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]
The answer of this problem is A because 3 times 21 is 63 and then you plus 1
which is 64
$25.12
Work:
23.99+3.60=27.59
27.59-4.51=23.08
23.08+0.95=24.03
24.03+2.17=26.20
26.20-1.08=25.12
Answer:x=20 y=70
Step-by-step explanation:
We have slope intercept form. Parallel means the same slope, and we get to choose the intercept to fit the new point (-6,0).
y = (1/3) x + b
0 = (1/3) (-6) + b
2 = b
Answer y = (1/3) x + 2 third choice
