The value of the determinant |2 3; 4 5| is:
- 22
- -2 Correct answer.
- 10
- 2
Explanation
For a 2×2 determinant |a b; c d| the value is ad − bc. Here that is (2)(5) − (3)(4) = 10 − 12 = −2.
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For a 2×2 determinant |a b; c d| the value is ad − bc. Here that is (2)(5) − (3)(4) = 10 − 12 = −2.
The modulus of a + bi is √(a² + b²). For 3 − 4i this gives √(9 + 16) = √25 = 5.
Slope equals the change in y divided by the change in x: (11 − 2)/(4 − 1) = 9/3 = 3.
By the chain rule, d/dx sin(u) = cos(u)·du/dx with u = 3x, so du/dx = 3. Hence the derivative is 3cos(3x).
An antiderivative of x² is x³/3. Evaluating from 0 to 1 gives 1/3 − 0 = 1/3.
The principal range of sin⁻¹ is [−π/2, π/2], and sin(π/6) = 1/2. Therefore the principal value is π/6.
Order is the highest derivative present, which is the second derivative, so order = 2. Degree is the power of that highest derivative after the equation is polynomial in derivatives, which is 3.
Using the standard limit lim(t→0)(sin t)/t = 1 with t = 5x, we get (sin 5x)/x = 5·(sin 5x)/(5x) → 5·1 = 5.
Separating variables gives dy/y = dx, and integrating both sides yields ln|y| = x + k. Exponentiating gives y = Ceˣ.
For an n×n matrix, |kA| = kⁿ|A|. Here n = 3 and k = 2, so |2A| = 2³ × 4 = 8 × 4 = 32.
The general term is C(6,r)xʳ, so the coefficient of x³ is C(6,3) = 6!/(3!3!) = 20.
Here the first term a = 1 and the common ratio r = 1/3, with |r| < 1. The sum is a/(1 − r) = 1/(1 − 1/3) = 1/(2/3) = 3/2.
Here a² = 25 and b² = 9, so a = 5, b = 3 and c = √(a² − b²) = √16 = 4. The eccentricity is e = c/a = 4/5.
f′(x) = 3x² − 3 = 0 gives x = ±1. Evaluating: f(−1) = 2, f(1) = −2, f(−2) = −2, f(2) = 2. The maximum value over the interval is 2.
Integrating by parts with u = x and dv = eˣ dx gives x·eˣ − ∫eˣ dx = x·eˣ − eˣ + C.
The dot product is the sum of products of corresponding components: (2)(1) + (3)(−1) + (−1)(2) = 2 − 3 − 2 = −3.
For a binomial distribution the mean is np. Here np = 10 × 0.3 = 3.
For an upper triangular matrix the determinant equals the product of the diagonal entries. The product 1 × 1 × 1 = 1.
The slope dy/dx = 2x equals 4 at x = 2. Using point-slope form: y − 4 = 4(x − 2), which simplifies to y = 4x − 4.
An antiderivative of sin x is −cos x. Evaluating −cos x from 0 to π/2 gives (−cos(π/2)) − (−cos 0) = 0 + 1 = 1.
There are 36 equally likely ordered outcomes. The pairs summing to 7 are (1,6),(2,5),(3,4),(4,3),(5,2),(6,1), six in all, so the probability is 6/36 = 1/6.
Matrix multiplication sums products of rows of A with columns of A. Row 1 of A² is (1×1+2×3, 1×2+2×4) = (7, 10); row 2 is (3×1+4×3, 3×2+4×4) = (15, 22).
Using 1 − cos 2x = 2sin²x, the quotient becomes 2(sin x/x)². Since sin x/x → 1, the limit is 2 × 1² = 2.
The curves meet where x² = x, i.e. x = 0 and x = 1. The area is ∫₀¹ (x − x²) dx = [x²/2 − x³/3]₀¹ = 1/2 − 1/3 = 1/6.
Let f(x) = x³ − 3x + 1. Its derivative 3x² − 3 vanishes at x = ±1, where f(−1) = 3 and f(1) = −1. Since the local maximum is positive and the local minimum is negative, the cubic crosses the x-axis three times, giving three distinct real roots.
For a 2×2 matrix [[a, b], [c, d]] the determinant is ad − bc. Here that gives (2)(3) − (1)(4) = 6 − 4 = 2.
