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Andru [333]
3 years ago
13

Similar polygons help

Mathematics
1 answer:
Nikolay [14]3 years ago
5 0

Answer:

5. x =6 | 6. x = 7 | 7. x = 35

Step-by-step explanation:

5. 8/12 = x/9 |   72 = 12x | x = 6

6. 10/5 = 4.5/x | 45 = 5x | x = 7

7. 12/18 = x+1/24 | 12x+12 = 432 | x = 35

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Find the least common denominator <img src="https://tex.z-dn.net/?f=%5Cfrac%7B3%7D%7B9%7D%20%5Cfrac%7B8%7D%7B4%7D" id="TexFormul
MA_775_DIABLO [31]

Answer:

36 is your least common denominator. Hope this helps

Step-by-step explanation:

8 0
2 years ago
Allan drew a polygon with 4 sides and 4 angles. All four sides are equal. None of the angles are right angles. What figure did A
Lady_Fox [76]

Answer: The figure drawn by Alan is a rhombus.


Step-by-step explanation:

Given: Allan drew a polygon with 4 sides and 4 angles with two properties:-

  • All four sides are equal.
  • None of the angles are right angles.

We know that only a rhombus is a flat shape with 4 equal straight sides and 4 angles .All sides have equal length. Its angles need not to be right angle.

Therefore, the figure drawn by Alan is a rhombus.


8 0
3 years ago
Read 2 more answers
????????????????????????????
frutty [35]

Answer:

B

Step-by-step explanation:

i took the same question hope this helps :D

5 0
3 years ago
Read 2 more answers
The length of a rectangle is 5 metres less than twice the breadth. If the perimeter is 50 meters,find the length and breadth
MAXImum [283]
<h3><u>S</u><u> </u><u>O</u><u> </u><u>L</u><u> </u><u>U</u><u> </u><u>T</u><u> </u><u>I</u><u> </u><u>O</u><u> </u><u>N</u><u> </u><u>:</u></h3>

As per the given question, it is stated that the length of a rectangle is 5 m less than twice the breadth.

Assumption : Let us assume the length as "l" and width as "b". So,

\twoheadrightarrow \quad\sf{ Length =2(Width)-5}

\twoheadrightarrow \quad\sf{ \ell=(2b-5) \; m}

Also, we are given that the perimeter of the rectangle is 50 m. Basically, we need to apply here the formula of perimeter of rectangle which will act as a linear equation here.

\\ \twoheadrightarrow \quad\sf{ Perimeter_{(Rectangle)} = 2(\ell +b) } \\

  • <em>l</em> denotes length
  • <em>b</em> denotes breadth

\\ \twoheadrightarrow \quad\sf{50= 2(2b-5+b)} \\

\\ \twoheadrightarrow \quad\sf{50= 2(3b-5)} \\

\\ \twoheadrightarrow \quad\sf{50= 6b - 10} \\

\\ \twoheadrightarrow \quad\sf{50+10= 6b} \\

\\ \twoheadrightarrow \quad\sf{60= 6b} \\

\\ \twoheadrightarrow \quad\sf{\cancel{\dfrac{60}{6}}=b} \\

\\ \twoheadrightarrow \quad\underline{\bf{10\; m = Width }} \\

Now, finding the length. According to the question,

\twoheadrightarrow \quad\sf{ \ell=(2b-5) \; m}

\twoheadrightarrow \quad\sf{ \ell=2(10)-5\; m}

\twoheadrightarrow \quad\sf{ \ell=20-5\; m}

\\ \twoheadrightarrow \quad\underline{\bf{15\; m = Length }} \\

<u>Therefore</u><u>,</u><u> </u><u>length</u><u> </u><u>and</u><u> </u><u>breadth</u><u> </u><u>of</u><u> </u><u>the</u><u> </u><u>r</u><u>ectangle</u><u> </u><u>is</u><u> </u><u>1</u><u>5</u><u> </u><u>m</u><u> </u><u>and</u><u> </u><u>10</u><u> </u><u>m</u><u>.</u><u> </u>

7 0
3 years ago
What is the area, in square centimeters, of the shaded part of the rectangle
mojhsa [17]

Answer:

I honestly dont know I just need the points

8 0
3 years ago
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