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lesantik [10]
3 years ago
13

CAN SOMEONE PLS TELL ME DOMAIN AND RANGE

Mathematics
2 answers:
lorasvet [3.4K]3 years ago
5 0

Answers:

  • Domain is (-4, 3]
  • Range is (-5, 5]

=====================================================

Explanation:

The domain is the set of allowed x input values, aka the set of all allowed x coordinates of the points. We see that -4 < x \le 3. It might help to draw vertical lines through the endpoints until you reach the x axis. Note the open hole at x = -4 to indicate we do not include this as part of the domain (hence the lack of "or equal to" for the first inequality sign).

The interval -4 < x \le 3 then can be condensed into the shorthand form (-4, 3] which is the domain in interval notation.

It says: x is between -4 and 3. It can't equal -4 but it can equal 3.

So the use of parenthesis versus square brackets tells the reader which endpoint is included or not.

--------------------------------------------------

The range describes all possible y outputs. We see that y = 5 is the largest it gets and y = -5 is the lower bound. It might help to draw horizontal lines through the endpoints until you reach the y axis. The open hole means -5 is not part of the range.

The range as a compound inequality is -5 < y \le 5. This condenses into the shorthand of (-5, 5] which is the range in interval notation.

Verbally, the range is the set of y values such that y is between -5 and 5. It can't equal -5 but it can equal 5.

riadik2000 [5.3K]3 years ago
5 0

Answer:

Hello,

Step-by-step explanation:

Domain(f)= ]-4;3]={x€R | -4<x≤3}

Rang(f)=]-5;5]={y€R | -5< y ≤5}

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What are the real zeroes of x3 + 6 x2 – 9x - 54?
o-na [289]

Answer:

Option B 3,-3,-6 is correct.

Step-by-step explanation:

We need to find real zeroes of x^3+6x^2-9x-54

Solving

x^3+6x^2-9x-54\\=(x^3+6x^2)+(-9x-54)

Taking x^2 common from first 2 terms and -9 from last two terms we get

=(x^3+6x^2)+(-9x-54)\\=x^2(x+6)-9(x+6)\\

Taking (x+6) common

(x+6)(x^2-9)\\

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=(x+6)((x)^2-(3)^2)\\=(x+6)(x+3)(x-3)

Putting it equal to zero,

(x+6)(x+3)(x-3) =0\\x+6 =0, x+3=0\,\, and\,\, x-3=0\\x=-6, x=-3\,\, and\,\,  x=3

So, Option B 3,-3,-6 is correct.

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Read 2 more answers
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