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Greeley [361]
2 years ago
10

GIVING OUT BRAINLIEST TO WHOEVER GETS ALL OF THEM RIGHT

Mathematics
1 answer:
Thepotemich [5.8K]2 years ago
6 0

Answer:

4) \frac{x}{7\cdot x +x^{2}} is equivalent to \frac{1}{7+x} for all x \ne -7. (Answer: A)

5) \frac{-14\cdot x^{3}}{x^{3}-5\cdot x^{4}} is equivalent to -\frac{14}{1-5\cdot x} for all x \ne \frac{1}{5}. (Answer: B)

6) \frac{x+7}{x^{2}+4\cdot x - 21} is equivalent to \frac{1}{x-3} for all x \ne 3. (Answer: None)

7) \frac{x^{2}+3\cdot x -4}{x+4} is equivalent to x - 1. (Answer: None)

8)  \frac{2}{3\cdot a}\cdot \frac{2}{a^{2}} is equivalent to \frac{4}{3\cdot a^{3}} for all a\ne 0. (Answer: A)

Step-by-step explanation:

We proceed to simplify each expression below:

4) \frac{x}{7\cdot x +x^{2}}

(i) \frac{x}{7\cdot x +x^{2}} Given

(ii) \frac{x}{x\cdot (7+x)} Distributive property

(iii) \frac{1}{7+x} \cdot \frac{x}{x} Distributive property

(iv) \frac{1}{7+x} Existence of multiplicative inverse/Modulative property/Result

Rational functions are undefined when denominator equals 0. That is:

7+x = 0

x = -7

Hence, we conclude that \frac{x}{7\cdot x +x^{2}} is equivalent to \frac{1}{7+x} for all x \ne -7. (Answer: A)

5) \frac{-14\cdot x^{3}}{x^{3}-5\cdot x^{4}}

(i) \frac{-14\cdot x^{3}}{x^{3}-5\cdot x^{4}} Given

(ii) \frac{x^{3}\cdot (-14)}{x^{3}\cdot (1-5\cdot x)} Distributive property

(iii) \frac{x^{3}}{x^{3}} \cdot \left(-\frac{14}{1-5\cdot x} \right) Distributive property

(iv) -\frac{14}{1-5\cdot x} Commutative property/Existence of multiplicative inverse/Modulative property/Result

Rational functions are undefined when denominator equals 0. That is:

1-5\cdot x = 0

5\cdot x = 1

x = \frac{1}{5}

Hence, we conclude that \frac{-14\cdot x^{3}}{x^{3}-5\cdot x^{4}} is equivalent to -\frac{14}{1-5\cdot x} for all x \ne \frac{1}{5}. (Answer: B)

6) \frac{x+7}{x^{2}+4\cdot x - 21}

(i) \frac{x+7}{x^{2}+4\cdot x - 21} Given

(ii) \frac{x+7}{(x+7)\cdot (x-3)} x^{2} -(r_{1}+r_{2})\cdot x +r_{1}\cdot r_{2} = (x-r_{1})\cdot (x-r_{2})

(iii) \frac{1}{x-3}\cdot \frac{x+7}{x+7} Commutative and distributive properties.

(iv) \frac{1}{x-3} Existence of multiplicative inverse/Modulative property/Result

Rational functions are undefined when denominator equals 0. That is:

x-3 = 0

x = 3

Hence, we conclude that \frac{x+7}{x^{2}+4\cdot x - 21} is equivalent to \frac{1}{x-3} for all x \ne 3. (Answer: None)

7) \frac{x^{2}+3\cdot x -4}{x+4}

(i) \frac{x^{2}+3\cdot x -4}{x+4} Given

(ii) \frac{(x+4)\cdot (x-1)}{x+4}  x^{2} -(r_{1}+r_{2})\cdot x +r_{1}\cdot r_{2} = (x-r_{1})\cdot (x-r_{2})

(iii) (x-1)\cdot \left(\frac{x+4}{x+4} \right) Commutative and distributive properties.

(iv) x - 1 Existence of additive inverse/Modulative property/Result

Polynomic function are defined for all value of x.

\frac{x^{2}+3\cdot x -4}{x+4} is equivalent to x - 1. (Answer: None)

8) \frac{2}{3\cdot a}\cdot \frac{2}{a^{2}}

(i) \frac{2}{3\cdot a}\cdot \frac{2}{a^{2}}

(ii) \frac{4}{3\cdot a^{3}} \frac{a}{b}\cdot \frac{c}{d} = \frac{a\cdot b}{c\cdot d}/Result

Rational functions are undefined when denominator equals 0. That is:

3\cdot a^{3} = 0

a = 0

Hence, \frac{2}{3\cdot a}\cdot \frac{2}{a^{2}} is equivalent to \frac{4}{3\cdot a^{3}} for all a\ne 0. (Answer: A)

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