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2 years ago

Consider the steps for determining the quotient of 12/x-4 / 4/x-4

Mathematics

2 answers:

Ierofanga [76]2 years ago

5 0

Answer:

Quotient = 3

Step-by-step explanation:

We need to solve the given expression :

$\dfrac{\dfrac{12}{x-4}}{\dfrac{4}{x-4}}$

It can be done as follows :

As, $\dfrac{\dfrac{a}{b}}{\dfrac{c}{d}}=\dfrac{a}{b}\times \dfrac{d}{c}$

So,

$\dfrac{\dfrac{12}{x-4}}{\dfrac{4}{x-4}}=\dfrac{12}{x-4}\times \dfrac{x-4}{4}\\\\=\dfrac{12}{4}\\\\=3$

Hence, the required quotient is 3.

Lelu [443]2 years ago

4 0

The answer is

a rational

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3 years ago

How is 3072 written in expanded form??

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3 years ago

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3 years ago

An art club is going to a museum. Six adults are going. 18 more kids than adults are going. Adults tickets cost $24 each, which

dybincka [34]

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2 years ago

The polynomial of degree 5, P ( x ) has leading coefficient a=1, has roots of multiplicity 2 at x = 3 and x = 0 , and a root of

ser-zykov [4K]

Answer:

$p_{5} (t) = x^{5} - 5\cdot x^{4} + 3\cdot x^{3} +9\cdot x^{2}$ , for $r_{1} = 0$

Step-by-step explanation:

The general form of quintic-order polynomial is:

$p_{5}(t) = a\cdot x^{5} + b\cdot x^{4} + c\cdot x^{3} + d\cdot x^{2} + e \cdot x + f$

According to the statement of the problem, the polynomial has the following roots:

$p_{5} (t) = (x - r_{1})\cdot (x-3)^{2}\cdot x^{2} \cdot (x+1)$

Then, some algebraic handling is done to expand the polynomial:

$p_{5} (t) = (x - r_{1}) \cdot (x^{3}-6\cdot x^{2}+9\cdot x) \cdot (x+1)\\p_{5} (t) = (x - r_{1}) \cdot (x^{4}-5\cdot x^{3} + 3 \cdot x^{2} + 9 \cdot x)$

$p_{5} (t) = x^{5} - (5+r_{1})\cdot x^{4} + (3 + 5\cdot r_{1})\cdot x^{3} +(9-3\cdot r_{1})\cdot x^{2} - 9 \cdot r_{1}\cdot x$

If $r_{1} = 0$ , then:

$p_{5} (t) = x^{5} - 5\cdot x^{4} + 3\cdot x^{3} +9\cdot x^{2}$

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2 years ago