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Colt1911 [192]
3 years ago
15

CAN SOMEONE SOLVE BOTH OF THESE PROBLEMS ILL GIVE BRAINLIEST

Mathematics
2 answers:
raketka [301]3 years ago
7 0

Answer:

1) HL=15

2)EF=23

Step-by-step explanation:

fredd [130]3 years ago
6 0

Answer:

(I did the 2nd one first), , HL=21, EF=31

Step-by-step explanation:

FInd the midsegment of a trapezoid by adding the bases and dividing by two

(9x+3+5)/2=4x+7, now solve for x

9x+3+5=8x+14

x=6, so ef is 31

Do the first one the same way;

(2.5(12)+12)/2=HL

42/2=HL, HL=21

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antoniya [11.8K]

Answer:

-3

Step-by-step explanation:

Subtract 25x             125=-25x+50

Subtract 50                125-50= -25x

Simplify                       75= -25x

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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
4 years ago
How many students in the marching band are girls
Hoochie [10]

Answer:

Hello,

If 7/10 boys represent 28 boys so, 3/10 girls ? how many girls ?

cross product

7/10 > 28

3/10 > ?

3/10*28/(7/10)=12

There are 12 girls


Step-by-step explanation:


3 0
4 years ago
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