The modulus is √(3² + 4²) = √25 = 5. Since both real and imaginary parts are positive, the argument is tan⁻¹(4/3), measured from the positive real axis.
The first term is a = 3 and the common difference is d = 4. The nth term is a + (n − 1)d, so the 5th term is 3 + 4·4 = 19.
Differentiating gives 2y·dy/dx = 8, so dy/dx = 4/y, which equals 1 at (2, 4). The normal is perpendicular to the tangent, so its slope is the negative reciprocal, −1.
Substituting u = 2x + 3 gives du = 2 dx, so the integral becomes ½∫u⁴ du = u⁵/10 + C. Replacing u restores (2x + 3)⁵/10 + C.
The dot product is (1)(2) + (2)(1) + (2)(−2) = 0. Since neither vector is zero, a zero dot product means the vectors are perpendicular, so the angle is 90°.
Two-digit numbers run from 10 to 99, giving 90 in total. The multiples of 7 in this range are 14, 21, …, 98, which number 13. The probability is therefore 13/90.
The mean is 6, and the squared deviations are 16, 4, 0, 4, 16, whose sum is 40. The variance is 40/5 = 8, so the standard deviation is √8 = 2√2.
Using tan⁻¹a + tan⁻¹b = tan⁻¹((a + b)/(1 − ab)) for ab < 1 gives tan⁻¹(1) + tan⁻¹(2) = tan⁻¹(−3), taken as 3π/4 in the principal range. Adding tan⁻¹(3) then gives 3π/4 + tan⁻¹(3) = π.
Consecutive integer roots differ by 1, so the discriminant (difference of roots)² equals 1. Since the discriminant is p² − 4q, its value is 1.
The radius is the distance from (2, −3) to (0, 0), namely √13. Expanding (x − 2)² + (y + 3)² = 13 and simplifying gives x² + y² − 4x + 6y = 0, consistent with passing through the origin.
By the chain rule, differentiate the outer sine to get cos(log x) and multiply by the derivative of log x, which is 1/x. The result is cos(log x)/x.
The antiderivative of 1/x is log x, so the definite integral equals log e − log 1 = 1 − 0 = 1.
Treat the two L's as one block, giving 4 units: the block, E, V, E. These can be arranged in 4!/2! = 12 ways because the two E's are identical, and the L's inside the block have only one order.
For [[a, b], [c, d]] the determinant is ad − bc = (2)(4) − (3)(1) = 8 − 3 = 5.
Setting f′(x) = 2x − 4 = 0 gives x = 2, and f″(x) = 2 > 0 confirms a minimum. Substituting gives f(2) = 4 − 8 + 7 = 3, so the minimum point is (2, 3).
Divide numerator and denominator by x² to get (3 + 2/x + 1/x²)/(1 − 5/x²). As x → ∞, the terms with x in the denominator vanish, leaving 3/1 = 3.
Substitute u = x², so du = 2x dx and x dx = du/2. The integral becomes ½∫₀¹ e^u du = ½(e − 1).
The vector AB equals the position vector of B minus that of A: (1 − 2)i + (1 − (−1))j + (2 − 1)k = −i + 2j + k.
For a binomial distribution, mean = np = 4 and variance = np(1 − p) = 3. Dividing gives 1 − p = 3/4, so p = 1/4, and then n = 4/(1/4) = 16.
The principal value of sin⁻¹ lies in [−π/2, π/2]. Since sin(5π/6) = sin(π/6) = ½, the principal value is π/6.
The auxiliary equation is m² − 5m + 6 = 0, with roots m = 2 and m = 3. Distinct real roots give the solution y = C₁e^(2x) + C₂e^(3x).
The given line has slope −2/3, so a perpendicular line has slope 3/2. Using point-slope form with (1, 2): y − 2 = (3/2)(x − 1), which simplifies to 3x − 2y + 1 = 0.
The general term is C(7, r)x^(7−r)2^r. For the power x⁵ take r = 2, giving C(7, 2)·2² = 21·4 = 84.
The determinant is 1·4 − 2·3 = −2, which is nonzero, so A is invertible. The inverse is (1/−2)·[[4, −2], [−3, 1]] = [[−2, 1], [3/2, −1/2]].