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]
Answer:
−
2(
5
y
+
3
)
(
y
−
6
)
Step-by-step explanation:
hope it helps
14(4)÷(7)+5(17)-3 please help
218
X +(-5y)= 13
-x+ 9y =(-17)
0+<u>4y</u>=<u>-4</u>
4 4
y=1
x+-5(1)=13
x-5=13
+5 <u>+5
</u> x=18
y=1
Answer:
(7/2, 1)
Step-by-step explanation:
use the midpoint formula
