L=Lim tan(x)^2/x x->0
Since both numerator and denominator evaluate to zero, we could apply l'Hôpital rule by taking derivatives.
d(tan^2(x))/dx=2tan(x).d(tan(x))/dx = 2tan(x)sec^2(x)
d(x)/dx = 1
=>
L=2tan(x)sec^2(x)/1 x->0
= (2(0)/1^2)/1
=0/1
=0
Another way using series,
We know that tan(x) = x+x^3/3+2x^5/15+.....
then tan^2(x), using binomial expansion gives
x^2+2*x^4/3+.... (we only need two terms)
and again apply l'Hôpital's rule, we have
L=d(x^2+2x^4/3+...)/d(x) = (2x+8x^3/3+...)/1
=0 as x->0
Answer:
(
, 3)
Step-by-step explanation:
I got 5/2 as my x variable and 3 for my y variable to being the midpoint. In which I added both my x variables provided ( 3 and 2 ) and then I divided them by two to get half of that (since you are looking for in between those two digits technically so you are wanting to find the midpoint or the middle of that line segment). I also did the same with the two y variables as well which were 7 and -1, then I divided both by two to get the variable 3.
Twenty seven minus ten plus four
Answer:
False
Step-by-step explanation:
first step is
x+y = 30
x-y = 12
1/3 divide by 1/4
flip the 2nd number and the sign from division to multiplication
1/3 x 4/1
multiply
4/3