What is the solution set of the quadratic inequality x^2-5< _0

-sqrt(5) ≤ x ≤ sqrt(5)

Step-by-step explanation:

x^2-5≤0

Add 5 to each side

x^2-5+5≤0+5

x^2 ≤5

Take the square root of each side, remembering to flip the inequality for the negative sign.  Since this is less than we use and in between

sqrt(x^2) ≤ sqrt(5)   and sqrt(x^2) ≥ -sqrt(5)  

x ≤ sqrt(5) and x ≥- sqrt(5)  

-sqrt(5) ≤ x ≤ sqrt(5)

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step-by-step explanation:

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

answer:

f(5/2) = 10/3

i hope this you!