Mathematics, 24.02.2020 14:53 jose0765678755

How can I solve algebras
(2b3c/a3b2c2)4/6

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Answer from: ligittiger12806

" $\frac{8b^4}{3a^1^2}$ " .

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Step-by-step explanation:

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We are asked to simplify the following given expression:

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$\frac{(\frac{2b^3c}{a^3b^2c^2})^4}{6}$ ;

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Let us start with the "numerator" ; which functions as a separate "fraction"—in and of itself—with its own numerator and denominator:

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$(\frac{2b^3c}{a^3b^2c^2})^4$ ;

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Note the following "division" property of exponents:

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→ $(\frac{a}{b})^c = \frac{a^c}{b^c}$ ; $b\neq 0$ .

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Then:

→ (\frac{2b^3c}{a^3b^2c^2})^4 ;

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= $\frac{(2b^3c)^4}{(a^3b^2c^2)^4}$ ; $a\neq 0$ ; $b\neq 0$ ; $c\neq 0$ .

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Then, start by simplifying "this particular numerator":

→ (2b^3c)^4 = ? ;

Note the following "multiplication" property of exponents:

→ (a^b)^c = a^(^b^*^c^) ; $b\neq 0$ .

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Then:

→ (2b^3c)^4 = 2^4 * b^(^3^*^4^)(c^4) = (2*2*2*2) * b^1^2 * c^4 = 16 b^1^2c^4 ;

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The continue by simplifying "this particular denominator":

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→ (a^3b^2c^2)^4 = a^(^3^*^4^)b^(^2^*^4^)c^(^2^*^4^) = a^1^2b^8c^8

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And we have:

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→ $\frac{16b^1^2c^4}{a^1^2b^8c^8}$ .

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Note the following properties of exponents:

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→ $a^-^b = \frac{1}{a^b}$ ; $a\neq 0$ .

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→ $\frac{a^c}{a^b} = a^(^c^-^b^)$ ; $a \neq 0$ .

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So, we have:

→ \frac{16b^1^2c^4}{a^1^2b^8c^8} ;

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→ 16 ÷ 1 = 16 ; ("1" is the "implied coefficient") ; in the numerator.

→ a^1^2 ; stays in the denominator;

→ $\frac{b^1^2}{b^8} = b^(^1^2^-^8^)=b^4$ ; replaces the original: " b^1^2 " [in the numerator]; b^8 is eliminated in the denominator;

→ $\frac{c^4}{c^8} = c^(^4^-^8^) = c^-^4 = \frac{1}{c^4}$ ; The original " c^4 " [in the numerator] is eliminated; the original " c^8 " [in the denominator] is replaced with " c^4 " . ________________

And the expression is rewritten as:

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→ $\frac{16b^4}{a^1^2c^4}$ ;

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→ Now, from the original given problem, we divide this value by "6" ;

which is the same value we get by multiplying this value by: " $\frac{1}{6}$ " ;

→ as follows:

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→ $\frac{16b^4}{a^1^2c^4} *\frac{1}{6}$ ;

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Considering the "16" and the "6" ; each of these can be divided by "2" ;

Specifically, " (16 ÷2 = 8) " ; and: " (6 ÷ 2 = 3) " .

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So, we can rewrite the expression — by substituting "8" in lieu of the "16" ; and "3" in lieu of the "6" ; as follows:

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→ $\frac{8b^4}{a^1^2} * \frac{1}{3}$ ;

And further simplify;

→ $\frac{8b^4}{a^1^2} * \frac{1}{3} = \frac{(8b^4 * 1)}{(a^1^2*3)} = \frac{8b^4}{3a^1^2} .$

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Hope this helps!

Best wishes!

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Answer from: Quest

i think it is everything has to equal 180 so, if i correct 4=c. 6=d. 8=a. 7=b. and the last question would be c. because they are parallel lines and they are congruent.

step-by-step explanation:

the reason everything has to equal to 180 is because that is how it works. i'm guessing that is its rule. 4 and 8 are congruent and parallel so, that would be 30+30 because they have the same number. it would equal 60. 6 and 7 are congruent and parallel so, that would be 30+30 because they have the same number. it would equal 60. 60+60 is 180. so, now you have your answer

Answer from: Quest

the answer is 99.

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How can I solve algebras
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