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Romashka [77]
4 years ago
10

What is the total surface area for #3

Mathematics
2 answers:
kifflom [539]4 years ago
8 0
The answer choice c.
Liono4ka [1.6K]4 years ago
7 0
I think the answer is C
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Answer:

Step-by-step explanation:

\sf \dfrac{1}{3}a^3-\dfrac{3}{4}a^2-\dfrac{5}{2}-\left[\dfrac{5}{2}a^2+\dfrac{3}{2}a^3+\dfrac{a}{3}-\dfrac{6}{5}\right]=

               \sf =  \dfrac{1}{3}a^3-\dfrac{3}{4}a^2-\dfrac{5}{2}-\dfrac{5}{2}a^2-\dfrac{3}{2}a^3-\dfrac{a}{3}+\dfrac{6}{5}\\\\\\( Combine \ like \ terms)\\\\= \dfrac{1}{3}a^3 -\dfrac{3}{2}a^3 -\dfrac{3}{4}a^2-\dfrac{5}{2}a^2-\dfrac{a}{3}-\dfrac{5}{2}+\dfrac{6}{5}\\\\=\left[\dfrac{1*2}{3*2}-\dfrac{3*3}{2*3}\right]a^3 + \left[-\dfrac{3}{4}-\dfrac{5*2}{2*2}\right]a^2-\dfrac{a}{3}+\left[-\dfrac{5*5}{2*5}+\dfrac{6*2}{5*2}\right]\\\\\\

               \sf ==\left[\dfrac{2}{6}-\dfrac{9}{6}\right]a^3+\left[-\dfrac{3}{4}-\dfrac{10}{4}\right]a^2-\dfrac{a}{3}+\left[-\dfrac{25}{10}+\dfrac{12}{10}\right]\\\\=\dfrac{2-9}{6}a^3+\dfrac{(-3-10)}{4}a^2-\dfrac{a}{3}+\dfrac{(-25+12)}{15}\\\\

               \sf = \dfrac{-7}{6}a^3+ \dfrac{(-13)}{4}a^2-\dfrac{a}{3}+\dfrac{(-13)}{15}\\\\=-\dfrac{7}{6}a^3-\dfrac{13}{4}a^2-\dfrac{a}{3}-\dfrac{13}{15}

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