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belka [17]
3 years ago
7

5/9 rounded to the nearest half?

Mathematics
1 answer:
Archy [21]3 years ago
6 0

0.6.

5/9 = 0.5555555... which rounds to 0.6

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How does the method for solving equations with fractional or decimal coefficients and constants compare with the method for solv
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Answer:

Step-by-step explanation:

When solving equations with fractional or decimal coefficients, the equations needs to be multiplied by the multiple of denominator such that the equations have integer coefficients and constants

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2 years ago
Can you guys help me with this i think its d or a
liraira [26]

Answer:

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Step-by-step explanation:

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3 years ago
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Isaak is writing an explicit formula to represent the sequence.
gulaghasi [49]
In a geometric progression:
a, b, c, d...
The common ratio can be obtained using:
b/a = c/b = d/c
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2 years ago
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Explain how to form a linear combination to eliminate the variable y for this system. 2x - 3y = 3 5x + 2y = 17A)Multiply the fir
jonny [76]
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7 0
3 years ago
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

5 0
3 years ago
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