√ 5 + √ 125 + 25 [original problem - I assume 25 is not under a root]
√5 + √ 25·5 + 25 [find perfect square factors of 125]
√5 + 5√5 +25 [simplify √125]
6√5 + 25 [add 1√5 + 5√5 together]
***If the 25 is under a root, simply take it's square root (5) and add it to the end of the problem. The answer would be 6√5 + 5***
The domain of the function given are:
y=cot x
{x∈R: πn<x<π(n+1)} where n any integer
y=csc x
{x∈R: πn<x<π(n+1)} where n any integer
y=cos x
R (all real numbers)
y=sec x
{x∈R:π(n+1/2)<x<π(n+3/2)} where n is all integers.
from above we see that the the function that has a domain of all real numbers except npi is y=cot x and y=csc x
Z=k/x
2=k/3
k=6
k is constant so it will remain 6 now subsitute 13 in equation and solve
z=6/13
2.7 is already rounded to the nearest tenth.
Answer:
Step-by-step explanation:
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