10. The Cartesian product A × A has 9 elements among which are found (–1, 0) and (0,1). Find the set A and the remaining elements of A × A.
10. The Cartesian product A × A has 9 elements among which are found (–1, 0) and (0,1). Find the set A and the remaining elements of A × A.
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1 Answer
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10. Given, n (A × A)=9
n (A) ×n (A) = 9.
n (A)2 = 32.
n (A) = 3 .
And (–1,0), (0,1) A × A i.e., A × A = { (x, y), x A, y B}
? A= {–1,0,1}
And A × A= {–1,0,1} × {–1,0,1}
= { (–1, –1), ( –1,0), ( –1,1), (0, –1), (0,0), (0,1), (1, –1), (1,0), (1,1)}
Similar Questions for you
Total number of possible relation =
Favourable relations =
Probability =
Circle S? : x² + y² - 10x - 10y + 41 = 0.
Center C? = (5, 5). Radius r? = √ (5² + 5² - 41) = √ (25 + 25 - 41) = √9 = 3.
Circle S? : x² + y² - 16x - 10y + 80 = 0.
Center C? = (8, 5). Radius r? = √ (8² + 5² - 80) = √ (64 + 25 - 80) = √9 = 3.
The solution checks if the center of one circle lies on the other.
Put C? (8, 5) into S? : 8² + 5² - 10 (8) - 10 (5) + 41 = 64 + 25 - 80 - 50 + 41 = 130 - 130 = 0. So C? lies on S?
Put C? (5, 5) into S? : 5² + 5² - 16 (5) - 10 (5) + 80 = 25 + 25 - 80 - 50 + 80 = 130 - 130 = 0. So C? lies on S?
This means bo
Kindly consider then following figure
36. Given, A={9,10,11,12,13}.
f(x)=the highest prime factor of n.
and f: A → N.
Then, f(9)=3 [? prime factor of 9=3]
f (10)=5 [? prime factor of 10=2,5]
f(11)=11 [? prime factor of 11 = 11]
f(12)=3 [? prime factor of 12 = 2, 3]
f(13)=13 [? prime factor of 13 = 13]
?Range of f=set of all image of f(x) = {3,5,11,13}.
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