Answer:
g(t) - h(t) = -2t - 1
Step-by-step explanation:
g(t) - h(t) = t-3-3t+2
 = -2t -1
 
        
             
        
        
        
No, the radius is not 3.14 .  The radius is 1/2 of the diameter . . .  That's 6 feet.
If you see 3.14 on the sheet, then that's the number you're supposed to use for pi .
The area of any circle is (pi) x (radius squared).
If the table were a full circle, its area would be (pi) x (6 squared) = 36 pi square feet.
But it's only half of that . . . 18 pi = (18) x (3.14) = <u>56.52 square feet</u>.
That's called the "area" of the table, not the "square feet" of the table.
And another thing:   I see you're asking for the "closest" number.   Don't ask me 
how I know this, but I'm pretty sure that right under this question wherever you 
copied it from, there's a list of choices, and when you posted the question, for 
some reason you decided not to share the list.
        
             
        
        
        
the answer it four and one sixths
 
        
                    
             
        
        
        
Answer:
 (-4, 0) U (1, ∞)
Step-by-step explanation:
Set each factor EQUAL to zero to find the zeroes (since it is not actually equal to zero, you will use an open circle when graphing and an open bracket when writing in interval notation).
x = 0           x-1 = 0             x + 4 = 0
                   x = 1                x = -4
Next, choose a value to the far left, between each of the zeroes, and to the far right to evaluate if it makes a true statement when input into the given inequality.
far left (I choose -5): -5(-5 - 1)(-5 + 4) > 0   →  (-)(-)(-) > 0  →   negative > 0  FALSE
- 4 to 0 (I choose -2): -2(-2 - 1)(-2 + 4) > 0   →  (-)(-)(+) > 0  →   positive > 0  TRUE
0 to 1 (I choose 0.5): .5(.5 - 1)(.5 + 4) > 0   →  (+)(-)(+) > 0  →   negative > 0  FALSE
far right (I choose 2): 2(2 - 1)(2 + 4) > 0   →  (+)(+)(+) > 0  →   positive > 0  TRUE
 
        
             
        
        
        
the net is pretty much the net of a long box, kinda like the one in the example in the picture below.  Due to that, we can pretty much assume the two sides sticking up and down, are just two small 6x3 rectangles, namely, they have a height of 3, reason why we assume that, if that if we fold the other sides to make out the box, those two sides sticking out, must be 3m to neatly snugfit.
now, if we close the box as it stands, the sides(laterals) will be on the left-right sides two 3x15 rectangles, and on the front-back sides, two 6x15 rectangles.
we're excluding the top and bottom sides, because those are not "laterals", or sides of the box.
