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New answer posted
8 months agoContributor-Level 10
Limit (θ→0) [tan(πcos²θ) / sin(2πsin²θ)]
Let θ → 0. Then cos²θ → 1 and sin²θ → 0.
Let u = πsin²θ. As θ → 0, u → 0.
cos²θ = 1 - sin²θ = 1 - u/π.
The expression becomes:
Limit (u→0) [tan(π(1 - u/π)) / sin(2u)]
= Limit (u→0) [tan(π - u) / sin(2u)]
= Limit (u→0) [-tan(u) / sin(2u)]
= Limit (u→0) [-tan(u) / (2sin(u)cos(u))]
= Limit (u→0) [-(sin(u)/cos(u)) / (2sin(u)cos(u))]
= Limit (u→0) [-1 / (2cos²(u))] = -1 / (2 * 1²) = -1/2.
New answer posted
8 months agoContributor-Level 10
cos(x)(3sin(x) + cos(x) + 3)dy = (1 + ysin(x)(3sin(x) + cos(x) + 3))dx
This seems mistyped. A more likely form is:
dy/dx - (sin(x)/(cos(x)))y = 1 / (cos(x)(3sin(x) + cos(x) + 3))
dy/dx - tan(x)y = sec(x) / (3sin(x) + cos(x) + 3)
The integrating factor (I.F.) is:
I.F. = e^∫(-tan(x))dx = e^(ln|cos(x)|) = cos(x).
Multiplying by I.F.:
d(y*cos(x))/dx = 1 / (3sin(x) + cos(x) + 3)
y*cos(x) = ∫ dx / (3sin(x) + cos(x) + 3)
Using Weierstrass substitution, let t = tan(x/2):
sin(x) = 2t/(1+t²), cos(x) = (1-t²)/(1+t²), dx = 2dt/(1+t²)
∫ (2dt/(1+t²)) / (3(2t/(1+t²)) + (1-t²)/(1+t²) + 3)
= ∫ 2dt / (6t + 1 - t² + 3 + 3t²) = ∫ 2dt / (2t² + 6t
New answer posted
8 months agoContributor-Level 10
Given the function:
f(x) = { x(2 - sin(1/x)), if x ≠ 0
{ 0, if x = 0
For x < 0: f(x) = x(2 - sin(1/x))
For x > 0: f(x) = x(2 - sin(1/x))
The derivative f'(x) for x ≠ 0 is:
f'(x) = 1*(2 - sin(1/x)) + x*(-cos(1/x))*(-1/x²) = 2 - sin(1/x) + (1/x)cos(1/x)
The text calculates the derivative differently:
For x < 0: f'(x) = -2 + sin(1/x) - (1/x)cos(1/x)
For x > 0: f'(x) = 2 - sin(1/x) + (1/x)cos(1/x)
To check if f'(0) is defined, we would need to use the limit definition of the derivative at x=0. As x approaches 0, the term (1/x)cos(1/x) oscillates and does not approach a finite limit. Therefore, f'(0) is undefined.
New answer posted
8 months agoContributor-Level 10
P(O at even place) = 1/2, P('O' at odd place) = 1/3
P(1 at even place) = 1 - 1/2 = 1/2
P(1 at odd place) = 1 - 1/3 = 2/3
The probability that 10 is followed by 01 is given by a sequence of probabilities, which seems to represent a specific arrangement or transition. The calculation is:
P(10 is followed by 01) = (2/3) * (1/2) * (1/3) * (1/2) * (1/2) * (1/3) * (1/2) * (2/3) = 1/18 + 1/18 = 1/9.
The calculation seems incomplete or context is missing.
New answer posted
8 months agoContributor-Level 10
- Given vectors OP = xi + yj - k and OQ = -i + 2j + 3xk.
- PQ = OQ - OP = (-1 - x)i + (2 - y)j + (3x + 1)k
- Given |PQ| = √20, so |PQ|² = 20.
(-1 - x)² + (2 - y)² + (3x + 1)² = 20
(1 + x)² + (2 - y)² + (3x + 1)² = 20 .(i) - Given OP ⊥ OQ, so OP · OQ = 0.
(x)(-1) + (y)(2) + (-1)(3x) = 0
-x + 2y - 3x = 0 ⇒ -4x + 2y = 0 ⇒ y = 2x .(ii)
Substitute (ii) into (i):
(1 + x)² + (2 - 2x)² + (3x + 1)² = 20
1 + 2x + x² + 4 - 8x + 4x² + 9x² + 6x + 1 = 20
14x² = 14 ⇒ x² = 1 ⇒ x = ±1.
When x = 1, y = 2. When x = -1, y = -2.
So, (x, y) can be (1, 2) or (-1, -2).- Given that OR, OP, and OQ are coplanar, their scalar triple product is 0: [OR OP OQ]
- Given OP ⊥ OQ, so OP · OQ = 0.
New answer posted
8 months agoContributor-Level 10
The truth table for the logical expression (p ∧ q) → (p → q) is as follows:
p | q | p ∧ q | p → q | (p ∧ q) → (p → q) |
T | T | T | T | T |
T | F | F | F | T |
F | T | F | T | T |
F | F | F | T | T |
The final column shows that the expression is a tautology, meaning it is always true regardless of the truth values of p and q.
New answer posted
8 months agoContributor-Level 10
The equation of a plane parallel to x - 2y + 2z - 3 = 0 is x - 2y + 2z + λ = 0.
The distance from the point (1, 2, 3) to this plane is 1.
|1 - 2 (2) + 2 (3) + λ| / √ (1² + (-2)² + 2²) = 1
|1 - 4 + 6 + λ| / √9 = 1
|3 + λ| / 3 = 1
|3 + λ| = 3
3 + λ = 3 or 3 + λ = -3
λ = 0 or λ = -6.
New answer posted
8 months ago
Contributor-Level 10
1 = (2-1)¹ (The n is likely 1).
3? = (7-4)³ (This seems to be a pattern matching (a-b)^c).
4²? = (12-8)? ! = 4²?
The blank space must be (5-3)² = 2² = 4.
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