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Varvara68 [4.7K]
3 years ago
13

What are the ranges of this relation?

Mathematics
1 answer:
ryzh [129]3 years ago
3 0

Answer:

The range of a relation is the set of the second coordinates from the ordered pairs.

Step-by-step explanation:

The range of a relation from A to B is a subset of B. For Example: If A = {2, 4, 6, 8) B = {5, 7, 1, 9}. Let R be the relation 'is less than' from A to B.

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What is the domain of the function y = √x + 4 ?
puteri [66]

Hello from MrBillDoesMath!

Answer:

Domain:  x >=0 for both cases


Discussion:

Assuming we are dealing with real valued functions, the domain of

y = x^(1/2) +4

is the set of all "x" such that x>= 0  (so we are taking the square root of a positive, or zero, real number)


The domain of  x^(1/2) +6 -7 is the same as for the last function  and for the same reason.


Thank you,

MrB

8 0
3 years ago
How can you use a right angle to decide how to name another angle
blondinia [14]
You can see if the other angles add up to 180 or 360 degrees (depending on shape) then add them to make it the number, for example, if a triangle has a right angle, then the other 2 angles are 45 degrees, knowing that EVERY triangle's angles add up to 180 degrees.
5 0
3 years ago
In the figare, Linet Intersects parallel Lines m and n.
otez555 [7]

Answer:

B.

Step-by-step explanation:

<2 and <3 are vertically opposite angles and are also congruent. Both the angles are hence called vertical angles.

5 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
Solve for x by factoring X2-2x=35
valkas [14]
By factoring, x equals both -5 and 7.
7 0
4 years ago
Read 2 more answers
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