Maths

The truth table for the logical expression (p ∧ q) → (p → q) is as follows:

The final column shows that the expression is a tautology, meaning it is always true regardless of the truth values of p and q.

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.

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.

Initial mean of 25 observations is 40.
X? = (Σx? )/25 = 40 => Σx? = 25 * 40 = 1000.

A teacher of age 60 retires.
The new sum of ages for the remaining 24 people is 1000 - 60 = 940.

A new teacher of age x joins.
The new sum for 25 people is 940 + x.
The new mean is 39.
(940 + x) / 25 = 39
940 + x = 39 * 25 = 975
x = 975 - 940 = 35.
The new teacher's age is 35.

Circle 1: x² + y² - 10x - 10y + 41 = 0
Center C? = (5,5).
Radius r? = √ (5² + 5² - 41) = √ (25+25-41) = √9 = 3.

Circle 2: x² + y² - 22x - 10y + 137 = 0
Center C? = (11,5).
Radius r? = √ (11² + 5² - 137) = √ (121+25-137) = √9 = 3.

Distance between centers d (C? , C? ) = √ (11-5)² + (5-5)²) = √ (6²) = 6.

Sum of radii r? + r? = 3 + 3 = 6.
Since the distance between the centers is equal to the sum of their radii, the circles touch externally at one point.

Given the equation y = 3 + 1/ (4 + 1/y).
y - 3 = 1 / (4y+1)/y)
y - 3 = y / (4y+1)
(y-3) (4y+1) = y
4y² + y - 12y - 3 = y
4y² - 11y - 3 = y
4y² - 12y - 3 = 0

Using the quadratic formula to solve for y:
y = [-b ± √ (b²-4ac)] / 2a
y = [12 ± √ (-12)² - 4*4* (-3)] / (2*4)
y = [12 ± √ (144 + 48)] / 8
y = [12 ± √192] / 8 = [12 ± 8√3] / 8 = 3/2 ± √3.
y = 1.5 ± √3.

Since y > 0 (from the structure of the equation), both solutions are positive. The solution selects y = 1.5 + √3.

A function f (x) is continuous at x=1, so lim (x→1? ) f (x) = lim (x→1? ) f (x) = f (1).
Assuming a piecewise function like f (x) = { -x, x<1; ax+b, x1 } (structure inferred from derivative).
Continuity at x=1: f (1) = 1. a (1)+b = 1 => a+b=1.

The function is differentiable at x=1. The derivative of f (x) at x=1 from the left is -1. The derivative from the right is a.
So, a = -1. (The image has 2a = -1, which would imply a function like -x and ax²+b). Let's assume f' (x) = 2a for x>1.
2a = -1 => a = -1/2.

From a+b=1, b = 1 - a = 1 - (-1/2) = 3/2.
So, a = -1/2 and b = 3/2.

Evaluate the limit:
L = lim (x→0) [sin? ¹ (x) - tan? ¹ (x)] / 3x³

Using Taylor series expansions around x=0:
sin? ¹ (x) = x + x³/6 + O (x? )
tan? ¹ (x) = x - x³/3 + O (x? )

L = lim (x→0) [ (x + x³/6) - (x - x³/3) ] / 3x³
L = lim (x→0) [ x³/6 + x³/3 ] / 3x³
L = lim (x→0) [ (1/6 + 1/3)x³ ] / 3x³
L = (1/2) / 3 = 1/6

The solution shows 3L = 1/2, which is correct. And 6L = 1, also correct.
The final line 6L+1=2 implies 6L=1, confirming the result.