<span>To minimize the perimeter you should always have a square.
sqrt(289) = 17
The dimensions should be 17 X 17
To see , try starting at length 1, and gradually increase the length.
The height decreases at a faster rate than the length increases, up until you reach a square.
Or if you want to use algebra, Say the width is 17-x
Then the length is 289/(17-x)
Now, this is bigger than 17+x, as shown here:
289/(17-x) > 17+x
289 > 289 - x^2
which is true.
so the perimeter would be bigger than 2 * (17- x + 17 + x) = 2 * (2 * 17) = 4 * 17
Again, the dimensions should be a square. 17 X 17.</span>
The answer is A) the height of students
Answer:
10 (positive 10)
Step-by-step explanation:
5 times -2 is equal to -10.
-10 times -1 is equal to 10.
We have two relations between length and width. One is given in the problem statement. The other is given by the formula for perimeter. We can solve the two equations in two unknowns using substitution.
Let w and l represent the width and length of the sign in feet, respectively.
... l = 2w -12 . . . . . the length is 12 ft less than twice the width
... p = 2(l +w) = 114 . . . . the perimeter is 114 ft
Using the first equation for l, we can substitute for l in the second equation.
... 114 = 2((2w -12) +w)
... 114 = 6w -24 . . . . . . . . simpify
... 138 = 6w . . . . . . . . . . . add 24
... 23 = w . . . . . . . . . . . . . divide by 6
... l = 2w -12 = 2·23 -12 = 34 . . . . use the equation for l to find l
The length and width of the sign are 34 ft and 23 ft, respectively.