-9 to the negative 1 power

1 answer:

8 0

Bear in mind that, (-9) and -9 are very different mammals, we're assuming, you meant (-9)⁻¹.

$\bf \left.\qquad \qquad \right.\textit{negative exponents}\\\\
a^{-{ n}} \implies \cfrac{1}{a^{ n}}
\qquad \qquad
\cfrac{1}{a^{ n}}\implies a^{-{ n}}
\qquad \qquad 
a^{{{ n}}}\implies \cfrac{1}{a^{-{{ n}}}}
\\\\
-------------------------------\\\\
(-9)^{-1}\implies \cfrac{1}{-9}\implies -\cfrac{1}{9}$

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How would I go about solving this problem???

schepotkina [342]

So.. if you take a peek at the picture below

the trunk is really just a half-cylinder on top of a square, with a depth of 2 meters

what's the volume? well, easy enough, take the volume of the cylinder, then half it
take the volume of the rectangular prism, and then add them both

$\bf \textit{volume of a cylinder}\\\\

\begin{array}{llll}
C=\pi r^2 h\\\\
\textit{half that}\\\\
\cfrac{\pi r^2 h}{2}
\end{array}\qquad 
\begin{cases}
r=radius\\
h=height\\
-------\\
r=\frac{1}{2}\\
h=2
\end{cases}\implies \cfrac{C}{2}=\cfrac{\pi \left( \frac{1}{2}\right)^2 2}{2}\\\\
-----------------------------\\\\
\textit{volume of a square}\\\\
V=lwh\qquad 
\begin{cases}
l=length\\
w=width\\
h=height
----------\\
l=1\\
w=1\\
h=2
\end{cases}\implies V=2$

now.. for the surface area... $\bf \textit{surface area of a cylinder}\\\\
\begin{array}{llll}
S=2\pi r(h+r)\\\\
\textit{half that}\\\\
\cfrac{2\pi r(h+r)}{2}
\end{array}\begin{cases}
r=radius\\
h=height\\
-------\\
r=\frac{1}{2}\\
h=2
\end{cases}$

now.. for the surface area of the prism... well

is really just 6 rectangles stacked up to each other at the edges

so... get the area of the lateral rectangles, and the one at the bottom, skip the rectangle atop, because is the one overlapping the cylinder, and is not outside, and thus is not surface area then

for the lateral ones, you have a front of 1x1, a back of 1x1 and a left of 1x2 and a right of 1x2, and then the one at the bottom, which is a 1x2

then add both surface areas, and that's the surface area of the trunk