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3 years ago

What do The restrictions represent in rational expressions

Mathematics

1 answer:

yan [13]3 years ago

5 0

Answer:

The real numbers that give a value of 0 in the denominator are not part of the domain. These values are called restrictions. restrictions in rational expressions make the rational expression undefined. So, the values that make the denominator zero, thus making the overall rational expression undefined represent the restrictions in rational expressions.

Step-by-step explanation:

Rational expressions usually are not defined for all real numbers. The real numbers that give a value of 0 in the denominator are not part of the domain. These values are called restrictions.

Not all rational expressions can be termed as defined for all real numbers. Certain real numbers that can make the denominator zero are not considered as part of domain. Hence, these values are said to be restrictions.

Let the rational expression be

$\frac{P}{Q}$ and Q ≠ 0

P and Q are termed as polynomials and Q can not be zero. The Q which is denominator here can not be equal to zero because we cannot divide by zero. The reason is that division by zero would make the overall rational expression undefined.

It is to be noted that numerator of rational expression can be any real number but denominator must not be zero as it would make the rational expression undefined.

Let suppose the rational expression as

$\frac{3x^{2}-4 }{x-7}$

⇒ $x - 7 = 0$

⇒ $x = 7$

It is clear that the real number 7 would make the denominator zero, hence making the overall rational expression defined. So, the restricted value here be x = 7. Hence, x ≠ 0.

So, it is clear that restrictions in rational expressions make the rational expression undefined. So, the values that make the denominator zero, thus making the overall rational expression undefined represent the restrictions in rational expressions.

Keywords: rational expression, restriction values

Learn more about rational expressions from brainly.com/question/1928496

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A quality control expert at Glotech computers wants to test their new monitors. The production manager claims they have a mean l

tankabanditka [31]

Answer:

The probability that the mean monitor life would be greater than 96.3 months in a sample of 84 monitors

P(X⁻ ≥ 96.3) = 0.0087

Step-by-step explanation:

Step(i):-

Given that the mean of the Population = 95

Given that the standard deviation of the Population = 5

Let 'X' be the random variable in a normal distribution

Let X⁻ = 96.3

Given that the size 'n' = 84 monitors

Step(ii):-

The Empirical rule

$Z = \frac{x^{-} -mean}{\frac{S.D}{\sqrt{n} } }$

$Z = \frac{96.3 - 95}{\frac{5}{\sqrt{84} } }$

Z = 2.383

The probability that the mean monitor life would be greater than 96.3 months in a sample of 84 monitors

P(X⁻ ≥ 96.3) = P(Z≥2.383)

= 1- P( Z<2.383)

= 1-( 0.5 -+A(2.38))

= 0.5 - A(2.38)

= 0.5 -0.4913

= 0.0087

Final answer:-

The probability that the mean monitor life would be greater than 96.3 months in a sample of 84 monitors

P(X⁻ ≥ 96.3) = 0.0087

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2 years ago

Scott wants to cover his patio with concrete pavers. If the patio is shaped like a trapezoid whose bases are 11 feet and 13 feet

GenaCL600 [577]

Answer:

Scott will need 168 ft² of pavers to cover his patio

Explanation:

Scott wants to cover a trapezoid-shaped patio

This means that:

To get the number of square feet of pavers he'll need, we need to get the area of his patio

Area of trapezium id calculated as follows:

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We are given that:

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Scott will need 168 ft² of pavers to cover his patio

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3 years ago

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dem82 [27]

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Step-by-step explanation:

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3 years ago

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jenyasd209 [6]

Answer:

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3 years ago

Read 2 more answers

On a coordinate plane, triangle A B C has points (negative 3, negative 1), (negative 2, negative 3), (1, negative 1) and is refl

andreev551 [17]

Answer:

A' = (-3, 3)

Step-by-step explanation:

If the points are as follows:

A = (-3, -1)

B = (-2, -3)

C = (1, -1)

When reflecting across a y axis the x values remain the same.

When reflecting across y = 1 the y values will remain equal distant from y = 1.

A' = (-3, 3) y value = |-1 - 1| = 2 + 1 = 3

B' = (-2, 5) y value = |-3 - 1| = 4 + 1 = 5

C' = (1, 3) y value = |-1 - 1| = 2 + 1 = 3

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3 years ago