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zimovet [89]
2 years ago
12

7 Point Question Tyler was trying to prove that all rhombuses are similar. He thought, “If I draw 2 rhombuses, I can find dilati

ons that will take one rhombus exactly onto the other.” Then he wrote a proof. However, all rhombuses are not similar. Which steps in Tyler’s proof have flaws? Explain what is wrong with each step that is wrong. Step 1: Let ABCD and WXYZ be rhombuses. By the definition of a rhombus, AB≅BC≅CD≅DA and WX≅XY≅YZ≅ZW. Step 2: Dilate rhombus ABCD by the scale factor given by $$ WX AB​. Side A'B' will be the same length as WX because of how I chose the scale factor. Because all sides of a rhombus are congruent, B'C' will be the same length as WX and therefore the same length as XY, C'D' will be the same length as YZ, and A'D' will be the same length as WZ. That means all the corresponding sides of A'B'C'D' and WXYZ will be the same length. Step 3: If all the sides in 2 figures are proportional, then those 2 figures are similar, so there must be a sequence of transformations that take ABCD onto WXYZ using dilations and rigid motions. Step 4: Translate A'B'C'D' by the directed line segment A'W so that A″ and W coincide. Step 5: Rotate A″B″C″D″ by angle B″WX so that B‴ and X coincide. Now segments A‴B‴ and WX coincide. Step 6: If needed, reflect A‴B‴C‴D‴ over segment WX so that D⁗ and Z are on the same side of WX. Now segments A⁗D⁗ and WZ coincide. Step 7: Once 3 vertices of the rhombus are lined up, the other vertex has to line up as well or else the shapes wouldn’t be rhombuses. So, we have shown we can use rigid motions and dilations to line up any 2 rhombuses, and therefore, all rhombuses are similar. (Select all that apply.) Step 1 is incorrect. The definition of a rhombus does not say that all the sides are equal. Step 2 is incorrect. The scale factor is not $$ . Step 3 is incorrect. It is only true for triangles that if all the side lengths in a figure are proportional, the figures must be similar. Step 4 is incorrect. Translating by the directed line segment
Mathematics
1 answer:
Viefleur [7K]2 years ago
7 0

A rhombus can have the same shape and dimension as a square or a

parallelogram, therefore, all rhombuses are not similar.

  • A step in the proof that have flaws is <u>step 3</u>.

Reasons:

By reading through the given steps, we have, the first step that is incorrect

is step 3.

Step 3: The requirement for the similarity of two figures based on the

proportionality of the corresponding sides of the figures is related with

triangles as given by the triangle proportionality theorem.

This is so because, the lengths of the sides of a triangle depends on the

opposite interior angle, while the side of a rhombus can have a

combination of large length and and acute angle.

Two rhombuses that have proportional sides may be dissimilar,

because they can have different interior angles.

Therefore;

<u>The step in Tyler's proof that have a flaw is Step 3</u>

Learn more about similarity between geometric figures here:

brainly.com/question/10270676

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HELP PLEASE ASAP
Pie

\bold{\huge{\underline{ Solution }}}

<u>We </u><u>have</u><u>, </u>

  • Line segment AB
  • The coordinates of the midpoint of line segment AB is ( -8 , 8 )
  • Coordinates of one of the end point of the line segment is (-2,20)

Let the coordinates of the end point of the line segment AB be ( x1 , y1 ) and (x2 , y2)

<u>Also</u><u>, </u>

Let the coordinates of midpoint of the line segment AB be ( x, y)

<u>We </u><u>know </u><u>that</u><u>, </u>

For finding the midpoints of line segment we use formula :-

\bold{\purple{ M( x,  y) = }}{\bold{\purple{\dfrac{(x1 +x2)}{2}}}}{\bold{\purple{,}}}{\bold{\purple{\dfrac{(y1 + y2)}{2}}}}

<u>According </u><u>to </u><u>the </u><u>question</u><u>, </u>

  • The coordinates of midpoint and one of the end point of line segment AB are ( -8,8) and (-2,-20) .

<u>For </u><u>x </u><u>coordinates </u><u>:</u><u>-</u>

\sf{  -8  = }{\sf{\dfrac{(- 2 +x2)}{2}}}

\sf{2}{\sf{\times{ -8  = - 2 + x2 }}}

\sf{ - 16 = - 2 + x2 }

\sf{ x2 = -16 + 2 }

\bold{ x2 = -14  }

<h3><u>Now</u><u>, </u></h3>

<u>For </u><u>y </u><u>coordinates </u><u>:</u><u>-</u>

\sf{  8  = }{\sf{\dfrac{(- 20 +x2)}{2}}}

\sf{2}{\sf{\times{ 8   = - 20 + x2 }}}

\sf{ 16 = - 20 + x2 }

\sf{ y2 = 16 + 20 }

\bold{ y2 = 36  }

Thus, The coordinates of another end points of line segment AB is ( -14 , 36)

Hence, Option A is correct answer

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3 years ago
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andrey2020 [161]
A liter is 0.264 gallons so 13 * 0.264 = 3.432 and round it up it is 3.4
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3 years ago
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NikAS [45]

Answer:

Isnt it where you write it in the easiest way to understand it? sorry if its wrong :/

Step-by-step explanation:

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gayaneshka [121]
Hi.
your answer is 1/3, or 2/6, or 3/9, or 4/12, etc.
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6 0
3 years ago
Gannon has deposited $742 in a savings account that earns interest at a rate of 3.4% compounded monthly. What will the account b
marta [7]
Firstly, solve the effective annual interest (ieff) with the equation,
 
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where i is the interest rate and m is the number of times the interest is compounded in a year. In this problem, m is 12

Substituting the values, 
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To solve for the future (F) amount of the present investment (P), 
                                       
                                        F = P x (1 + ieff)^n

where n is number of years.
 
                                        F = ($742) x (1 + 0.03453)^15

Thus, the answer is $1234.76. 


 
                

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