Mathematics, 01.11.2019 14:31 glander2

Abarrel of tomato sauce has spilled on a tile floor. the sauce flow can be expressed with the function r(t) = 2t, where t represents time in minutes and r represents how far the sauce is spreading.

the spilled sauce is creating a circular pattern on the tile. the area of the pattern can be expressed as a(r) = πr2.

part a: find the area of the circle of spilled sauce as a function of time, or a[r(t)]. show your work. (6 points)

part b: how large is the area of spilled sauce after 5 minutes? you may use 3.14 to approximate π in this problem. (4 points)

Answers: 1

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Answers

For the answer to the question above, I'll show my solution for the answer below.
Area A = pi r^2
The rate of change of area = d A/ d t = 2 pi r d r/d t
r= 2t, d r = 2 d t
This gives
d A/ d t = 2 pi 2t 2 d t = 8 pi t dt
Integrate it
A= 4 pi t^2.
After t=5, the area will be 100* 3.14 square= 314 square units.

Answer from: itsjoke5550

As per the statement:

The sauce flow can be expressed with the function is given by:

r(t) = 2t

where, t represents time in minutes and r represents how far the sauce is spreading.

The spilled sauce is creating a circular pattern on the tile.

The area of the pattern can be expressed as:

$A(r) = \pi r^2$

Part A.

To find the area of the circle of spilled sauce as a function of time.

A[r(t)]

then;

$A[r(t)] = \pi (r(t))^2$

Substitute the function r(t) we have;

$A[r(t)] = \pi (2t)^2 = \pi 4t^2 = 4 \pi t^2$ .....[1]

therefore, the area of the circle of spilled sauce as a function of time, is:

$A[r(t)] = 4 \pi t^2$ .

Part B:

We have to find how large is the area of spilled sauce after 5 minute.

use π = 3.14 and t = 5 minutes

Substitutes these in [1] we have;

$A[r(5)] = 4 \cdot 3.14 \cdot 5^2$

⇒ $A[r(5)] = 100 \cdot 3.14$

Simplify:

A[r(5)] = 314 square unit

therefore, 314 square unit large is the area of spilled sauce after 5 minute.

Answer from: Quest

10010010001010010101001001010010100101010010101010010010101010110010001011010010010001

step-by-step explanation:

Answer from: Quest

la parábola es el lugar

geométrico en el cual los puntos tienen una distancia idéntica, equidistantes,

de un punto denominado foco, f, y una recta denominada directriz

el radio vector de una

parábola es el segmento que existe entre un punto cualquiera y el foco.

por la ecuación se observa

que la parábola abre hacia la derecha, por lo cual podemos usar las siguientes fórmulas

la ecuación ordinaria de la

parábola es y^2 = 4px, donde p es la longitud desde el vértice de la parábola al

foco.

la fórmula de la directriz

es x + p = 0

resolviendo:

y^2-9x = 0

y^2 = 9x

4p = 9, entonces p = 9/4

entonces el foco tendrá la

coordenada f(9/4,0)

la directriz vendrá

expresada por

x + p = 0

x = -p

x = -9/4

el problema indica que la

ordenada es igual a 6, entonces

(6)^2=9x

x = 36/9

x = 4

entonces el punto al cual

hay que obtener la distancia será p(4,6)

calculando se tendrá: pf =

pf = 6,25 u

ver más en -

step-by-step explanation:

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