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Ilya [14]
3 years ago
8

Surface area helpp needed

Mathematics
1 answer:
Firlakuza [10]3 years ago
5 0

Answer:

144 un²

Step-by-step explanation:

Front Parallelogram = 5 x 11 = 55

Two triangles = 3 x 4 = 12

Back Parallelogram = 3 x 11 = 33

Bottom rectangle = 4 x 11 = 44

Total =

55 + 12 + 33 + 44 = 144

units = un²

144 un²

If my answer is incorrect, pls correct me!

If you like my answer and explanation, mark me as brainliest!

-Chetan K

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Answer:

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Step-by-step explanation:

The domain of a function is all possible values for the value of x.

The denominator in a rational function cannot equal to 0.

x+2\neq 0

Subtracting 2 from both sides.

x\neq -2

The domain of the function is all real numbers except -2.

7 0
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This is Grade 9 Mathematics so I kinda stuck here so, if somebody can help me Please sin30⁰+csc60⁰​
mamaluj [8]

\huge\mathrm{Answer࿐}

  • \sin(30)  +  \csc(60)

  • \dfrac{1}{2}  +  \dfrac{2}{ \sqrt{3} }

  • \dfrac{ \sqrt{3 } + 4 }{2 \sqrt{3} }

To remove √3 from denominator, we multiply both numerator as well as denominator with √3

  • \dfrac{ \sqrt{3 } + 4 }{2 \sqrt{3} }  \times  \dfrac{ \sqrt{3} }{ \sqrt{3} }

  • \dfrac{3 + 4 \sqrt{3} }{6}

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Step-by-step explanation:

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5 0
3 years ago
Find the inverse of the given​ matrix, if it exists.Aequals=left bracket Start 3 By 3 Matrix 1st Row 1st Column 1 2nd Column 0 3
BabaBlast [244]

Answer:

A^{-1}=\left[ \begin{array}{ccc} \frac{1}{9} & \frac{4}{27} & - \frac{2}{27} \\\\ \frac{8}{9} & \frac{5}{27} & \frac{11}{27} \\\\ - \frac{4}{9} & \frac{2}{27} & - \frac{1}{27} \end{array} \right]

Step-by-step explanation:

We want to find the inverse of A=\left[ \begin{array}{ccc} 1 & 0 & -2 \\\\ 4 & 1 & 3 \\\\ -4 & 2 & 3 \end{array} \right]

To find the inverse matrix, augment it with the identity matrix and perform row operations trying to make the identity matrix to the left. Then to the right will be inverse matrix.

So, augment the matrix with identity matrix:

\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 4&1&3&0&1&0 \\\\ -4&2&3&0&0&1\end{array}\right]

  • Subtract row 1 multiplied by 4 from row 2

\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ -4&2&3&0&0&1\end{array}\right]

  • Add row 1 multiplied by 4 to row 3

\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ 0&2&-5&4&0&1\end{array}\right]

  • Subtract row 2 multiplied by 2 from row 3

\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ 0&0&-27&12&-2&1\end{array}\right]

  • Divide row 3 by −27

\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ 0&0&1&- \frac{4}{9}&\frac{2}{27}&- \frac{1}{27}\end{array}\right]

  • Add row 3 multiplied by 2 to row 1

\left[ \begin{array}{ccc|ccc}1&0&0&\frac{1}{9}&\frac{4}{27}&- \frac{2}{27} \\\\ 0&1&11&-4&1&0 \\\\ 0&0&1&- \frac{4}{9}&\frac{2}{27}&- \frac{1}{27}\end{array}\right]

  • Subtract row 3 multiplied by 11 from row 2

\left[ \begin{array}{ccc|ccc}1&0&0&\frac{1}{9}&\frac{4}{27}&- \frac{2}{27} \\\\ 0&1&0&\frac{8}{9}&\frac{5}{27}&\frac{11}{27} \\\\ 0&0&1&- \frac{4}{9}&\frac{2}{27}&- \frac{1}{27}\end{array}\right]

As can be seen, we have obtained the identity matrix to the left. So, we are done.

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