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sukhopar [10]
3 years ago
12

The Sum of twice a number and 5 is at most 15. What are the possible values for the number?

Mathematics
2 answers:
NISA [10]3 years ago
6 0
Let's call that number x, twice the number will be 2x, sum of it and 5 is, 2x+5. So if this expression is at most 15, it means, it can be only less than 15 and equal to 15. Let's show this as an inequality.

2x+5\le 15

Let's solve.

2x+5\le 15\\ \\ 2x\le 15-5\\ \\ 2x\le 10\\ \\ x\le \frac { 10 }{ 2 } \\ \\ x\le 5

x\quad \in \quad \left[ -\infty ,5 \right]

If the number is a natural number. The values it can take is, x\quad \in \quad \left[ 0,1,2,3,4,5 \right]


Misha Larkins [42]3 years ago
5 0
5 10=15
3x5=15
so 3.5.10
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<img src="https://tex.z-dn.net/?f=%5Cbegin%7Bequation%7D%5Ctext%20%7B%20Question%3A%20Find%20the%20value%20of%20%7D%20%5Cfrac%7B
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\bold{Heya!}

Your answer to this is:

\sf{= \frac{1}{2} \: + \frac{2}{1 \: + \: x^2} \: + \: \frac{1}{1 \: + \: x^4} \: + \: \frac{2^1^0^0}{1 \: + \: x^2^0^0}

<h2>→ <u>EXPLANATION :-</u></h2>

<u />

<u />\sf{= \frac{1}{2} \: + \frac{2}{1 \: + \: x^2} \: + \: \frac{1}{1 \: + \: x^4} \: + \: \frac{2^1^0^0}{1 \: + \: x^2^0^0} \: (1)

\sf{Apply \: rule:} \: (a) = a

\sf{(1) = 1

\sf{= \frac{1}{2} \: + \frac{2}{1 \: + \: x^2} \: + \: \frac{1}{1 \: + \: x^4} \: + \: \frac{2^1^0^0}{1 \: + \: x^2^0^0} \: ^. \: 1

\sf{\frac{1}{1 \: + \:1} = \frac{1}{2}

\sf{\frac{2^1^0^0}{1 \: + \: x^2^0^0} ^.^ \: 1 \: = \: \frac{2^1^0^0}{1 \: + \: x^2^0^0}

\sf{= \frac{1}{2} \: + \frac{2}{1 \: + \: x^2} \: + \: \frac{1}{1 \: + \: x^4} \: + \: \frac{2^1^0^0}{1 \: + \: x^2^0^0}

Hopefully This Helps ! ~

#LearnWithBrainly

\underline{Answer :}

<em>Jaceysan ~</em>

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