Well, assuming that you need to round to the nearest whole number, what is six closer to? 0, or 10? The answer is 10. Therefore, you would round up to 33. Hope this helps!
Answer:
77
Step-by-step explanation:
percentage Students
78%-------------- -273
100%-------------- ---X
(100x273)/78 = 350 students in class so
how many did not pass?
350-273 = 77 students not pass
Answer: 24 & 72
Step-by-step explanation: 24 + 72 = 96
24 x 72 = 1728
If T and V are complementary angles, their sum is 90°.
V + T = 90°
48° + (2X+10)° = 90° . . . . . . . substitute given information
2X + 58 = 90 . . . . . . . . . . . . .. collect terms
2X = 32 . . . . . . . . . . . . . . . . .. subtract 58
X = 16 . . . . . . . . . . . . . . . . . .. divide by 2
The value of X is 16.
18. If f(x)=[xsin πx] {where [x] denotes greatest integer function}, then f(x) is:
since x denotes the greatest integers which could the negative or the positive values, also x has a domain of all real numbers, and has no discontinuous point, then x is continuous in (-1,0).
Answer: B]
20. Given that g(x)=1/(x^2+x-1) and f(x)=1/(x-3), then to evaluate the discontinuous point in g(f(x)) we consider the denominator of g(x) and f(x). g(x) has no discontinuous point while f(x) is continuous at all points but x=3. Hence we shall say that g(f(x)) will also be discontinuous at x=3. Hence the answer is:
C] 3
21. Given that f(x)=[tan² x] where [.] is greatest integer function, from this we can see that tan x is continuous at all points apart from the point 180x+90, where x=0,1,2,3....
This implies that since some points are not continuous, then the limit does not exist.
Answer is:
A]
