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New answer posted
2 months agoContributor-Level 10
Given the equations:
t? (A + 2B) = -1 which expands to t? (A) + 2t? (B) = -1 . (I)
t? (2A - B) = 3 which expands to 2t? (A) - t? (B) = 3 . (II)
Solving equations (I) and (II) simultaneously, we get:
t? (A) = 1
t? (B) = -1
Therefore, t? (A) - t? (B) = 1 - (-1) = 2.
New answer posted
2 months agoContributor-Level 10
The general equation of a circle is given by:
az z? + α? z + αz? + d = 0
This can be rewritten as:
z? + (α? /a)z + (α/a)z? + d/a = 0
From this, we can identify the centre and radius:
Centre = -α/a
Radius = √ (|-α/a|² - d/a)
For a real circle to exist, the term under the square root must be non-negative:
|-α/a|² - d/a ≥ 0
|α|²/|a|² - d/a ≥ 0
|α|² - ad ≥ 0, where a ∈ R - {0}.
New answer posted
2 months agoContributor-Level 10
f (x) = (cos (sin x) - cos x) / x? We need lim (x→0) f (x) = 1/k.
Using cos C - cos D = -2 sin (C+D)/2) sin (C-D)/2).
f (x) = -2 sin (sin x + x)/2) sin (sin x - x)/2) / x?
For small x, sin x ≈ x.
lim (x→0) f (x) = lim -2 * ( (sin x + x)/2 ) * ( (sin x - x)/2 ) / x?
Using series expansion: sin x = x - x³/3! + x? /5! - .
sin x + x = 2x - x³/6 + .
sin x - x = -x³/6 + x? /120 - .
f (x) ≈ -2 * ( (2x)/2 ) * ( (-x³/6)/2 ) / x?
≈ -2 * (x) * (-x³/12) / x?
≈ (2x? /12) / x? = 2/12 = 1/6.
So, 1/k = 1/6 ⇒ k = 6.
New question posted
2 months agoNew answer posted
2 months agoContributor-Level 10
The equation of the plane passing through the line of intersection is:
(2x - 7y + 4z - 3) + λ (3x - 5y + 4z + 11) = 0.
The plane passes through the point (-2, 1, 3).
(2 (-2) - 7 (1) + 4 (3) - 3) + λ (3 (-2) - 5 (1) + 4 (3) + 11) = 0
(-4 - 7 + 12 - 3) + λ (-6 - 5 + 12 + 11) = 0
(-2) + λ (12) = 0 ⇒ 12λ = 2 ⇒ λ = 1/6.
Substitute λ back into the equation:
(2x - 7y + 4z - 3) + (1/6) (3x - 5y + 4z + 11) = 0
Multiply by 6:
6 (2x - 7y + 4z - 3) + (3x - 5y + 4z + 11) = 0
12x - 42y + 24z - 18 + 3x - 5y + 4z + 11 = 0
15x - 47y + 28z - 7 = 0.
This is the equation ax + by + cz - 7 = 0.
So, a=15, b=-47, c=28.
We need to find the value of 2a + b
New question posted
2 months agoTaking an Exam? Selecting a College?
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