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strojnjashka [21]
3 years ago
5

A

Mathematics
1 answer:
Gnesinka [82]3 years ago
7 0

Answer:

A: 779 cm²

B: 1837 cm²

Step-by-step explanation:

For both problems, use the formula for surface area of a cylinder:

SA = 2πr² + 2πrh

2πr² is the two bases.

2πrh is the curved surface.

<u>PROBLEM A</u>

"the cylinder is 60 cm long" is h = 60.

If given diameter, you can find "r" by dividing it by 2. d = 2r

Given d = 4, then r = 2.

SA = 2πr² + 2πrh

SA = 2π2² + 2π2(60)

SA = 8π + 240π               Add

SA = 248π                       Exact answer

SA ≈ 779.114978             Answer on calculator

SA ≈ 779           Rounded answer

Remember to include the units.

The surface area is about 779 cm².

<u>PROBLEM B</u>

"80 cm long" h = 80.

"circumference of 22 cm". C = 22. Remember C = 2πr. Find "r".

C = 2πr

22 = 2πr

11 = πr

r ≈ 11/π

SA = 2πr² + 2πrh

SA = 2π(11/π)² + 22(80)       Substitute 2πr with the circumference.

SA ≈ 1837.030992            Answer on calculator

SA ≈ 1837               Rounded answer

Remember to include the units.

The surface area is about 1837 cm².

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Step-by-step explanation:

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Step-by-step explanation:

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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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