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2 years ago

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The table below shows the total number of hours, Jessica, Emily, and Trina, each studied after given number of weeks. Part A one

of the relationships between weeks and total hours studied is proportional. Whose number of total hours studied is proportional to the number of weeks studied ? For the proportional relationship, write an equation to represent y, the total number of hours studied, in terms of x, the number of weeks. Explain how u determined your equation. Pls help

2 answers:

pishuonlain [190]2 years ago

8 0

Answer:Y times X = 5

Step-by-step explanation:

Naily [24]2 years ago

3 0

Answer:

See attached

Step-by-step explanation:

Refer to attached

Only column 3 - Trina's numbers represent proportional relationship

Equation for it is: y = 2.5x

Calculation is provided on the table

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Write 127.90 in expanded form

yan [13]

127.90 in expanded form:

100 + 20 + 7 + 0.9

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2 years ago

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) |

Bas_tet [7]

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.

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

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

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

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

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

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

$\bar{x}=\frac{21}{168}*40=5$

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

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

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

$\bar{y}=\frac{k}{168k}\int^{9}_{1}\int^{4}_{1}y^{3} dydx$

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

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

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

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

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

Therefore the center of mass is:

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

I hope it helps you!

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2 years ago

Circle D is shown. Angle A C B intercepts arc A B. Arc A B has a measure of 76 degrees. What is the measure of AngleC ? 38° 76°

Valentin [98]

I think the answer is 76 ...

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2 years ago

Read 2 more answers

An entrepreneur estimates his total profit (total revenue minus total cost) for his proposed company as p(x) = x3 − 4x2 + 5x − 2

RideAnS [48]

Answer:

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Step-by-step explanation:

We are given:

p(x) = x^3-4x^2+5x-20

We would find the value of x

Solving:

x^3-4x^2+5x-20 = 0

(x^3-4x^2)+(5x-20) = 0

x^2(x-4)+5(x-4) = 0

(x-4)(x^2+5)=0

=> x-4 = 0 and x^2+5 =0

x = 4 and x^2 = -5

x = 4 and x = ±√-5 or ±√5 i (not real solutions)

So, x = 4

So, year in which the entrepreneur break even is 4

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2 years ago

The summer camp Jessica attends is four hundred twenty-three and four tenths miles from her home. A detour adds 10 miles to the

juin [17]

Given that Jessica attends summer camp at a distance of four hundred twenty-three and four tenth mile = $423\frac{4}{10} = 423 + \frac{4}{10} = 423+0.4 = 423.4 miles$

And a detour adds 10 miles to the distance.

That means we need to add 10 miles to the given distance.

So new distance = 423.4 + 10 = 433.4 miles

Hence Jessica needs to travel 433.4 miles for summer camp.

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2 years ago