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IRINA_888 [86]
3 years ago
5

If the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram

Mathematics
2 answers:
diamong [38]3 years ago
4 0

Answer:

property of parallelograms

scoundrel [369]3 years ago
3 0
True - specifically, that parallelogram could be a parallelogram, rhombus, rectangle or a square.
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Find the Area and Perimeter of a rectangle where length and width are given as 12 and 8 cm respectively.​
sertanlavr [38]
  • Length=L=12cm
  • Breadth=B=8cm

Area:-

\\ \rm\hookrightarrow LB

\\ \rm\hookrightarrow 12(8)

\\ \rm\hookrightarrow 96cm^2

6 0
2 years ago
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A train leaves Chicago traveling west at 60 miles per hour.An hour later,at 12 noon,another train leaves Chicago traveling east
KatRina [158]

60t +60 = 80t

subtract 60t from both sides of=

20t = 60

t = 3

The two trains are the same distance from Chicago at 3pm, at that time both trains will have covered 240 miles

7 0
3 years ago
A pipe that is 120 cm long resonates to produce sound of wavelengths 480 cm, 160 cm, and 96 cm but does not resonate at any wave
erma4kov [3.2K]

Answer:

A Pipe that is 120 cm long resonates to produce sound of wavelengths 480 cm, 160 cm and 96 cm but does not resonate at any wavelengths longer than these. This pipe is:

A. closed at both ends

B. open at one end and closed at one end

C. open at both ends.

D. we cannot tell because we do not know the frequency of the sound.

The right choice is:

B. open at one end and closed at one end .

Step-by-step explanation:

Given:

Length of the pipe, L = 120 cm

Its wavelength \lambda_1 = 480 cm

                         \lambda_2 = 160 cm and \lambda_3 = 96 cm

We have to find whether the pipe is open,closed or open-closed or none.

Note:

  • The fundamental wavelength of a pipe which is open at both ends is 2L.
  • The fundamental wavelength of a pipe which is closed at one end and open at another end is 4L.

So,

The fundamental wavelength:

⇒ 4L=4(120)=480\ cm

It seems that the pipe is open at one end and closed at one end.

Now lets check with the subsequent wavelengths.

For one side open and one side closed pipe:

An odd-integer number of quarter wavelength have to fit into the tube of length L.

⇒  \lambda_2=\frac{4L}{3}                                   ⇒  \lambda_3=\frac{4L}{5}

⇒ \lambda_2=\frac{4(120)}{3}                              ⇒  \lambda_3=\frac{4(120)}{5}

⇒ \lambda_2=\frac{480}{3}                                  ⇒  \lambda_3=\frac{480}{5}

⇒ \lambda_2=160\ cm                           ⇒   \lambda_3=96\ cm  

So the pipe is open at one end and closed at one end .

6 0
3 years ago
Simplify<br> 5 x 5 squared
schepotkina [342]
Answer:
125

Explanation:
5^2 = 25
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6 0
3 years ago
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What is the distance between the points (-1,4) and (7,4)?
user100 [1]

Answer:

8

Step-by-step explanation:

=(7−(−1))2+(4−4)2‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾√

=(8)2+(0)2‾‾‾‾‾‾‾‾‾‾√

=64+0‾‾‾‾‾‾√

=6‾√4

=8

8 0
3 years ago
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