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New answer posted
a year agoContributor-Level 9
Two drawn cards are spades. There are 50 cards left.
The missing card could be a spade or not a spade.
P (missing card is spade) = 11/50 (since 11 spades remain out of 50 cards).
P (missing card is not spade) = 1 - 11/50 = 39/50
New answer posted
a year agoContributor-Level 10
Digits are 1, 3, 5, 7, 9. We need to form a 6-digit number where exactly one digit is repeated.
Choose the digit to be repeated:? C? ways.
Choose the positions for these two repeated digits:? C? ways.
Arrange the remaining 4 distinct digits in the remaining 4 places:? P? = 4! ways.
Total numbers =? C? *? C? * 4! = 5 * 15 * 24 = 1800.
The solution in the image 5/2 (6!) seems to follow a different logic which is unclear. 5 * (6!/2) = 5 * 360 = 1800. This logic is: choose one of 5 digits to repeat. Arrange the 6 digits, and since two are identical, divide by 2!
New answer posted
a year agoContributor-Level 9
A (-3, -6,1), B (2,4, -3). Point P divides AB in ratio k:1.
P = [ (2k-3)/ (k+1), (4k-6)/ (k+1), (-3k+1)/ (k+1)]
P lies on the plane lx + my + nz = 0.
l (2k-3) + m (4k-6) + n (-3k+1) = 0
k (2l + 4m - 3n) = 3l + 6m - n
⇒ k = (3l + 6m - n) / (2l + 4m - 3n)
Plane contains the line (x-1)/-1 = (y+4)/2 = (z+2)/3.
The plane passes through (1, -4, -2) and its normal is perpendicular to the line's direction vector.
-l + 2m + 3n = 0
l (1) + m (-4) + n (-2) = 0 ⇒ l - 4m - 2n = 0
Solving these gives l/-8 = m/-1 = n/-2. Let l=8, m=1, n=2.
k = (3 (8) + 6 (1) - 2) / (2 (8) + 4 (1) - 3 (2) = (24+6-2)/ (16+4-6) = 28/14 = 2.
New answer posted
a year agoContributor-Level 9
S? = ∑ tan? ¹ (6? / (2²? ¹ + 3²? ¹) from r=1 to k. (Assuming n in image is r)
t? = tan? ¹ (6? / (2²? ¹ + 3²? ¹)
= tan? ¹ ( (3/2) * (3/2)^ (2r) / ( (9/4) + (3/2)^ (2r+2) (This seems overly complex. Let's re-examine the image's simplification).
t? = tan? ¹ (6? / (2 * 4? + 3 * 9? ). The image simplifies the denominator to 2²? ¹ + 3²? ¹, which is different. Following the image's next step:
t? = tan? ¹ [ 6? / ( 1 + (3/2)^ (2r+1) ] (This denominator is incorrect).
The image seems to simplify t? into:
t? = tan? ¹ (3/2)? ¹) - tan? ¹ (3/2)? )
S? = [tan? ¹ (3/2)²) - tan? ¹ (3/2)] + [tan? ¹ (3/2)³) - tan? ¹ (3/2)²)] + . + [t
New answer posted
a year agoContributor-Level 10
Expression = (49)¹²? - 1) / 48
This uses the sum of a geometric series or a? - b? factorization.
(x? - 1) / (x - 1) = 1 + x + x² + . + x? ¹.
Let x = 49. (49¹²? - 1)/48 is an integer.
The solution shows (49? ³-1) (49? ³+1) / 48. This is correct factorization. Since 49 is odd, 49? ³ is odd. So 49? ³-1 and 49? ³+1 are consecutive even numbers. One is divisible by 2, the other by 4, so their product is divisible by 8. Also, 49 ≡ 1 (mod 3), so 49? ³-1 is divisible by 3. Hence the numerator is divisible by 24. It is also divisible by 48.
New answer posted
a year agoContributor-Level 9
(1 - x + x²)³? = ∑ a? x? (from j=0 to 3n)
= a? + a? x + a? x² + . + a? x³? (I)
Let A = a? + a? + a? + .
Let B = a? + a? + a? + .
In (I) put x = 1: (1 - 1 + 1)³? = 1.
1 = a? + a? + a? + a? + . (A + B = 1)
In (I) put x = -1: (1 - (-1) + (-1)²)³? = 3³?
3³? = a? - a? + a? - a? + . (A - B = 3³? )
(This seems incorrect based on the provided solution. Following the image:)
In (I) put x = -1, (1+1+1)^n = 1. (There must be a typo in the original problem, probably (1-x+x²)^n).
Assuming (1-x+x²)^n. Put x=-1 gives 3^n.
The provided text says putting x=-1 gives 1.
1 = a? - a? + a? - a? + .
Adding the two equations: 2 = 2 (a? + a? + a? + .) = 2A
New answer posted
a year agoContributor-Level 9
A = [i, -i], [-i, i]
A² = [-2, 2], [2, -2]
A? = [8, -8], [-8, 8]
A? = [-128, 128], [128, -128]
A? [x, y]? =?
-128x + 128y = 8 ⇒ -16x + 16y = 1 ⇒ x - y = -1/16 (I)
128x - 128y = 64 ⇒ 16x - 16y = 8 ⇒ x - y = 1/2 (II)
System is inconsistent hence No solution
New answer posted
a year ago
Contributor-Level 10
The angle bisector a is parallel to λ ( b? +? ) or μ ( b? -? ).
b? = (i+j)/√2 and? = (i-j+4k)/√ (1+1+16) = (i-j+4k)/ (3√2).
Case 1: a = λ ( (i+j)/√2 + (i-j+4k)/ (3√2) )
a = λ/√2 * (3 (i+j) + (i-j+4k)/3 = λ/ (3√2) * (4i + 2j + 4k).
a is given as αi + 2j + βk.
Comparing the j-component: 2 = λ/ (3√2) * 2 ⇒ λ = 3√2.
So, a = 1 * (4i + 2j + 4k) = 4i + 2j + 4k.
Comparing with αi + 2j + βk, we get α = 4 and β = 4.
(The image has a second case that needs evaluation as well).
Case 2: a = μ ( b? -? )
a = μ/ (3√2) * (3 (i+j) - (i-j+4k) = μ/ (3√2) * (2i + 4j - 4k).
Comparing the j-component: 2 = μ/ (3√2) * 4 ⇒ 4μ
New answer posted
a year agoContributor-Level 9
Equation of chord of x² + y² = 25 with mid point (h, k) is xh + yk = h² + k².
Or, y = (-h/k)x + (h² + k²)/k.
If this touches the ellipse x²/9 + y²/16 = 1, then the condition for tangency c² = a²m² + b² must be satisfied.
Here, m = -h/k, c = (h²+k²)/k, a²=9, b²=16.
(h² + k²)/k)² = 9 (-h/k)² + 16
(h² + k²)²/k² = 9h²/k² + 16
⇒ (h² + k²)² = 9h² + 16k²
∴ Required locus (x² + y²)² = 9x² + 16y².
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