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Kazeer [188]
4 years ago
6

TexFormula1" title=" \frac{2}{15y} + \frac{3}{5y {}^{3} }" alt=" \frac{2}{15y} + \frac{3}{5y {}^{3} }" align="absmiddle" class="latex-formula">
Mathematics
1 answer:
Xelga [282]4 years ago
3 0

Answer:

2y²   +   9

---------------

      15y³

Step-by-step explanation:

Start by identifying the LCD, and then change each fraction so that its denominator is the LCD.

Here the LCD is 15y³, which is evenly divisible by 15y and 5y³.

Focus now on the first fraction:  2 / (15y).  Multiplying numerator and denominator of this fraction by y² results in

y²·2          2y²

--------- → ----------

y²·15y       15y³           ←This is the correct LCD

Multiplying numerator and denominator of the second fraction by 3 results in:

   3·3            9

------------ → ---------

 3·5y³         15y³          ←This is the correct LCD

So now those two original terms look like:

 2y²         9

--------- + --------

 15y³       15y³

and this can be written in simpler form as:

2y²   +   9

---------------

      15y³

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A pentagonal prism has a height of 9 feet, and each base edge is 3 feet. Find the volume. Round your answer to the nearest hundr
OLga [1]

Height of the pentagon = 9 ft

Edge of base = 3 ft

area of pentagon = \frac{s^{2}n}{4 tan\left ( \frac{180}{n} \right )}

s is the length of any side = 3

n is the number of sides = 5

tan is the tangent function calculated in degrees

\\ \\\\area of pentagon = \frac{3^{2}(5)}{4 tan\left ( \frac{180}{5} \right )}\\\\\\area of pentagon = \frac{9(5)}{4 tan\left ( 36 \right )}\\\\area of pentagon = \frac{45}{4  ( 0.7265 )}\\\\area of pentagon = \frac{45}{2.906}\\\\area of pentagon = 15.4852

Volume of the pentagon = area of base (height )

Volume of the pentagon = 15.4852 (9)

Volume of the pentagon = 139.366

Volume of the pentagon = 139.37 cu.ft


5 0
3 years ago
Read 2 more answers
Given the circle with the equation (x + 1)2 + y2 = 36, determine the location of each point with respect to the graph of the cir
kati45 [8]
\bf \textit{equation of a circle}\\\\ 
(x- h)^2+(y- k)^2= r^2
\qquad 
center~~(\stackrel{}{ h},\stackrel{}{ k})\qquad \qquad 
radius=\stackrel{}{ r}\\\\
-------------------------------\\\\
(x+1)^2+y^2=36\implies [x-(\stackrel{h}{-1})]^2+[y-\stackrel{k}{0}]^2=\stackrel{r}{6^2}~~~~
\begin{cases}
\stackrel{center}{(-1,0)}\\
\stackrel{radius}{6}
\end{cases}

so, that's the equation of the circle, and that's its center, any point "ON" the circle, namely on its circumference, will have a distance to the center of 6 units, since that's the radius.

\bf ~~~~~~~~~~~~\textit{distance between 2 points}
\\\\
(\stackrel{x_1}{-1}~,~\stackrel{y_1}{0})\qquad 
A(\stackrel{x_2}{-1}~,~\stackrel{y_2}{1})\qquad \qquad 
%  distance value
d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}
\\\\\\
\stackrel{distance}{d}=\sqrt{[-1-(-1)]^2+(1-0)^2}\implies d=\sqrt{(-1+1)^2+1^2}
\\\\\\
d=\sqrt{0+1}\implies d=1

well, the distance from the center to A is 1, namely is "inside the circle".

\bf ~~~~~~~~~~~~\textit{distance between 2 points}
\\\\
(\stackrel{x_1}{-1}~,~\stackrel{y_1}{0})\qquad 
B(\stackrel{x_2}{-1}~,~\stackrel{y_2}{6})\\\\\\
\stackrel{distance}{d}=\sqrt{[-1-(-1)]^2+(6-0)^2}\implies d=\sqrt{(-1+1)^2+6^2}
\\\\\\
d=\sqrt{0+36}\implies d=6

notice, the distance to B is exactly 6, and you know what that means.

\bf ~~~~~~~~~~~~\textit{distance between 2 points}
\\\\
(\stackrel{x_1}{-1}~,~\stackrel{y_1}{0})\qquad 
C(\stackrel{x_2}{4}~,~\stackrel{y_2}{-8})
\\\\\\
\stackrel{distance}{d}=\sqrt{[4-(-1)]^2+[-8-0]^2}\implies d=\sqrt{(4+1)^2+(-8)^2}
\\\\\\
d=\sqrt{25+64}\implies d=\sqrt{89}\implies d\approx 9.43398

notice, C is farther than the radius 6, meaning is outside the circle, hiking about on the plane.
3 0
3 years ago
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Complete the following statement <br> (Picture above)
SOVA2 [1]
For this problem, you would replace x with 7 then solve.

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3/9 - sqrt(4) =

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6 0
3 years ago
Find the range for the set of data 24, 30, 17, 22, 22
OverLord2011 [107]

\huge\text{Hey there!}

\huge\textbf{Question reads....}

\text{Find the range for the set of data 24, 30, 17, 22, 22}

\huge\textbf{What does \boxed{range} mean in math?}

\boxed{Range}\rightarrow\text{is the DIFFERENCE between the biggest number and the}\\\text{smallest number.}

\huge\textbf{How do you find the \boxed{range}?}

\text{You find the biggest number \& subtract it from the smallest number.}

\huge\textbf{Equation:}

\text{24, 30, 17, 22, 22}

\huge\textbf{The \boxed{\mathsf{\mathsf{biggest}}} number }\huge\boxed{\downarrow}

\text{30}

\huge\textbf{The \boxed{\mathsf{\mathsf{smallest}}} number }\huge\boxed{\downarrow}

\text{17}

\huge\textbf{Equation:}

\rm{30 - 17}

\huge\textbf{Simplify it:}

\large\text{Start at 30 and go DOWN 17 spaces to the \boxed{left} and you will}\\\large\text{have your answer. }

\huge\textbf{Therefore, your answer should be:}

\huge\boxed{\mathsf{13}}\huge\checkmark

\huge\text{Good luck on your assignment \& enjoy your day!}

<h3>~\frak{Amphitrite1040:)}</h3>
6 0
2 years ago
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Helga [31]
5v+8=93

Subtract 8 from both sides.

5v=85

Divide both sides by 5.

v=17
5 0
4 years ago
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