Find the length of UF
1 answer:
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Answer:
Justice lived in Germany for 15 months and he lived in Holland for 9 months.
Step-by-step explanation:
1. You must make a system of equations:
Where is the number on months he lived in Germany and
is the number on months he lived in Holland.
2. You can solve this system of equations by using the Elimination Method. If you multiply the first equation by -120, you can cancel the variable when you add both equations, and then you can calculate
. When you obtain the value of
, you can substitute it into one of the original equations and solve for
. Then:
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
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