x) = 1·ln x + x·(1/x) = ln x + 1. So the derivative is 1 + ln x.
The principal range of sin⁻¹ is [−π/2, π/2]. Since sin(−π/6) = −1/2 and −π/6 lies in this range, the principal value is −π/6.
For an n×n matrix, |kA| = kⁿ|A|. With k = 2 and n = 3, we get |2A| = 2³·2 = 8·2 = 16.
The expression |x − 2| + |x + 3| equals the distance from x to 2 plus the distance from x to −3. For any x between −3 and 2 inclusive, this sum equals exactly 5, the distance between the two points. Hence every x in [−3, 2] is a solution, giving infinitely many.
The area equals the definite integral of x² from 0 to 3. Since the antiderivative is x³/3, evaluating gives 27/3 − 0 = 9.
For the parabola y² = 4ax with a = 1, the tangency condition for y = mx + c is c = a/m. Here c = 1 and a = 1, so 1 = 1/m, giving m = 1.
The total of the five observations is 5 × 12 = 60. Removing 20 leaves a sum of 40 over four observations, so the new mean is 40/4 = 10.
Separating variables gives dy/y = dx/x. Integrating both sides yields ln|y| = ln x + ln|c|, so y = cx, where c is an arbitrary constant.
If λ is an eigenvalue of A with eigenvector v, then A²v = A(λv) = λ²v. Hence the eigenvalues of A² are the squares of those of A, namely 1² = 1 and 3² = 9.
This is a geometric series with first term a = 1 and common ratio r = 1/2. Since |r| < 1, the sum is a/(1 − r) = 1/(1 − 1/2) = 2.
Differentiating gives f'(x) = 3x² − 6x = 3x(x − 2). Setting f'(x) = 0 gives x = 0 and x = 2. Since f' changes sign at both points, each is a local extremum.
Total committees without restriction: C(8, 3) = 56. Committees containing both of the two particular people: choosing 1 more from the remaining 6 gives C(6, 1) = 6. Subtracting gives 56 − 6 = 50. Wait-recomputing: 56 − 6 = 50. So the answer is 50.
Write the multinomial sum over non-negative a, b, c with a + b + c = 5, where the x-power is b + 2c. For x⁴ we need b + 2c = 4. The valid triples are (1, 4, 0), (2, 2, 1), (3, 0, 2), and (0, 1, 2) is invalid since b = 0? Checking (0, 0, 2) gives sum 2 ≠ 5. Summing multinomial terms 5!/(a!b!c!) over the valid triples gives 5 + 30 + 10 = 45.
The condition is the focus–directrix definition of a conic with focus (2, 0) and directrix x = 1. The ratio of distance to the focus over distance to the directrix is the eccentricity, which is given as 2. Since e = 2 > 1, the conic is a hyperbola.
For ax² + bx + c = 0 the sum of roots equals −b/a. Here a = 2, b = −7, so the sum is 7/2. The product would be c/a = 3/2, which is a different quantity.
The dot product is the sum of componentwise products: (2)(1) + (3)(−1) = 2 − 3 = −1. The sign matters because the j-components have opposite signs.
An antiderivative of cos x is sin x, so the definite integral equals sin(π/2) − sin(0) = 1 − 0 = 1. The integrand is positive on the interval, so a negative answer is impossible.
The word has 5 letters with L repeated twice and E repeated twice. The count is 5!/(2!·2!) = 120/4 = 30, dividing by the factorials of the repeated-letter counts.
The modulus of a quotient is the quotient of moduli: |3 + 4i| = 5 and |1 + i| = √2, so the modulus is 5/√2, which equals 5√2/2. Simplifying the quotient first gives (7 + i)/2, whose modulus is √50/2 = 5/√2, confirming the result.
In the third quadrant both sine and cosine are negative. With tan θ = 3/4 the reference triangle has sides 3, 4, 5, so |sin θ| = 3/5 and hence sin θ = −3/5.
The term containing x³ is C(6,3)·2³·(−x)³ = 20·8·(−1)·x³ = −160x³. Both the binomial coefficient and the sign from (−x)³ must be included.
Rewrite the second line as 3x + 4y + 5/2 = 0 so both have the same coefficients. The distance is |−7 − 5/2|/√(9 + 16) = (19/2)/5 = 19/10.
Using the standard limit (e^{t} − 1)/t → 1 as t → 0, substitute t = 3x to get (e^{3x} − 1)/x = 3·(e^{3x} − 1)/(3x) → 3. Equivalently, L'Hôpital's rule gives 3e^{3x} at x = 0, which is 3.
For a 2×2 matrix, A⁻¹ = (1/det A)·[[d, −b], [−c, a]]. Here det A = 4 − 6 = −2, so A⁻¹ = (−1/2)[[4, −2], [−3, 1]] = [[−2, 1], [3/2, −1/2]].
The number of heads follows a binomial distribution with n = 4 and p = 1/2. The probability is C(4,2)(1/2)⁴ = 6/16 = 3/8.
The mean is (n + 1)/2 and the mean of squares is (n + 1)(2n + 1)/6. Subtracting the square of the mean gives variance (n² − 1)/12.
For y² = 4ax with a = 2, the ends of the latus rectum are (2, 4) and (2, −4). The tangents there are y·4 = 4(x + 2) and y·(−4) = 4(x + 2), which meet at (−2, 0).
Integrating by parts with u = x and dv = e^{x} dx gives x·e^{x} − ∫e^{x} dx = x·e^{x} − e^{x} + C. Differentiating this result returns x·e^{x}, confirming it.
Separating variables gives dy/y = −cot x dx, and integrating yields ln|y| = −ln|sin x| + k. Hence y = C/sin x is not right; rechecking with the integrating factor sin x gives d(y·sin x)/dx = 0, so y·sin x = C and y = C·cosec x.
Dividing by 144 gives x²/16 − y²/9 = 1, so a² = 16 and b² = 9. For a hyperbola e = √(a² + b²)/a = √25/4 = 5/4.
For a cubic x³ + px² + qx + r = 0, the sum of pairwise products of roots equals q. Here q = 11, so αβ + βγ + γα = 11. The roots are 1, 2, 3 and indeed 2 + 6 + 3 = 11.
Using inclusion–exclusion, the count is 3⁴ − C(3,1)·2⁴ + C(3,2)·1⁴ = 81 − 48 + 3 = 36. Equivalently, one target element is hit twice and the others once, giving C(4,2)·3! = 36.
The second row is exactly twice the first row, so the rows are linearly dependent. A determinant with two proportional rows is always zero.
f′(x) = 3x² ≥ 0 with equality only at x = 0, and f is continuous. A function with nonnegative derivative that vanishes only at isolated points is strictly increasing, and here indeed x₁ < x₂ implies x₁³ < x₂³.
The coefficient of variation is (standard deviation/mean) × 100% = (4/20) × 100% = 20%. It expresses dispersion as a percentage of the mean and is unit-free.
The perpendicular distance is |4(1) + 3(2) − 5|/√(16 + 9) = |5|/5 = 1. The numerator is the absolute value of the line expression at the point.
For |r| < 1 the sum is a/(1 − r) = 8/(1 + 1/2) = 8/(3/2) = 16/3. The negative ratio increases the denominator rather than decreasing it.
From A² = A, any eigenvalue λ satisfies λ² = λ, so λ is 0 or 1. A nontrivial idempotent matrix has trace equal to its rank, so both eigenvalues 0 and 1 occur.
Write sin x + cos x = √2·sin(x + π/4), whose maximum is √2 since the sine factor peaks at 1. Squaring and using sin 2x ≤ 1 gives the same bound.