I'm going to assume that the room is a rectangle.
The area of a rectangle is A = lw, where l=length of the rectangle and w=width of the rectangle.
You're given that the length, l = (x+5)ft and the width, w = (x+4)ft. You're also told that the area, A = 600 sq. ft. Plug these values into the equation for the area of a rectangle and FOIL to multiply the two factors:

Now subtract 600 from both sides to get a quadratic equation that's equal to zero. That way you can factor the quadratic to find the roots/solutions of your equation. One of the solutions is the value of x that you would use to find the dimensions of the room:

Now you know that x could be -29 or 20. For dimensions, the value of x must give you a positive value for length and width. That means x can only be 20. Plugging x=20 into your equations for the length and width, you get:
Length = x + 5 = 20 + 5 = 25 ft.
Width = x + 4 = 20 + 4 = 24 ft.
The dimensions of your room are 25ft (length) by 24ft (width).
Answer: In particular, let’s focus our attention on the behavior of each graph at and around . 2 and x= -1 for x < 2. There are open circles at both endpoints (2, 1) and (-2, 1). The third is h (x) = 1 / (x-2)^2, in which the function curves asymptotically towards y=0 and x=2 in quadrants one and two."
Step-by-step explanation: I think this is the problem ur on
a+b+c=9 and a2+b2+c2=35 then by using identity
(a+b+c)2=a2+b2+c2+2ab+2bc+2ca,we get the answer 9x9=35+2(ab+bc+ca)
2(ab+bc+ca)=9x9-35
2(ab+bc+ca) =81-35then we get ab+bc+ca=46/2 so ab+bc+ca=23
Answer:
2.5
Step-by-step explanation:
i did it and got it right