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Westkost [7]
3 years ago
5

What is the volume of a rectangular prism with the dimensions: base 3 1 2 cm, height 1 1 2 cm, and length 5 1 2 cm?

Mathematics
1 answer:
aleksandrvk [35]3 years ago
3 0

Answer:

The volume of a rectangular prism is 28\frac{7}{8}\ cm^{3}

Step-by-step explanation:

we know that

The volume of the rectangular prism is equal to

V=BHL

Convert the given dimensions to an improper fractions

B=3\frac{1}{2}\ cm=\frac{3*2+1}{2}=\frac{7}{2}\ cm

H=1\frac{1}{2}\ cm=\frac{1*2+1}{2}=\frac{3}{2}\ cm

L=5\frac{1}{2}\ cm=\frac{5*2+1}{2}=\frac{11}{2}\ cm

substitute in the formula

V=(\frac{7}{2})(\frac{3}{2})(\frac{11}{2})=\frac{231}{8}\ cm^{3}

Convert to mixed number

\frac{231}{8}=\frac{224}{8}+\frac{7}{8}=28\frac{7}{8}\ cm^{3}

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scZoUnD [109]

Answer:

It is an arithmetic sequence and the common difference is 5

Step-by-step explanation:

4+5 = 9; 9+5=14 (adding is arithmetic, multiplying is geometric)

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3 years ago
5×10+3÷4×5+8-45×6×23÷4=
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Mkey [24]

Answer:

X = \begin{bmatrix}1&3\\ 2&4\end{bmatrix}

Step-by-step explanation:

The question we have at hand is, in other words,

\begin{bmatrix}4&2\\ \:1&1\end{bmatrix}\left(X\right)=\begin{bmatrix}8&20\\ \:3&7\end{bmatrix} - where we have to solve for the value of X

If we have to isolate X here, then we would have to take the inverse of the following matrix ...

\begin{bmatrix}4&2\\ \:1&1\end{bmatrix} ... so that it should be as follows ... \begin{bmatrix}4&2\\ \:1&1\end{bmatrix}^{-1}

Therefore, we can conclude that the equation as to solve for " X " will be the following,

X=\begin{bmatrix}4&2\\ 1&1\end{bmatrix}^{-1}\begin{bmatrix}8&20\\ 3&7\end{bmatrix} - First find the 2 x 2 matrix inverse of the first portion,

\begin{bmatrix}4&2\\ 1&1\end{bmatrix}^{-1} = \frac{1}{\det \begin{pmatrix}4&2\\ 1&1\end{pmatrix}}\begin{pmatrix}1&-2\\ -1&4\end{pmatrix}= \frac{1}{2}\begin{bmatrix}1&-2\\ -1&4\end{bmatrix} = \begin{bmatrix}\frac{1}{2}&-1\\ -\frac{1}{2}&2\end{bmatrix}

At this point we have to multiply the rows of the first matrix by the rows of the second matrix,

X = \begin{bmatrix}\frac{1}{2}&-1\\ -\frac{1}{2}&2\end{bmatrix}\begin{bmatrix}8&20\\ 3&7\end{bmatrix} ,

X = \begin{pmatrix}\frac{1}{2}\cdot \:8+\left(-1\right)\cdot \:3&\frac{1}{2}\cdot \:20+\left(-1\right)\cdot \:7\\ \left(-\frac{1}{2}\right)\cdot \:8+2\cdot \:3&\left(-\frac{1}{2}\right)\cdot \:20+2\cdot \:7\end{pmatrix} - Simplifying this, we should get ...

\begin{bmatrix}1&3\\ 2&4\end{bmatrix} ... which is our solution.

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