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prohojiy [21]
3 years ago
7

What's a range and domain?

Mathematics
2 answers:
Fofino [41]3 years ago
7 0

Hey!!

Answer: The domain of a function is the complete set of possible values of the independent variable. In plain English, this definition means: The domain is the set of all possible x-values which will make the function "work", and will output real y-values.

Hope this helps!

~LENA~

kirill115 [55]3 years ago
3 0

Hey there! Welcome to Brainly!

The domain is all of the possible x-values, or inputs of a function.

The range is all of the possible y-values, or outputs of a function.

I hope this helps!

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The top and bottom margins of a poster 66 cm each, and the side margins are 44 cm each. If the area of the printed material on t
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Answer:

  • width: 24 cm
  • height: 36 cm

Step-by-step explanation:

When margins are involved, the smallest area will be the one that has its dimensions in the same proportion as the margins. If x is the "multiplier", the dimensions of the printed area are ...

  (4x)(6x) = 384 cm^2

  x^2 = 16 cm^2 . . . . . divide by 24

  x = 4 cm

The printed area is 4x by 6x, so is 16 cm by 24 cm. With the margins added, the smallest poster will be ...

  24 cm by 36 cm

_____

<em>Comment on margins</em>

It should be obvious that if both side margins are 4 cm, then the width of the poster is 8 cm more than the printed width. Similarly, the 6 cm top and bottom margins make the height of the poster 12 cm more than the height of the printed area.

_____

<em>Alternate solution</em>

Let w represent the width of the printed area. Then the printed height is 384/w, and the total poster area is ...

  A = (w+8)(384/w +12) = 384 +12w +3072/w +96

Differentiating with respect to w gives ...

  A' = 12 -3072/w^2

Setting this to zero and solving for w gives ...

  w = √(3072/12) = 16 . . . . matches above solution.

__

<em>Generic solution</em>

If we let s and t represent the side and top margins, and we use "a" for the printed area, then the above equation becomes the symbolic equation ...

  A = (w +s)(a/w +t)

  A' = t - sa/w^2

For A' = 0, ...

  w = √(sa/t)

and the height is ...

  a/w = a/√(sa/t) = √(ta/s)

Then the ratio of width to height is ...

  w/(a/w) = w^2/a = (sa/t)/a

  width/height = s/t . . . . . . the premise we started with, above

6 0
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