-keeping a thorough record of the steps and outcomes of a scientific experiment; -conducting an experiment to examine the impacts of gravity on Earth; -solving an expression in stages by adhering to the order of operations.
Respuesta:
Por lo tanto, la integral de superficie es
.
Explicación paso a paso:
La función dada es,

Para encontrar,
donde S=A=superficie del cilindro elíptico debemos aplicar el teorema de divergencia, así que,





- Si el vector unitario
está dirigido en dirección positiva (hacia afuera), entonces z=c y,

- Si el vector unitario
está dirigido en dirección negativa (hacia adentro), entonces z=-c y,

Por lo tanto, la integral de superficie sin el vector unitario de la superficie es,

Response:
Step-by-step explanation:
We need to form an equation that illustrates the area that Felicia has covered, denoted as y, in relation to the number of tiles she utilized, represented by x.
Let x= Number of tiles
y=Area occupied by Felicia's tiles





The relationship between x and y is constant. Hence, it follows a direct proportion.
The equation for direct proportion is shown as

Where
k=The value of the ratio of x and y
x and y are variable factors
Here, we have k=9
Substituting the known values into the equation
leads us to



This represents the necessary equation depicting the area Felicia covered, y, in terms of the number of tiles utilized, x.
Response:
Step-by-step clarification:
Reply:
a) y-8 = (y₀-8), b) 2y -5 = (2y₀-5)
Clarification:
To address these equations, using direct integration is the simplest approach.
a) The equation provided is
dy / dt = -y + 8
dy / (y-8) = dt
We substitute variables
y-8 = u
dy = du
Substituting and integrating gives us
∫ du / u = ∫ dt
Ln (y-8) = t
Evaluating at the lower limits t = 0 for y = y₀
ln (y-8) - ln (y₀-8) = t-0
Simplifying the equation results in
ln (y-8 / y₀-8) = t
y-8 / y₀-8 =
y-8 = (y₀-8)
b) The equation here is
dy / dt = 2y -5
u = 2y -5
du = 2 dy
du / 2u = dt
Integrating now
½ Ln (2y-5) = t
Evaluating at limits gives
½ [ln (2y-5) - ln (2y₀-5)] = t
Ln (2y-5 / 2y₀-5) = 2t
2y -5 = (2y₀-5)
c) The equation here bears a strong resemblance to the previous one
u = 2y -10
du = 2 dy
∫ du / 2u = dt
ln (2y-10) = 2t
We evaluate this to get
ln (2y-10) –ln (2y₀-10) = 2t
2y-10 = (2y₀-10)