What is (81(2)^5(2+4))/(2(2(2)+4)^2(2+2)^2) ?

The solution to (81(2)^5(2+4))/(2(2(2)+4)^2(2+2)^2) is 243/32

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simplify (81(2)^5(2+4))/(2(2(2)+4)^2(2+2)^2)

simplify

\frac{81 ( 2 ) ^{5} ( 2 + 4 )}{2 ( 2 ( 2 ) + 4 ) ^{2} ( 2 + 2 ) ^{2}}

\frac{243}{32}

Decimal Notation

7.59375

Solution steps

\frac{81 \cdot 2 ^{5} ( 2 + 4 )}{2 ( 2 ( 2 ) + 4 ) ^{2} ( 2 + 2 ) ^{2}}

Apply rule:

(a) = a

(2) = 2

= \frac{81 \cdot 2 ^{5} ( 2 + 4 )}{2 ( 2 \cdot 2 + 4 ) ^{2} ( 2 + 2 ) ^{2}}

\frac{81 \cdot 2 ^{5} ( 2 + 4 )}{2 ( 2 \cdot 2 + 4 ) ^{2} ( 2 + 2 ) ^{2}} = \frac{81 \cdot 2 ^{5} \cdot 6}{2 ( 2 \cdot 2 + 4 ) ^{2} \cdot 4 ^{2}}

= \frac{81 \cdot 2 ^{5} \cdot 6}{2 ( 2 \cdot 2 + 4 ) ^{2} \cdot 4 ^{2}}

\frac{2 ^{5}}{2} = 2^{4}

= \frac{81 \cdot 2 ^{4} \cdot 6}{( 2 \cdot 2 + 4 ) ^{2} \cdot 4 ^{2}}

Factor

4^{2} : 2^{4}

= \frac{81 \cdot 2 ^{4} \cdot 6}{( 2 \cdot 2 + 4 ) ^{2} \cdot 2 ^{4}}

Cancel the common factor:

2^{4}

= \frac{81 \cdot 6}{( 2 \cdot 2 + 4 ) ^{2}}

\frac{81 \cdot 6}{( 2 \cdot 2 + 4 ) ^{2}} = \frac{243}{32}

= \frac{243}{32}

y, y x^{2} - 2 x^{3} = 36

5 (\frac{1}{3})^{3}

- 5 > 6 (c - 4) + 7

(2 m - 3 + 7 m^{2}) - (3 - 9 m^{2} - 2 m)

2 x - 1 \geq 0

What is (81(2)^5(2+4))/(2(2(2)+4)^2(2+2)^2) ?
The solution to (81(2)^5(2+4))/(2(2(2)+4)^2(2+2)^2) is 243/32