Answer:
The approximate solution to the system is (1.2, 4.4)
x = 1.2 and y = 4.4
Step-by-step explanation:
The solution of the system of linear equations equation y = –0.25x + 4.7 and y = 4.9x – 1.64 is shown in the attached graph. The red line represents the equation y = –0.25x + 4.7 and the blue line represents the equation = 4.9x – 1.64.
The solution of the system of equations is their point of intersection shown on the graph.
The point of intersection is (1.231, 4.392). To the nearest tenth, it is (1.2, 4,4). So x = 1.2 and y = 4.4.
So the approximate solution to the system is (1.2, 4.4)
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,
= 120 cm
Its wavelength
= 480 cm
= 160 cm and
= 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:
⇒ 
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.
⇒
⇒ 
⇒
⇒ 
⇒
⇒ 
⇒
⇒
So the pipe is open at one end and closed at one end
.
Denise is constructing A square.
Note: A square has all sides equal.
We already given two vertices M and N of the square.
And another edge of the square is made by from N.
Because a square has all sides of equal length, the side NO should also be equal to MN side of the square.
Therefore, <em>Denise need to place the point of the compass on point N and draw an arc that intersects N O, using MN as the width for the opening of the compass. That would make the NO equals MN.</em>
Therefore, correct option is :
D) place the point of the compass on point N and draw an arc that intersects N O, using MN as the width for the opening of the compass.