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3 years ago

-4x2+8x-3+12-8x2-2x-3+6

Mathematics

1 answer:

beks73 [17]3 years ago

8 0

Answer: -24

Step-by-step explanation:

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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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3 years ago

La matemática financiera experimentó grandes avances con la creación de la banca y los negocios creados por los emprendedores. E

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Answer:

c = $\frac{1}{10}$

f= 1500

Step-by-step explanation:

Tenemos la siguiente ecuación

$C = f+cm$

Tenemos los siguientes valores para m=15000, c = 3000 y para m = 20000, c=3500. Es decir, tenemos las siguientes ecuaciones

$3000=f+c15000, 3500=f+c20000$

Si restamos la segunda ecuación de la primera, tenemos que

$3500-3000 = f-f+(20000c-15000c)$

Luego

$500 = 5000c$

Entonces $c = \frac{500}{5000} = \frac{1}{10}$

Si tomamos la primera ecuación, tenemos que $3000-\frac{15000}{10} = f$ . Es decir,

$3000-1500 = f = 1500$

Esta solución, al plantear 2 ecuaciones y dos incógnitas, plantea un sistema dos por dos.

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3 years ago

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