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:
x= 16 ft.
Step-by-step explanation:
Let 'x' and 'y' are the lengths of the two pieces then:
1.) x + y = 36
2.) x : y = 4 : 5 cross multiply
3.) 5x = 4y
by solving the system of equations
x + y = 36
5x = 4y
we find
x = 16 ft
y = 20 ft
It would be nonsense, as if the beetle were at normal size, then it would be considered 1/1 as a scale. If you wanted to increase it, you would increase the numerator.
116 degrees. It’s simple.
56 is what the low was, but the high was 60 more.
Therefore,
56+60=116