Russell runs 9/10 of a mile in five minutes at this rate how many miles can he run in one minute

-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.

I hope this provides you with insight.

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)