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vagabundo [1.1K]
2 years ago
13

Which is the rate of change for the interval between 3 and 6 on the x-axis? 0-2

Mathematics
1 answer:
Hunter-Best [27]2 years ago
8 0

Image is missing, so i have attached it.

Answer:

Rate of change = 2

Step-by-step explanation:

From the image attached, we are given a parabola which opens upwards with many points marked on the curve.

We are asked to find the rate of change for the interval between 3 and 6 on the x axis.

Now, rate of change of any function f(x) from a to b is given by the formula ;

Rate of change = [f(b) - f(a)]/(b - a)

For this question, a = 3 and b = 6

f(3) is simply the y-value when x is 3. On the graph, f(3) = -2

Similarly,f(6) = 4

Thus;

Rate of change = [4 - (-2)]/(6 - 3)

Rate of change = 6/3

Rate of change = 2

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distribute-   8x+4=20

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answer-   x=2.5


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Expand 4x(7x-11)<br>expand 3x(7-5x)<br>Can someone help me with tge questions above?
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<span>(    7 x    -    11    )

<span>(   7   -   5x   )</span></span>
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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) |
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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