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ahrayia [7]
3 years ago
6

Casey can buy 3 sandwiches and 5 cups of coffee for $26. Eric can buy 4

Mathematics
1 answer:
ioda3 years ago
3 0

Answer:

Step-by-step explanation:

Let's identify what we are looking for in terms of variables.  Sandwiches are s and coffee is c.  Casey buys 3 sandwiches, which is represented then by 3s, and 5 cups of coffee, which is represented by 5c.   Those all put together on one bill comes to 26.  So Casey's equation for his purchases is 3s + 5c = 26.  Eric buys 4 sandwiches, 4s, and 2 cups of coffee, 2c, and his total purchase was 23.  Eric's equation for his purchases then is 4s + 2c = 23.  In order to solve for c, the cost of a cup of coffee, we need to multiply both of those bolded equations by some factor to eliminate the s's.  The coefficients on the s terms are 4 and 3.  4 and 3 both go into 12 evenly, so we will multiply the first bolded equation by 4 and the second one by -3 so the s terms cancel out.  4[3s + 5c = 26] means that 12s + 20c = 104.  Multiplying the second  bolded equation   by -3:  -3[4s + 2c = 23] means that -12s - 6c = -69.  The s terms cancel because 12s - 12s = 0s.  We are left with a system of equations that just contain one unknown now, which is c, what we are looking to solve for.  20c = 104 and -6c = -69.  Adding those together by the method of elimination (which is what we've been doing all this time), 14c = 35.  That means that c = 2.5 and a cup of coffee is $2.50.  There you go!

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Step-by-step explanation:

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2 years ago
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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:

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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

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

Note:

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So,

The fundamental wavelength:

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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 .

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