Volume of the parallelepiped spanned by ⟨1,0,0⟩, ⟨0,1,0⟩, ⟨1,1,1⟩ equals:
- 1 Correct answer.
- 0
- 2
- √3
Explanation
⟨0,1,0⟩×⟨1,1,1⟩=⟨1,0,−1⟩; then ⟨1,0,0⟩·⟨1,0,−1⟩=1. So the absolute value of the scalar triple product (parallelepiped volume) is 1.
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⟨0,1,0⟩×⟨1,1,1⟩=⟨1,0,−1⟩; then ⟨1,0,0⟩·⟨1,0,−1⟩=1. So the absolute value of the scalar triple product (parallelepiped volume) is 1.
A square root function is defined over real numbers when its radicand is non-negative: 2x - 8 >= 0 => 2x >= 8 => x >= 4. In interval notation, this is [4, infinity).
Complete the square: f(x) = (x - 3)^2 - 9 + 14 = (x - 3)^2 + 5. Since (x - 3)^2 >= 0 for all real x, the minimum value occurs at x = 3 with f(3) = 5. Therefore, the range is [5, infinity).
Set y = (2x + 1) / 3. Solve for x: 3y = 2x + 1 => 2x = 3y - 1 => x = (3y - 1) / 2. Swapping variables gives f^(-1)(x) = (3x - 1) / 2.
Direct substitution yields 0/0 indeterminate form. Factor the numerator: x^2 - 9 = (x - 3)(x + 3). Cancel (x - 3) for x != 3: lim_(x -> 3) (x + 3) = 3 + 3 = 6.
Apply the power rule d/dx[x^n] = n*x^(n-1) to each term: d/dx[4x^3] = 12x^2; d/dx[-5x^2] = -10x; d/dx[7x] = 7; d/dx[-9] = 0. Combining these gives f'(x) = 12x^2 - 10x + 7.
Find the antiderivative F(x): int (3x^2 - 2x + 1) dx = x^3 - x^2 + x. Apply the Fundamental Theorem of Calculus from 0 to 2: F(2) = (2)^3 - (2)^2 + 2 = 8 - 4 + 2 = 6. F(0) = 0. Therefore, the integral evaluates to 6 - 0 = 6.
At x = 3, y = (3)^2 - 4(3) + 1 = 9 - 12 + 1 = -2, so the point of tangency is (3, -2). The derivative is y' = 2x - 4. At x = 3, the slope is m = 2(3) - 4 = 2. Using point-slope form: y - (-2) = 2(x - 3) => y + 2 = 2x - 6 => y = 2x - 8.
Compute the first derivative: f'(x) = 6x^2 - 6x - 12 = 6(x^2 - x - 2) = 6(x - 2)(x + 1). Critical points are x = -1 and x = 2. Second derivative: f''(x) = 12x - 6. At x = -1: f''(-1) = -18 < 0, indicating a local maximum. Evaluate: f(-1) = 2(-1)^3 - 3(-1)^2 - 12(-1) + 5 = -2 - 3 + 12 + 5 = 12. (At x = 2, f''(2) = 18 > 0 gives the local minimum f(2) = -15).
Use the fundamental trigonometric limit lim_(u -> 0) (sin(u) / u) = 1. Rewrite the expression: (sin(5x) / (3x)) = (5/3) * (sin(5x) / (5x)). Taking the limit as x -> 0 gives (5/3) * 1 = 5/3.
Use the quotient rule: f'(x) = [(2x)(x - 1) - (x^2 + 1)(1)] / (x - 1)^2. Simplify the numerator: 2x^2 - 2x - x^2 - 1 = x^2 - 2x - 1. Substitute x = 2: f'(2) = ((2)^2 - 2(2) - 1) / (2 - 1)^2 = (4 - 4 - 1) / 1 = -1.
Use the half-angle formula cos^2(x) = (1 + cos(2x)) / 2: int_0^(pi/2) cos^2(x) dx = int_0^(pi/2) (1/2 + (1/2)*cos(2x)) dx = [x/2 + (1/4)*sin(2x)]_0^(pi/2). Evaluating at the limits: (pi/4 + (1/4)*sin(pi)) - (0 + (1/4)*sin(0)) = pi/4 + 0 - 0 = pi/4.
