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svlad2 [7]
3 years ago
5

Select the best answer for the question

Mathematics
1 answer:
Klio2033 [76]3 years ago
5 0

Answer:

C. \left[\begin{array}{ccc}-24&0&-3\\8&-48&56\\25&-6&10\end{array}\right]

Step-by-step explanation:

The given matrices are;

F=\left[\begin{array}{cc}-2&0\\0&8\\2&1\end{array}\right]

and

C=\left[\begin{array}{ccc}12&0&\frac{3}{2}\\1&-6&7\end{array}\right]

FC=\left[\begin{array}{cc}-2&0\\0&8\\2&1\end{array}\right]\left[\begin{array}{ccc}12&0&\frac{3}{2}\\1&-6&7\end{array}\right]

FC=\left[\begin{array}{ccc}12(-2)+0(1)&0(-2)+-6(0)&\frac{3}{2}(-2)+7(0)\\12(0)+8(1)&0(0)+8(-6)&0(\frac{3}{2})+8(7)\\2(12)+1(1)&2(0)+1(-6)&2(\frac{3}{2})+1(7)\end{array}\right]

This simplifies to;

FC=\left[\begin{array}{ccc}-24&0&-3\\8&-48&56\\25&-6&10\end{array}\right]

The correct answer is C

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Find the right quotient. 36m 5 n 5 ÷ (12m 3)
polet [3.4K]
ANSWER

The right quotient is

3  {m}^{2} {n}^{5}

EXPLANATION


The given expression is :

\frac{36 {m}^{5} {n}^{5}  }{12 {m}^{3} }



\frac{36 {m}^{5} {n}^{5}  }{12 {m}^{3} }  =  \frac{36}{12}  \times  \frac{ {m}^{5} }{ {m}^{3}} \times  {n}^{5}


\frac{36 {m}^{5} {n}^{5}  }{12 {m}^{3} }  =  3\times  \frac{ {m}^{5} }{ {m}^{3}} \times  {n}^{5}


Recall that,


\frac{ {a}^{m} }{ {a}^{n} }  =  {a}^{m - n}
We apply this property to obtain;



\frac{36 {m}^{5} {n}^{5}  }{12 {m}^{3} }  =  3 \times   {m}^{5 - 3}  \times  {n}^{5}


\frac{36 {m}^{5} {n}^{5}  }{12 {m}^{3} }  =  3  {m}^{2} {n}^{5}

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