Plug in 0 for z
so it would change to -2<6
Answer:
The largest total area that can be enclosed will be a square of length 272 yards.
Step-by-step explanation:
First we get the perimeter of the large rectangular enclosure.
Perimeter of a rectangle =2(l + w)
Perimeter of the large rectangular enclosure= 1088 yard
Therefore:
2(L+W)=1088
The region inside the fence is the area
Area: A = LW
We need to solve the perimeter formula for either the length or width.
2L+ 2W= 1088 yd
2W= 1088– 2L
W = 
W = 544–L
Now substitute W = 544–L into the area formula
A = LW
A = L(544 – L)
A = 544L–L²
Since A is a quadratic expression, we re-write the expression with the exponents in descending order.
A = –L²+544L
Next, we look for the value of the x coordinate


L=272 yards
Plugging L=272 yards into the calculation for area:
A = –L²+544L
A(272)=-272²+544(272)
=73984 square yards
Thus the largest area that could be encompassed would be a square where each side has a length of 272 yards and a width of:
W = 544 – L
= 544 – 272
= 272 yards
Answer:
A. m^2+7m+10=0
Step-by-step explanation:
This is a problem in pattern matching, and in substituting a variable for a pattern.
(x^2+3)^2 +7x^2 +21 = -10 . . . . . . given
(x^2 +3)^2 +7(x^2 +3) = -10 . . . . . factor the last two terms
m^2 +7m = -10 . . . . . . . . . . . . subsitute m for x^2 +3
m^2 +7m +10 = 0 . . . . . . . . add 10 to both sides; matches A
There’s not actually a question. But I might be able to help if u show the full problem