Answer:
-6r + 5
Step-by-step explanation:
-4r - 2r + 5
= (-4 -2)r + 5
= -6r + 5
We are given the equation <span>an = (n^2/ sqrt(n^3+4n)) and asked to determine if the function is divergent or convergent. In this case, we find the limit of the function as n approaches infinity.
an = </span><span> (n^2/ sqrt(n^3+4n))
lim (n to infinity ) = infinity / infinty: ;indeterminate
Using L'hopitals rule, we derive
</span><span>lim (n to infinity ) = 2 n / 0.5* ( </span><span>n^3+4n) ^-0.5 * (3 n2 +4) : infinity / infinity
again, we derive
</span>lim (n to infinity ) = 2 (0.25) (( n^3+4n) ^-0.5))*(3 n2 +4) / 0.5* ( 6n + 4) :infinity / infinity
<span>
again,
</span>lim (n to infinity ) = 2 (0.25) (6n + 4) / 0.5* ( 6)* 0.5 <span>(( n^3+4n) ^-0.5))</span>
this goes on and the function is divergent
The answer to your problem is 66
Answer:
Step-by-step explanation:
we can write -1 for;
s(-1)= 4^3(-1)
4^-3 = 1/64
