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marishachu [46]
4 years ago
9

Polygon D is a scaled copy of polygon C using a scale factors of 6

Mathematics
1 answer:
Vladimir79 [104]4 years ago
8 0

Answer: The area of the Polygon D is 36 times larger than the area of the Polygon C.

Step-by-step explanation:

<h3> The complete exercise is: "Polygon D is a scaled copy of Polygon C using a scale factor of 6. How many times larger is the area of Polygon D than the area Polygon C"?</h3>

 In order to solve this problem it is important to analize the information provided in the exercise.

You know that the Polygon D was obtained by multiplying the lengths of the Polygon C by the scale factor of 6.

Then, you can identify that the Length scale factor used is:

Length\ scale\ factor=k=6

Now you have to find the Area scale factor.

Knowing that the Length scale factos is 6, you can say that the Area scale factor is:

Area \ scale\ factor=k^2=6^2

Finally, evaluating, you get that this is:

Area \ scale\ factor=36

Therefore, knowing the Area scale factor, you can determine that the area of the Polygon D is 36 times larger than the area of the Polygon C.

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a. 38.19m/s

b. 38.605m/s

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Step-by-step explanation:

The average velocity is defined as the relationship between the displacement that a body made and the total time it took to perform it. Mathematically is given by the next formula:

v_a_v_g = \frac{\Delta x}{\Delta t} =\frac{x_f-x_i}{t_f-t_i}

Where:

x_f=Final\hspace{3}distance\hspace{3}traveled\\x_i=Initial\hspace{3}distance\hspace{3}traveled\\t_f=Final\hspace{3}time\hspace{3}interval\\t_i=Initial\hspace{3}time\hspace{3}interval

a. Let's find h(3) and h(4) using the data provided by the problem:

h(3)=44(3)-0.83(3^2)=124.53=x_i\\h(4)=44(4)-0.83(4^2)=162.72=x_f

The average velocity over the interval [3, 4] is :

v_a_v_g=\frac{162.72-124.53}{4-3} =38.19m/s

b. Let's find h(3.5) using the data provided by the problem:

h(3.5)=44(3.5)-0.83(3.5^2)=143.8325=x_f

The average velocity over the interval [3, 3.5] is :

v_a_v_g=\frac{143.8325-124.53}{3.5-3} =38.605m/s

c. Let's find h(3.1) using the data provided by the problem:

h(3.1)=44(3.1)-0.83(3.1^2)=128.4237=x_f

The average velocity over the interval [3, 3.1] is :

v_a_v_g=\frac{128.4237-124.53}{3.1-3} =38.937m/s

d. Let's find h(3.01) using the data provided by the problem:

h(3.1)=44(3.01)-0.83(3.01^2)=124.920117=x_f

The average velocity over the interval [3, 3.01] is :

v_a_v_g=\frac{124.920117-124.53}{3.01-3} =39.0117m/s

e. Let's find h(3.001) using the data provided by the problem:

h(3.001)=44(3.001)-0.83(3.001^2)=124.5690192=x_f

v_a_v_g=\frac{124.5690192-124.53}{3.001-3} =39.01917m/s

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For this case we have the following polynomial:
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