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

c. 27 units.

step-by-step explanation:

we are told that in circle a, ∠bae ≅ ∠dae.

we can see from our given diagram that in and ;

, as these are radii of our given circle.

therefore, by sas congruence postulate.

hence, side be will be equal to side de as corresponding parts of congruent triangles are congruent. so we can set an equation to find the value of x as:

now let us substitute x=17 in the expression for the length of be.

therefore, length of be will be 27 units and option c is the correct choice.

he answer is x=1x3 because x=1 to x=3

he is going to fill two shelves and the third shelf will only have five video games

show us the triangle so we can answer it