Respuesta :
I've seen this problem before, many days ago
A.
p(t)=5t
A[p(t)]=A[5t]=π(5t)²=25πt²
A[p(t)]=25πt²
B.
t=5
A[p(t)]=25πt²
A[p(5)]=25π(5)²
A[p(5)]=25π25²
A[p(t)]=625π
A[p(t)]=625(3.14)=1962.5 square units
A.
p(t)=5t
A[p(t)]=A[5t]=π(5t)²=25πt²
A[p(t)]=25πt²
B.
t=5
A[p(t)]=25πt²
A[p(5)]=25π(5)²
A[p(5)]=25π25²
A[p(t)]=625π
A[p(t)]=625(3.14)=1962.5 square units
Answer:
Given the paint flow can be expressed with the function as:
[tex]p(t) = 5t[/tex]
where, t represents time in minutes and p represents how far the paint is spreading.
It is also given that: The flowing paint is creating a circular pattern on the tile
The area of the pattern can be expressed as:
[tex]A(p) = \pi p^2[/tex]
A
To find the area of the circle of spilled paint as a function of time.
[tex]A[p(t)] = A[5t][/tex]
⇒[tex]A[5t] = \pi \cdot (5t)^2 = 25 \pi t^2[/tex]
therefore, the area of the circle of spilled paint as a function of time [tex]A[p(t)] = 25 \pi t^2[/tex]. ......[1]
B
To find how large is the area of spilled paint after 2 minutes.
Substitute t = 2 minutes and [tex]\pi = 3.14[/tex] in [1] we have;
[tex]A[p(2)] = 25 \cdot 3.14 \cdot 2^2 = 100 \cdot 3.14 = 314[/tex] square units.
Therefore, 314 square units large is the area of spilled paint after 2 minutes.