Question 193752: Could you please help me with these problems. Thank you
Find each function value, if it exists.
f(t)=sqrt(t^2+1)
f(0)=
Simplify your answer. Type and exact answer, using radicals as needed. Type N if the square root is not a real number.
f(-2)=
Simplify your answer.
f(6)=
Simplify your answer
2. Simplify your answer. Type an exact answer, using radicals as needed.
root:3(a)(root:3(2a^2)+root:3(16a^2)
Answer by RAY100(1637)   (Show Source):
You can put this solution on YOUR website! y = sqrt(t^2+1)
,
y(0) = sqrt(0+1)=sqrt (1) = +/- 1
,
y(-2)= sqrt ( (-2)^2 +1) = sqrt (4+1) = sqrt (5)
,
y(6) = sqrt ( 6^2 +1) = sqrt (36 +1 ) = sqrt (37)
,
cube rt (a) * cube rt (2 a^2) * cube rt (16 a ^2)
,
cube rt ( a * 2a^2 * 16 a^2 )
,
cube rt ( 32 a^5)
,
cube rt (8*4*a^3*a^2)
,
cube rt (8*a^3) * cube rt (4a^2)
2 (a )cube rt (4a^2)
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