The complete question in the attached figure
we know that
area each pasture=15000 ft²
area each pasture=x*y
so
15000=x*y-----> equation 1
perimeter two rectangular pastures=1050 ft
perimeter two rectangular pastures=2*[2x+y]+y----> 4x+2y+y---> 4x+3y
so
1050=4x+3y----> divide by 3----> 1050/3=(4/3)x+y
clear variable y
y=350-(4/3)x-----> equation 2
the answer part a) y=350-(4/3)x------> <span>
it was proved</span>
Part b)Find the possible lengths and widths of each pasture.
substitute equation 2 in equation 1
15000=x*y-----> 15000=x*[350-(4/3)x]
15000=350x-(4/3)x²----> multiply by 3----> 45000=1050x-4x²
4x²-1050x+45000=0
using a graph tool----> to resolve the second order equation
see the attached figure
the solution is
x=53.942 ft
x=208.558 ft
15000=x*y-----> y=15000/x
for x=53.94 ft
y=278.09 ft
for x=208.56 ft
y=71.92 ft
the possible lengths and widths of each pasture are
case 1
x=53.94 ft
y=278.09 ft
case 2
x=208.56 ft
y=71.92 ft
1 lemonade.
2x+2=3x+1
-x+2=1
-x=-1
x=1 lemonade
Answer: 7.5
Step-by-step explanation:
First, converting R percent to r a decimal
r = R/100 = 7.5%/100 = 0.075 per year,
then, solving our equation
I = 1200 × 0.075 × 2 = 180
I = $ 180.00
You would be able to buy 7 seats because you would divide 10.50 by 80 and you would get 7.6.... and you don't have a 6th of a person so it would then be just 7.
The min or max of a parabola/quadratic function is the vertex
for
y=a(x-h)²+k
the vertex is (h,k)
so
vertex/min is at (-1,2)
h=-1
k=2
y=a(x-(-1))²+2
y=a(x+1)²+2
find a
given, (2,20) is on the graph
20=a(2+1)²+2
20=a(3)²+2
20=9a+2
minus 2 both sides
18=9a
divide by 9
2=a
y=2(x+1)²+2 is da equation
3rd one
f(x)=2(x+1)²+2