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2 years ago

6

Given the formula b=rs/t solve for s

1 answer:

Gekata [30.6K]2 years ago

7 0

Answer:

$s = \frac{bt}{r}$

Step-by-step explanation:

To solve for s, you need to follow these steps.

First multiply by t both sides of the equation.

$b = \frac{rs}{t}$

$bt = \frac{rst}{t}$

We get $bt = rs$

Now we divide by r, both sides of the equation.

$\frac{bt}{r} = \frac{rs}{r}$

Finallly we obtain:

$\frac{bt}{r} = s$ or

$s = \frac{bt}{r}$

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A freezer has a temperature of 14 degrees Fahrenheit. An ice-cube tray full of water is placed in the freezer. The function f(t)

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Answer:

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Step-by-step explanation:

You can use f(15) = 40 to solve for C, then find f(0), the initial temperature.

40 = f(15)

40 = Ce^(-0.045·15) +14 = .50916C +14

26 = .50916C

26/.50916 = C ≈ 51.065

Then f(0) is ...

f(0) = 51.065·e^0 +14 = 65.065 ≈ 65

The initial temperature of the water was 65 degrees Fahrenheit.

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1 year ago

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I think the answer is 3.47

Step-by-step explanation:

Terry starts out with 2.25 gallons, and he uses 1.33 so you subtract these so he has .92 gallons left. He adds 4.8 gallons so you add 4.8 to .92 and get 5.72 gallons total in the bucket... If the question is asking how much change between the beginning and end I would say 5.72- 2.25= 3.47

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Find the mass and center of mass of the lamina that occupies the region D and has the given density function rho. D = {(x, y) |

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Answer:

$M=168k$

$(\bar{x},\bar{y})=(5,\frac{85}{28})$

Step-by-step explanation:

Let's begin with the mass definition in terms of density.

$M=\int\int \rho dA$

Now, we know the limits of the integrals of x and y, and also know that ρ = ky², so we will have:

$M=\int^{9}_{1}\int^{4}_{1}ky^{2} dydx$

Let's solve this integral:

$M=k\int^{9}_{1}\frac{y^{3}}{3}|^{4}_{1}dx$

$M=k\int^{9}_{1}\frac{y^{3}}{3}|^{4}_{1}dx$

$M=k\int^{9}_{1}21dx$

$M=21k\int^{9}_{1}dx=21k*x|^{9}_{1}$

So the mass will be:

$M=21k*8=168k$

Now we need to find the x-coordinate of the center of mass.

$\bar{x}=\frac{1}{M}\int\int x*\rho dydx$

$\bar{x}=\frac{1}{M}\int^{9}_{1}\int^{4}_{1}x*ky^{2} dydx$

$\bar{x}=\frac{k}{168k}\int^{9}_{1}\int^{4}_{1}x*y^{2} dydx$

$\bar{x}=\frac{1}{168}\int^{9}_{1}x*\frac{y^{3}}{3}|^{4}_{1}dx$

$\bar{x}=\frac{1}{168}\int^{9}_{1}x*21 dx$

$\bar{x}=\frac{21}{168}\frac{x^{2}}{2}|^{9}_{1}$

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Now we need to find the y-coordinate of the center of mass.

$\bar{y}=\frac{1}{M}\int\int y*\rho dydx$

$\bar{y}=\frac{1}{M}\int^{9}_{1}\int^{4}_{1}y*ky^{2} dydx$

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$\bar{y}=\frac{1}{168}\int^{9}_{1}\frac{y^{4}}{4}|^{4}_{1}dx$

$\bar{y}=\frac{1}{168}\int^{9}_{1}\frac{255}{4}dx$

$\bar{y}=\frac{255}{672}\int^{9}_{1}dx$

$\bar{y}=\frac{255}{672}8=\frac{2040}{672}$

$\bar{y}=\frac{85}{28}$

Therefore the center of mass is:

$(\bar{x},\bar{y})=(5,\frac{85}{28})$

I hope it helps you!

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