The natural logarithm is defined strictly for positive arguments: (x - 2) / (5 - x) > 0. Multiply numerator and denominator by -1 (or rewrite): (x - 2) / (x - 5) < 0. The critical numbers are x = 2 and x = 5. Testing intervals: for x < 2, (-)/(-) = + > 0; for 2 < x < 5, (+)/(-) = - < 0 (satisfying the inequality); for x > 5, (+)/(+) = + > 0. Thus the domain is (2, 5).
The slope of the line y = 4x - 5 is m = 4. Two lines are parallel when their slopes are equal, so we set the derivative y' = 2x equal to 4: 2x = 4 => x = 2. The corresponding y-coordinate on the parabola is y = (2)^2 = 4. Thus the point is (2, 4).
Find critical points by setting the derivative to zero: f'(x) = 1 - 4/x^2 = 0 => x^2 = 4 => x = 2 (since x in [1, 4]). Evaluate f at the critical point and endpoints: f(1) = 1 + 4/1 = 5; f(2) = 2 + 4/2 = 4; f(4) = 4 + 4/4 = 5. The absolute minimum value on [1, 4] is 4.
Find the x-intercepts: 4 - x^2 = 0 => x = -2 and x = 2. Since the parabola is symmetric and non-negative on [-2, 2], the area is A = int_(-2)^2 (4 - x^2) dx = 2 * int_0^2 (4 - x^2) dx = 2 * [4x - x^3/3]_0^2 = 2 * (8 - 8/3) = 2 * (16/3) = 32/3.
Apply the product rule: f'(x) = d/dx[e^(2x)] * cos(x) + e^(2x) * d/dx[cos(x)] = 2*e^(2x)*cos(x) - e^(2x)*sin(x). Evaluating at x = 0: f'(0) = 2*e^0*cos(0) - e^0*sin(0) = 2(1)(1) - (1)(0) = 2.
Let x and y denote the side lengths: 2x + 2y = 24 => x + y = 12 => y = 12 - x. The area is A(x) = x(12 - x) = 12x - x^2. To find the maximum, take the derivative: A'(x) = 12 - 2x = 0 => x = 6 cm. Then y = 12 - 6 = 6 cm (a square). The maximum area is A(6) = 6 * 6 = 36 cm^2.
Use the identity 1 - cos(4x) = 2*sin^2(2x): lim_(x -> 0) 2*sin^2(2x) / x^2 = 2 * lim_(x -> 0) [sin(2x) / x]^2 = 2 * [2 * lim_(x -> 0) (sin(2x) / (2x))]^2 = 2 * (2)^2 = 2 * 4 = 8. (Alternatively, apply L'Hôpital's rule twice: lim 4*sin(4x)/(2x) = lim 16*cos(4x)/2 = 8).
Use substitution: let u = 1 - x^2, so du = -2x dx => x dx = -du / 2. When x = 0, u = 1; when x = 1, u = 0. The integral becomes: int_1^0 sqrt(u) * (-du / 2) = (1/2) * int_0^1 u^(1/2) du = (1/2) * [ (2/3) * u^(3/2) ]_0^1 = (1/2) * (2/3) = 1/3.
Differentiate implicitly with respect to x: 3x^2 + 3y^2 * y' - 3(y + x * y') = 0. Divide by 3: x^2 + y^2 * y' - y - x * y' = 0 => y'(y^2 - x) = y - x^2 => y' = (y - x^2) / (y^2 - x). Substitute (x, y) = (1, 2): y' = (2 - 1^2) / (2^2 - 1) = (2 - 1) / (4 - 1) = 1/3.
There are 6 equally likely outcomes. The favorable outcomes greater than 4 are 5 and 6, so P = 2/6 = 1/3.
Order does not matter in a committee, so use combinations: C(8, 3) = 8! / (3! * 5!) = (8 * 7 * 6) / (3 * 2 * 1) = 336 / 6 = 56.
The expected value is E(X) = sum of x * P(X = x) = 0 * 0.2 + 1 * 0.5 + 2 * 0.3 = 0 + 0.5 + 0.6 = 1.1.
E(X) = 1 * 0.5 + 2 * 0.25 + 4 * 0.25 = 0.5 + 0.5 + 1 = 2. E(X^2) = 1 * 0.5 + 4 * 0.25 + 16 * 0.25 = 0.5 + 1 + 4 = 5.5. Therefore Var(X) = 5.5 - 2^2 = 1.5.
Use the complement. Both miss with probability (1 - 0.8)(1 - 0.7) = 0.2 * 0.3 = 0.06, because the shots are independent. So P(at least one hit) = 1 - 0.06 = 0.94.
