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wolverine [178]
3 years ago
12

Anybody got an answer?

Mathematics
1 answer:
Anna11 [10]3 years ago
8 0
For a regular n-agon with side length "s" and apothem "h", the area will be
.. A = n*(1/2)*s*h
Your area is
.. A = 10*(1/2)*(3.25 m)*(5 m) ≈ 81.3 m^2
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3. Solve the system using elimination (not substitution or matrices). negative 2 x plus y minus 2 z equals negative 8A N D7 x pl
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Elimination Method

\begin{gathered} -2X+Y-2Z=-8 \\ 7X+Y+Z=-1 \\ 5X+2Y-Z=-9 \end{gathered}

If we multiply the equation 3 by (-1) we obtain this:

\begin{gathered} -2X+Y-2Z=-8 \\ 7X+Y+Z=-1 \\ -5X-2Y+Z=9 \end{gathered}

If we add them we obtain 0, therefore there are infinite solutions. So, let's write it in terms of Z

1. Using the 3rd equation we can obtain X(Y,Z)

\begin{gathered} 5X=-9-2Y+Z \\ X=\frac{-9-2Y+Z}{5} \\  \end{gathered}

2. We can replace this value of X in the 1st and 2nd equations

\begin{gathered} -2\cdot(\frac{-9-2Y+Z}{5})+Y-2Z=-8 \\ 7\cdot(\frac{-9-2Y+Z}{5})+Y+Z=-1 \end{gathered}

3. If we simplify:

\begin{gathered} \frac{-9Y+12Z-63}{5}=-1 \\ \frac{9Y-12Z+18}{5}=-8 \end{gathered}

4. We can obtain Y from this two equations:

\begin{gathered} Y=-\frac{-12Z+58}{9} \\  \end{gathered}

5. Now, we need to obtain X(Z). We can replace Y in X(Y,Z)

\begin{gathered} X=\frac{-9-2Y+Z}{5} \\ X=\frac{-9-2(-\frac{-12Z+58}{9})+Z}{5} \end{gathered}

6. If we simplify, we obtain:

X=\frac{-3Z+7}{9}

7. In conclusion, we obtain that

(X,Y,Z) =

(\frac{-3Z+7}{9},-\frac{-12Z+58}{9},Z)

8 0
1 year ago
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