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stepladder [879]
3 years ago
7

Please help me with this question.​

Mathematics
1 answer:
Evgen [1.6K]3 years ago
8 0

Answer:

\purple{ \boxed{\frac{d^{2} y}{dx ^{2} }   =    \frac{132}{ {x}^{13} } }}

Step-by-step explanation:

y =  \frac{1}{ {x}^{11} }  \\ y =  {x}^{ - 11}  \\  \frac{dy}{dx}  =  \frac{d}{dx}  {x}^{ - 11}  \\ \frac{dy}{dx}  =   - 11{x}^{ - 11 - 1}  \\ \frac{dy}{dx}  =   - 11{x}^{ - 12}  \\  \\  \frac{d}{dx}  \bigg(\frac{dy}{dx}  \bigg) =  \frac{d}{dx}  ( - 11 {x}^{ - 12} ) \\  \\ \frac{d^{2} y}{dx ^{2} }   =   - 11\frac{d}{dx}  ( {x}^{ - 12} ) \\  \\ \frac{d^{2} y}{dx ^{2} }   =   - 11(  - 12{x}^{ - 13} ) \\  \\ \frac{d^{2} y}{dx ^{2} }   =   132{x}^{ - 13} \\  \\  \huge \red{ \boxed{\frac{d^{2} y}{dx ^{2} }   =    \frac{132}{ {x}^{13} } }}

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Answer:

The given expression  (\frac{5}{6} a^{9}p^{5})  ^{3} = \frac{125}{216}  \times a^{(27) } \times p^{(15)}

Step-by-step explanation:

Here, the given expression is:  (\frac{5}{6} a^{9}p^{5})  ^{3}

Now, starting from the outer most bracket.

As we know :

(abc)^{n}   = (a)^{n} \times (b)^{n}  \times (c)^{n}

and (a^m)^{n} = a^{(m \times n)}

⇒ (\frac{5}{6} a^{9}p^{5})  ^{3} = (\frac{5}{6})^{3} \times (a^{9})^{3}   \times (p^{5}) ^{3}\\

=\frac{125}{216}  \times (a)^{(9\times3) } \times (p)^{(5 \times 3)}\\= \frac{125}{216}  \times a^{(27) } \times p^{(15)}

Hence, the given expression  (\frac{5}{6} a^{9}p^{5})  ^{3} = \frac{125}{216}  \times a^{(27) } \times p^{(15)}

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??

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