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blondinia [14]
3 years ago
13

Compare and contrast the effects that a proportional dimension change has on perimeter and area of figures

Mathematics
2 answers:
zloy xaker [14]3 years ago
6 0

Answer with explanation:

1.In terms of Perimeter

For any 2 Dimensional shape, having fixed dimension, (Perimeter is calculated=S), and if the shape is dilated with Dilation factor k,that is the similar shape has the perimeter equal to Dilation factor multiplied by Perimeter of Pre -Image=k S).

If Dilation Factor <1,

Perimeter of  image will be less than Pre image.

If , Dilation Factor >1

Perimeter of Image will be greater than Pre-Image.

Actual Perimeter of Shape = S

Dilation factor or Proportionality Constant = k

New Perimeter of Shape = k*S

2. In terms of Area

Let me explain this concept by taking a four sided Quadrilateral named square.

Suppose, Side of Square = a units

Area of Square = a²

If side is dilated by a factor of m units

Area of new Square obtained = (m a)²= m²a²

Area increases or decreases by m² units, if dilation factor<1,the area Decreases, and if Dilation Factor >1, area increases.

→So, when Shape is Dilated by a factor m ,or you can say constant of proportionality is m,then area of new Shape is (m²,)square of × area of original shape.

Kipish [7]3 years ago
3 0
A proportional dimension will change the perimeter by the factor of itself. And will change the area by the factor of the dilation squared.
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For the piecewise function, find the values h( - 6), h(0), h(5), and h(9).- 5x – 13, for x &lt; -3h(x) = { 5, for - 35x&lt;5for
sleet_krkn [62]

Given the piecewise function h(x):

h(x)=\begin{cases}-5x-13,x

-When x is less than "-3", h(x)=-5x-13

-When x is between -3 and 5, h(x)=5

-When x is greater than or equal to 5, h(x)=x+1

1) For h(-6), this notation indicates that you have to determine the value of h(x) when x=-6

-6 is less than -3, which means that for this value of x, the function has the following shape

h(x)=-5x-13

Replace the expression with x=-6 and calculate the corresponding value of x:

\begin{gathered} h(-6)=-5(-6)-13 \\ h(-6)=30-13 \\ h(-6)=17 \end{gathered}

2) For h(0), you have to determine the value of h(x) when x=0. Zero is between -3 and 5, for this value of x, the function h(x) has the following shape:

h(x)=5

This equation represents a horizontal line, which means that for every value within the interval of definition -3≤x<5, the function always has the same value h(x)=5

We can conclude that:

h(0)=5

3) For h(5), you have to determine the value of h(x) for x=5, for values of x greater than or equal to 5, h(x) has the following shape:

h(x)=x+1

Replace the expression with x=5 and calculate the corresponding value of h(x):

\begin{gathered} h(5)=5+1 \\ h(5)=6 \end{gathered}

4) For h(9), you have to determine the value of h(x) when x=9, 9 is greater than 5, for this value of x, the function has the following shape:

h(x)=x+1

Replace the expression with x=9, and calculate the corresponding value of h(x):

\begin{gathered} h(9)=9+1 \\ h(9)=10 \end{gathered}

So, to sum up:

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5 0
1 year ago
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