By the law of total probability, P(defective) = P(A) * P(defective | A) + P(B) * P(defective | B) = 0.6 * 0.02 + 0.4 * 0.05 = 0.012 + 0.020 = 0.032.
By the law of total probability, P(positive) = 0.01 * 0.9 + 0.99 * 0.05 = 0.009 + 0.0495 = 0.0585. By Bayes' formula, P(disease | positive) = P(positive | disease) * P(disease) / P(positive) = 0.009 / 0.0585 = 2/13, about 0.15. Because the disease is rare, most positive results come from the much larger healthy group.
First sort the data: 3, 4, 5, 7, 9, 12. With an even number of values (6), the median is the mean of the two middle values: (5 + 7) / 2 = 6.
Fill the four positions in order: 5 choices for the first digit, 4 for the second, 3 for the third, and 2 for the fourth. By the multiplication principle, 5 * 4 * 3 * 2 = 120 numbers (the number of arrangements A(5, 4)).
There are 2^4 = 16 equally likely sequences. The number of sequences with exactly 2 heads is C(4, 2) = 6, so P = 6/16 = 3/8. (Binomial formula: C(4, 2) * (1/2)^2 * (1/2)^2 = 6/16.)
Use the complement. The probability of no 6 on either roll is (5/6) * (5/6) = 25/36, so P(at least one 6) = 1 - 25/36 = 11/36.
Weight each mean by its class size: total points = 20 * 70 + 30 * 80 = 1,400 + 2,400 = 3,800. The combined mean is 3,800 / 50 = 76.
The mean is (2 + 4 + 4 + 4 + 5 + 5 + 7 + 9) / 8 = 40 / 8 = 5. The squared deviations are 9, 1, 1, 1, 0, 0, 4, 16, which sum to 32. The variance is 32 / 8 = 4.
P(first red) = 4/10. After one red card is removed, P(second red | first red) = 3/9. So P(both red) = (4/10) * (3/9) = 12/90 = 2/15. Equivalently, C(4, 2) / C(10, 2) = 6/45 = 2/15.
The probabilities sum to 1, so a + b = 0.8. The expected value gives 1 * a + 2 * b + 5 * 0.2 = 2.4, so a + 2b = 1.4. Subtracting the first equation from the second gives b = 0.6, and then a = 0.8 - 0.6 = 0.2. Check: 0.2 + 1.2 + 1.0 = 2.4.
Use the power rule: d/dx x^n = n x^(n - 1). Thus d/dx 3x^2 = 3 x 2x = 6x.
The tangent slope is the derivative value at the point. Since f'(x) = 2x, f'(3) = 6.
The basic trigonometric derivative is d/dx(sin x) = cos x. This is one of the standard derivatives used throughout differential calculus.
Use the product rule: f'(x) = 2x e^x + x^2 e^x. At x = 1, this becomes 2e + e = 3e.
Write f(x) = ln x x^{-1}. Then f'(x) = (1/x)x^{-1} + ln x(-x^{-2}) = (1 - ln x) / x^2. Since ln e = 1, f'(e) = 0.
At x = 2, the point is (2, 8). The derivative is 3x^2, so the slope at x = 2 is 12. Using y - 8 = 12(x - 2) gives y = 12x - 16.
Use the chain rule. The derivative of sin u is cos u times u', and for u = 3x, u' = 3, so the derivative is 3cos(3x).
The derivative of arctan x is 1/(1 + x^2). At x = 1, this equals 1/(1 + 1) = 1/2.
Each differentiation of e^{2x} multiplies by 2. After n differentiations, the factor is 2^n, so the nth derivative is 2^n e^{2x}.
Velocity is s'(t) = 3t^2 - 3, and acceleration is s''(t) = 6t. At t = 2, the acceleration is 12.
For f(x) = sqrt(x), f(4) = 2 and f'(x) = 1/(2sqrt(x)), so f'(4) = 1/4. With dx = 0.1, the approximation is 2 + (1/4)(0.1) = 2.025.
At x = e^2, the point is (e^2, 2). The derivative of ln x is 1/x, so the slope there is 1/e^2. Thus the tangent line is y - 2 = (x - e^2)/e^2.
The width is 2a and the height is 12 - a^2, so A(a) = 2a(12 - a^2) = 24a - 2a^3. A'(a) = 24 - 6a^2, so the positive critical point is a = 2. The maximum area is 2(2)(12 - 4) = 32.
An antiderivative of 2x is x^2. Evaluating from 0 to 3 gives 3^2 - 0^2 = 9.