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marishachu [46]
3 years ago
8

How do you solve 24 divided by 3/8?

Mathematics
1 answer:
ICE Princess25 [194]3 years ago
4 0
24/1 divided by 3/8 would be 64. That can be found by finding the reciprocal (inverse) of the second fraction, multiplying the numerators and denominators to get 192/3. Then to simplify, you can divide both numerator and denominator by 3 to get 64/1 or 64. Hope that helps!
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<img src="https://tex.z-dn.net/?f=%20%5Clarge%5Cbegin%7Bbmatrix%7D%20%5Cbegin%7Barray%7D%20%7B%20l%20l%20%7D%20%7B%202%20%7D%20%
SVETLANKA909090 [29]

\huge \boxed{\mathbb{QUESTION} \downarrow}

\begin{bmatrix} \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \end{bmatrix} \begin{bmatrix} \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { - 1 } & { 1 } & { 5 } \end{array} \end{bmatrix}

\large \boxed{\mathbb{ANSWER\: WITH\: EXPLANATION} \downarrow}

\begin{bmatrix} \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \end{bmatrix} \begin{bmatrix} \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { - 1 } & { 1 } & { 5 } \end{array} \end{bmatrix}

In matrix multiplication, the number of columns in the 1st matrix is equal to the number of rows in the 2nd matrix.

\left(\begin{matrix}2&3\\5&4\end{matrix}\right)\left(\begin{matrix}2&0&3\\-1&1&5\end{matrix}\right)

Multiply each element of the 1st row of the 1st matrix by the corresponding element of the 1st column of the 2nd matrix. Then add these products to obtain the element in the 1st row, 1st column of the product matrix.

\left(\begin{matrix}2\times 2+3\left(-1\right)&&\\&&\end{matrix}\right)

The remaining elements of the product matrix are found in the same way.

\left(\begin{matrix}2\times 2+3\left(-1\right)&3&2\times 3+3\times 5\\5\times 2+4\left(-1\right)&4&5\times 3+4\times 5\end{matrix}\right)

Simplify each element by multiplying the individual terms.

\left(\begin{matrix}4-3&3&6+15\\10-4&4&15+20\end{matrix}\right)

Now, sum each element of the matrix.

\large\boxed{\boxed{\left(\begin{matrix}1&3&21\\6&4&35\end{matrix}\right) }}

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