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

-3

step-by-step explanation:

a drawing really does here:

so i draw the horizontal line y=2 and the vertical line x=-2 at this intersection i put b (this is where my right angle is going to be).   so point b is at (-2,2).   point a is 5 units long and is also on y=2 so this point has to be (3,2) or (-7,2) -doesn't matter i will go with (3,2).

now we know c is at (-2,y) we don't know the y yet bet we know that c is on x=-2.

we have the area of the triangle is .5bh=12.5

we know the base length is 5 (ab)

we have the height is (2-y) if you look at your

plug in

.5(5)(2-y)=12.5

solve for y to get the y-coordinate for c

2.5(2-y)=12.5

divide both sides by 2.5 giving:

2-y=5

now subtract 2 on both sides

-y=3

y=-3

answer: the distance

step-by-step explanation:

to calculate the distance d of a line segment with endpoints (x1, y1) and (x2, y2) use the formula d (x2 x1)2 (y2 y1)2. to calculate the midpoint of a line segment with endpoints (x1, y1) and (x2, y2) use the formula

b, i believe, but not sure.

lol, i have been using this product for a while and the second half of the year for a lon and the first time in a long time ago that the government has not yet received enough votes to be a good the first place in the world to be able to do it in my world life, i