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Reika [66]
3 years ago
13

All side lengths in quadrilateral PQRS measure 4 units. A transformation maps PQRS to P'Q'R'S'. All sides of P'Q'R'S' measure 1

unit. Is the transformation an isometric transformation? Explain. Yes, the side lengths in the two figures are proportional. Yes, one figure maps to the other. No, the side lengths are not preserved. No, the angles are not preserved.
Mathematics
2 answers:
VladimirAG [237]3 years ago
6 0
The answer is no, the side lengths are not persevered. <span />
Dmitry [639]3 years ago
5 0

Answer:

No, the side lengths are not preserved.    

Step by step explanation:

We have been given that all side lengths in quadrilateral PQRS measure 4 units. A transformation maps PQRS to P'Q'R'S'. All sides of P'Q'R'S' measure 1 unit.

We can see that the transformation that maps PQRS to P'Q'R'S' is dilation as after dilating our pre-image PQRS with a factor of \frac{1}{4} we get our image P'Q'R'S'.  

Since we know that isometry does not change the shape or size of the figure and isometric figures are said to be geometrically congruent. In our case our side lengths are shrunken so the given transformation is not isometric.  

Therefore, our transformation is not isometric as the side lengths are not preserved.

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Answer:

31

Step-by-step explanation:

A=2(wl+hl+hw)

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w-Width  

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A=2(wl+hl+hw)=2·(1·2+4.5·2+4.5·1)=31

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3 years ago
Which function , A or B has a greater rate of change ? Be sure to include the values for the rates of change in your answer . Ex
arsen [322]
<h2>Answer:</h2>

A rate of change tells us  how one quantity changes in relation to another quantity. In a mathematical language, we can write this as follows:

Rate \ of \ Change=\frac{Change \ in \ y}{Change \ in \ x}

For linear functions. the rate of change is the slope of the line. Thus:

FOR THE TABLE:

By taking two points we can get the rate of change, so let's take (1,5) \ and \ (2,7):

ROC=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} \\ \\ ROC=\frac{7-5}{2-1} \\ \\ ROC=\frac{2}{1} \\ \\ \boxed{ROC=2}

FOR THE GRAPH:

Let's take (1,1) \ and \ (2,4):

ROC=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} \\ \\ ROC=\frac{4-1}{2-1} \\ \\ ROC=\frac{3}{1} \\ \\ \boxed{ROC=3}

As you can see, 3>2 so <em>the ROC of the function given by the graph is greater than the ROC of the function given by the table.</em>

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3 years ago
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emmasim [6.3K]

Answer:

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Step-by-step explanation:

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3 years ago
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Charra [1.4K]
Two is equivalent to twenty and 84
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Read 2 more answers
Match each function
Elis [28]

Answer:

* The degree of the function is 4 and the leading coefficient is positive

 f(x) = (x + 6)(2x - 3)(x - 1)²

* The degree of the function is 5 and the leading coefficient is negative

 f(x) = (x - 2)²(-2x - 1)²(-x + 1)

* The degree of the function is 6 and the leading coefficient is negative

 f(x) = (-x + 1)³(x + 2)²(x - 3)

* The degree of the function is 5 and the leading coefficient is positive

 f(x) = (-2x + 1)²(x - 3)²(x + 1)

Step-by-step explanation:

∵ f(x) = (x + 6)(2x - 3)(x - 1)²

∵  (x)(2x)(x²) = 2x^4

∴ The degree of the function is 4

∴ The leading coefficient is positive ⇒ (2)

∵ f(x) = (x - 2)²(-2x - 1)²(-x + 1)

∵ (x)²(-2x)²(-x) = (x²)(4x²)(-x) = -4x^5 ⇒ (neglect -ve with even power)

∴ The degree of the function is 5

∴ The leading coefficient is negative ⇒ (-4)

∵ f(x) = (-x + 1)³(x + 2)²(x - 3)

∵ (-x)³(x)²(x) = (-x³)(x²)(x) = -x^6

∴ The degree of the function is 6

∴ The leading coefficient is negative ⇒ (-1)

∵ f(x) = (-2x + 1)²(x - 3)²(x + 1)

∵ (-2x)²(x)²(x) = (4x²)(x²)(x) = 4x^5

∴ The degree of the function is 5

∴ The leading coefficient is positive ⇒ (4)

6 0
4 years ago
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