Hilary's Horse Hideaway is buying dirt to fill their circular training area. The training circle is 28 feet across. What is the
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Use Law of Cooling:
T0 = initial temperature, TA = ambient or final temperature
First solve for k using given info, T(3) = 42
Substituting k back into cooling equation gives:
At some time "t", it is brought back inside at temperature "x".
We know that temperature goes back up to 71 at 2:10 so the time it is inside is 10-t, where t is time that it had been outside.
The new cooling equation for when its back inside is:
Solve for x:
Sub back into original cooling equation, x = T(t)
Solve for t:
This means the exact time it was brought indoors was about 2.5 seconds before 2:05 PM
T0 = initial temperature, TA = ambient or final temperature
First solve for k using given info, T(3) = 42
Substituting k back into cooling equation gives:
At some time "t", it is brought back inside at temperature "x".
We know that temperature goes back up to 71 at 2:10 so the time it is inside is 10-t, where t is time that it had been outside.
The new cooling equation for when its back inside is:
Solve for x:
Sub back into original cooling equation, x = T(t)
Solve for t:
This means the exact time it was brought indoors was about 2.5 seconds before 2:05 PM
Let
x-------> the width of the rectangular area
y------> the length of the rectangular area
we know that
y=x+15------> equation 1
perimeter of a rectangle=2*[x+y]
2x+2y <= 150-------> equation 2
substitute 1 in 2
2x+2*[x+15] <=150--------> 2x+2x+30 <=150----> 4x <=150-30
4x <= 120---------> x <= 30
the width of the rectangular area is at most 30 ft
y=x+15
for x=30
y=30+15------> y=45
the length of the rectangular area is at most 45 ft
see the attached figure
the solution is<span> the shaded area</span>
x-------> the width of the rectangular area
y------> the length of the rectangular area
we know that
y=x+15------> equation 1
perimeter of a rectangle=2*[x+y]
2x+2y <= 150-------> equation 2
substitute 1 in 2
2x+2*[x+15] <=150--------> 2x+2x+30 <=150----> 4x <=150-30
4x <= 120---------> x <= 30
the width of the rectangular area is at most 30 ft
y=x+15
for x=30
y=30+15------> y=45
the length of the rectangular area is at most 45 ft
see the attached figure
the solution is<span> the shaded area</span>