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3 years ago

Any number that CAN be divided by 2 without having remainder is considered an _______ number

Mathematics

1 answer:

azamat3 years ago

8 0

Step-by-step explanation:

Any number that can be divided by 2 without having remainder is considered an even number.

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Which could be the first step in simplifying this expression? Check all that apply. (x cubed x Superscript negative 6 Baseline)

PtichkaEL [24]

Answer:

$(x^{-3} )^{2}$

$x^6 x^{-12}$

Step-by-step explanation:

$(x^{3} x^{-6} )^{2}$ is the expression given to be solved.

First of all let us have a look at 3 formulas:

$1.\ p^a \times p^b = p^{(a+b)}\\2.\ (p^a \times q^b)^c = (p^{a})^c \times (q^{b})^c\\3.\ (p^a)^b = p^{a\times b}$

Both the formula can be applied to the expression( $(x^{3} x^{-6} )^{2}$ ) during the first step while solving it.

Applying formula (1):

$(x^{3} x^{-6} )^{2}$

Comparing the terms of $(x^{3} x^{-6} )$ with $p^a \times p^b$

$p=x, a =3, b=-6$

$\Rightarrow x^{3+(-6)}\\\Rightarrow x^{3-6}\\\Rightarrow x^{-3}$

So, $(x^{3} x^{-6} )^{2}$ is reduced to $(x^{-3} )^{2}$

Applying formula (2):

Comparing the terms of $(x^{3} x^{-6} )^{2}$ with $(p^a \times q^b)^c$

$p=q=x, a =3, b=-6, c=2$

$\Rightarrow (x^{3})^2\times (x^{-6})^2\\\text{Applying Formula (3)}\\x^6 x^{-12}$

So, $(x^{3} x^{-6} )^{2}$ is reduced to $x^6 x^{-12}$ .

So, the answers can be:

$(x^{-3} )^{2}$

$x^6 x^{-12}$

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