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3 years ago

Is the number 5 is a prime or a composite

Mathematics

2 answers:

Aleksandr-060686 [28]3 years ago

8 0

Answer:

prime

Step-by-step explanation:

A prime number has only two factors: 1 and itself.

A composite number has more than two factors.

The number 1 is neither prime nor composite.

The prime numbers between 2 and 31 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 and 31 since each of these numbers has only two factors, itself and 1.]

Therefore 5 is prime

iris [78.8K]3 years ago

7 0

Answer:

It’s prime

Step-by-step explanation:

Numbers between 2 and 31 are since these only have 1 and itself.

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The cost function in a computer manufacturing plant is C(x) = 0.28x^2-0.7x+1, where C(x) is the cost per hour in millions of dol

Licemer1 [7]

9514 1404 393

Answer:

$562,500 per hour

Step-by-step explanation:

The cost will be a minimum where C'(x) = 0.

C'(x) = 0.56x -0.7 = 0

x = 0.7/0.56 = 1.25

The cost at that production point is ...

C(1.25) = (0.28×1.25 -0.7)1.25 +1 = -0.35×1.25 +1 = 0.5625

The minimum production cost is $562,500 per hour for production of 1250 items per hour.

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Additional comment

This is different than the minimum cost per item. This level of production gives a per-item cost of $450. The minimum cost per item is $358.30 at a production level of 1890 per hour.

5 0

2 years ago

<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

ankoles [38]

$\bold{Heya!}$

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>→ EXPLANATION :-</h2>

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

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

$\sf{(1) = 1$

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

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

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

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

Hopefully This Helps ! ~

#LearnWithBrainly

$\underline{Answer :}$

Jaceysan ~

7 0

1 year ago

Help me to solve it.

Artemon [7]

Answer:

That is my answer to your question

4 0

2 years ago

What is the solution for the equation StartFraction 5 Over 3 b cubed minus 2 b squared minus 5 EndFraction = StartFraction 2 Ove

wolverine [178]

Answer:

The solutions are:

$b=0,\:b=4$

Step-by-step explanation:

Considering the expression

$\frac{5}{3b^3-2b^2-5}=\frac{2}{b^3-2}$

Solving the expression

$\frac{5}{3b^3-2b^2-5}=\frac{2}{b^3-2}$

$\mathrm{Apply\:fraction\:cross\:multiply:\:if\:}\frac{a}{b}=\frac{c}{d}\mathrm{\:then\:}a\cdot \:d=b\cdot \:c$

$5\left(b^3-2\right)=\left(3b^3-2b^2-5\right)\cdot \:2$

$5b^3-10=6b^3-4b^2-10$

$\mathrm{Switch\:sides}$

$6b^3-4b^2-10=5b^3-10$

$6b^3-4b^2-10+10=5b^3-10+10$

$6b^3-4b^2=5b^3$

$\mathrm{Subtract\:}5b^3\mathrm{\:from\:both\:sides}$

$6b^3-4b^2-5b^3=5b^3-5b^3$

$b^3-4b^2=0$

$Using\:the\:Zero\:Factor\:Principle:$ $if\:\mathrm ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)$