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3 years ago

Which polygons can be mapped onto each other by similarity transformations?

Mathematics

2 answers:

Fittoniya [83]3 years ago

8 0

Solution:

Two polygons are said to be similar if ratio of their corresponding sides are same and their interior angles are congruent.

Consider Polygon 1 and Polygon 4

AB= DE= 8 units, AE= 6 units,

PT=QR= 4 Units, PQ= 3 Units

$\frac{AB}{PT}=\frac{DE}{QR}=\frac{AE}{PQ}=\frac{BC}{SR}=\frac{CD}{TS}$

So, Polygon 1 ≅ Polygon 4

Now, Consider Polygon 2 and Polygon 3

LM=KO=6 units, LK=4 units

GF= 4 units, GH=FJ =4 Units

As you can see that polygon 2 and polygon 4 doesn't have corresponding proportional sides i.e $\frac{6}{4}\neq \frac{4}{4}$ .

So, Polygon 2 is not congruent to Polygon 4.

Julli [10]3 years ago

3 0

Polygon A,B,C,D,E and polygon L,M,N.O.K

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3 years ago

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ivanzaharov [21]

Answer:

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3 years ago

What is the inverse of the function shown( in the form of a graph)

NemiM [27]

Answer:

$y = - 1x + b$

Step-by-step explanation:

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3 years ago

compute the projection of → a onto → b and the vector component of → a orthogonal to → b . give exact answers.

Nina [5.8K]

$\text { Saclar projection } \frac{1}{\sqrt{3}} \text { and Vector projection } \frac{1}{3}(\hat{i}+\hat{j}+\hat{k})$

We have been given two vectors $\vec{a}$ and $\vec{b}$ , we are to find out the scalar and vector projection of $\vec{b}$ onto $\vec{a}$

we have $\vec{a}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{b}=\hat{i}-\hat{j}+\hat{k}$

The scalar projection of $\vec{b}$ onto $\vec{a}$ means the magnitude of the resolved component of $\vec{b}$ the direction of $\vec{a}$ and is given by

The scalar projection of $\vec{b}$ onto

$\vec{a}=\frac{\vec{b} \cdot \vec{a}}{|\vec{a}|}$

$\begin{aligned}&=\frac{(\hat{i}+\hat{j}+\hat{k}) \cdot(\hat{i}-\hat{j}+\hat{k})}{\sqrt{1^2+1^1+1^2}} \\&=\frac{1^2-1^2+1^2}{\sqrt{3}}=\frac{1}{\sqrt{3}}\end{aligned}$

The Vector projection of $\vec{b}$ onto $\vec{a}$ means the resolved component of $\vec{b}$ in the direction of $\vec{a}$ and is given by

The vector projection of $\vec{b}$ onto

$\vec{a}=\frac{\vec{b} \cdot \vec{a}}{|\vec{a}|^2} \cdot(\hat{i}+\hat{j}+\hat{k})$

$\begin{aligned}&=\frac{(\hat{i}+\hat{j}+\hat{k}) \cdot(\hat{i}-\hat{j}+\hat{k})}{\left(\sqrt{1^2+1^1+1^2}\right)^2} \cdot(\hat{i}+\hat{j}+\hat{k}) \\&=\frac{1^2-1^2+1^2}{3} \cdot(\hat{i}+\hat{j}+\hat{k})=\frac{1}{3}(\hat{i}+\hat{j}+\hat{k})\end{aligned}$

To learn more about scalar and vector projection visit:brainly.com/question/21925479

#SPJ4

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1 year ago

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Talja [164]

Answer:

Step-by-step explanation:

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3 years ago

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