Answer:
2 and I do not play it.
Step-by-step explanation:
......
5/6 is closest to 1 because it it 1/6 away from 1 and is 5/6 away from 0 and is 2/6 away from 1/2 (referring 1/2 as 3/6) Hope this helps!
<span> -x^2 + -14x = 49
-(x^2 +14) = 49
-[(x+7)^2 -49] = 49
</span>
Answer:
A = 58.7 degrees
B = 66.9 degrees
C = 34.1 degrees
Step-by-step explanation:
<u><em>For <A:</em></u>
Tan A = 
Tan A = 
Tan A = 1.6
A = 
A = 58.7 degrees
<u>For <B:</u>
Sin B = 
Sin B = 
Sin B = 0.92
B = 
B = 66.9 degrees
<em><u>For <C:</u></em>
Sin C = 
Sin C = 
Sin C = 0.56
C = 
C = 34.1 degrees
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]