The area of the conference table in Mr. Nathan’s office must be no more than 175 ft2. If the length of the table is 18 ft more than the width, x, which interval can be the possible widths?

0 ft < w ≤ 7 ft

Step-by-step explanation:

A=wl,  where w, l >0

A≤175 ft²

l= w+18

w(w+18)≤175

w²+18w-175≤0

zeros of are -25 and 7

(w + 25)*(w - 7) ≤ 0

w+25≤0, w-7≥0 ⇒ w ⊂ ∅

w+25≥0, w-7≤0 ⇒ -25 ≤ w ≤ 7,

as w>0, we have 0 < w ≤ 7

mark other person brainliest

-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