Over the closed interval [3, 8], for which function can the extreme value theorem be applied? h (x) = StartFraction negative 2 Over 5 (x minus 4) squared EndFraction
h (x) = StartFraction (x minus 5) (x minus 1) Over x squared minus 25 EndFraction
h (x) = StartLayout Enlarged left-brace first row StartFraction 9 x Over 10 minus x EndFraction, x less-than 4 second row x + 2, x greater-than-or-equal-to 4 EndLayout
h (x) = StartLayout Enlarged left-brace first row negative x, x less-than 5 second row x squared minus 20, x greater-than-or-equal-to 5 EndLayout

x =±sqrt(1/3)

step-by-step explanation:

x+4/6x=1/x

multiply each side by 6x to get rid of the fractions

6x*(x+4/6x)=1/x*6x

distribute

6x^2 +4 = 6

subtract 4 from each side

6x^2 +4-4 = 6-4

6x^2 =2

divide by 6

6x^2/6 = 2/6

x^2 = 1/3

take the square root of each side

sqrt(x^2) = ±sqrt(1/3)

x =±sqrt(1/3)

9 x 8 divided by (5-3)=36

c.

step-by-step explanation:

she should of added 9^2 and 12^2. that is because it is a^2+b^2=c^2. c^2 is the hypotenuse and should always be bigger than the other two numbers. which is why she should of added 9 squared and 12 squared together instead of 15 squared.