Match the following.
| Column I | Column II |
| (i) The additive inverse of 2 |
(A) 0 |
| (ii) The greatest negative integer |
(B) - 2 |
| (iii) The smallest integer greater than every negative integer |
(C) 2 |
| (iv) Sum of predecessor and successor of 1 |
(D) - 1 |
A (i)→(C), (ii)→(D), (iii)→(A), (iv)→(B)
B (i)→(B), (ii)→(D), (iii)→(A), (iv)→(C)
C (i)→(C), (ii)→(D), (iii)→(B), (iv)→(A)
D (i)→(B), (ii)→(C), (iii)→(A), (iv)→(D)
Ans 1:
Class : Class 7
answer is B. it is actually a bit trcky as A and B are in same colour So, we need to focus a bit
Ans 3:
Class : Class 6
IT IS A. GREATEST NEGATIVE INTEGER IS 1, 0 2=2 PREDECESSOR SUCCESOR OF 1 , ADDITIVE INVERSE OF 2 IS -2 AND SMALLEST NUMBER GREATER THAN EVERY NEGATIVE INTEGER IS 0
Ans 7:
Class : Class 6
The problem is that A and B Are both in orange.. Is it just me or is it for everyone?
Post Your Answer
Post Your Answer
Post Your Answer
=
Ans 3:
Class : Class 5
this question is not formed correctly, I mean that it has no question. Silly soffffff
Post Your Answer
is equal to ______.
Post Your Answer
Ans 1:
Class : Class 6
3 ∠A = 4 ∠B = 6 ∠CL.C.M. of 3, 4 and 6 = 12We know, ∠A ∠B ∠C =180°12 ∠A 12 ∠B 12 ∠C = 12 × 180°12 ∠A 3 (4 ∠B) 2 (6 ∠C) = 12 × 180°12 ∠A 3 (3 ∠A) 2 (3 ∠A) = 12 × 180°12 ∠A 9 ∠A 6 ∠A = 12 × 180°27 ∠A = 12 × 180°∠A = 12 × 180° / 27 = 80°∠B = 3⁄4 ∠A = 3⁄4 of 80° = 60°∠C = 3⁄6 ∠A = 1⁄2 of 80° = 40°Therefore, ∠A ∠B - 3 ∠C = 80° 60° - 3 × 40°= 140° - 120°= 20°Answer-(C) 20°
Post Your Answer
equals _______.
Post Your Answer