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lana [24]
3 years ago
11

Find the volume of the cylinder. Use 3.14 for

Mathematics
1 answer:
dexar [7]3 years ago
4 0

Answer:

volume = 16956 meters

Step-by-step explanation:

volume = area * height

= pi * r^2 * height

= 3.14 * (30m/2)^2 * 27

volume = 16956 meters

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Is this correct? Or are the numbers wrong
Taya2010 [7]

Answer:

I think that's right. But the angle in the corner actually might be the 28 degrees. One sec this question is a little confusing

Step-by-step explanation:

8 0
2 years ago
The volume of clay used to build a cylindrical pillar with a height of 9 centimeters is 324π cubic centimeters. The area of the
Kitty [74]
The volume of a cylinder is written as V = πr²h. The volume is 324π cm³ and the height is 9 cm. Evaluating it, we will arrive at the value of radius as 6 cm. Area of the base can be obtained using the formula A = πr². If r = 6cm, area of the base will be 36π cm
3 0
3 years ago
Read 2 more answers
Number 1d please help me analytical geometry
lesantik [10]
For a) is just the distance formula

\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%  (a,b)
A&({{ x}}\quad ,&{{ 1}})\quad 
%  (c,d)
B&({{ -4}}\quad ,&{{ 1}})
\end{array}\qquad 
%  distance value
\begin{array}{llll}

d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}
\\\\\\
\sqrt{8} = \sqrt{({{ -4}}-{{ x}})^2 + (1-1)^2}
\end{array}
-----------------------------------------------------------------------------------------
for b)  is also the distance formula, just different coordinates and distance

\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%  (a,b)
A&({{ -7}}\quad ,&{{ y}})\quad 
%  (c,d)
B&({{ -3}}\quad ,&{{ 4}})
\end{array}\ \ 
\begin{array}{llll}

d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}
\\\\\\
4\sqrt{2} = \sqrt{(-3-(-7))^2+(4-y)^2}
\end{array}
--------------------------------------------------------------------------
for c)  well... we know AB = BC.... we do have the coordinates for A and B
so... find the distance for AB, that is \bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%  (a,b)
A&({{ -3}}\quad ,&{{ 0}})\quad 
%  (c,d)
B&({{ 5}}\quad ,&{{ -2}})
\end{array}\qquad 
%  distance value
\begin{array}{llll}

d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}\\\\
d=\boxed{?}

\end{array}

now.. whatever that is, is  = BC, so  the distance for BC is

\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%  (a,b)
B&({{ 5}}\quad ,&{{ -2}})\quad 
%  (c,d)
C&({{ -13}}\quad ,&{{ y}})
\end{array}\qquad 
%  distance value
\begin{array}{llll}

d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}\\\\
d=BC\\\\
BC=\boxed{?}

\end{array}

so... whatever distance you get for AB, set it equals to BC, BC will be in "y-terms" since the C point has a variable in its ordered points

so.. .solve AB = BC for "y"
------------------------------------------------------------------------------------

now d)   we know M and N are equidistant to P, that simply means that P is the midpoint of the segment MN

so use the midpoint formula

\bf \textit{middle point of 2 points }\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%  (a,b)
M&({{-2}}\quad ,&{{ 1}})\quad 
%  (c,d)
N&({{ x}}\quad ,&{{ 1}})
\end{array}\qquad
%   coordinates of midpoint 
\left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right)=P
\\\\\\


\bf \left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right)=(1,4)\implies 
\begin{cases}
\cfrac{{{ x_2}} + {{ x_1}}}{2}=1\leftarrow \textit{solve for "x"}\\\\
\cfrac{{{ y_2}} + {{ y_1}}}{2}=4
\end{cases}

now, for d), you can also just use the distance formula, find the distance for MP, then since MP = PN, find the distance for PN in x-terms and then set it to equal to MP and solve for "x"


7 0
3 years ago
Write an equation in slope intercept form (y=mx+b) for the graph below
mojhsa [17]

Answer:

y=-x+2

Step-by-step explanation:

Y intercept = 2

Slope = -1/1 = 1

y=-1x+2

Simplified

y=-x+2

4 0
3 years ago
Ralph owns a circular farm as shown. He is planning to fence a triangular pigpen ABC in the farm. Point M is the midpoint of the
Flauer [41]

Answer:

Total = \$942

Step-by-step explanation:

Given

A = (0,0)

C = (3,4)

See attachment for farm

Required

The cost of fencing the farm

First, calculate the radius of the farm.

The radius is represented with AC.

So, we have:

AC = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2} ---- distance formula

So, we have:

AC = \sqrt{(0 - 3)^2 + (0 - 4)^2}

AC = \sqrt{(- 3)^2 + ( - 4)^2}

AC = \sqrt{9+16}

AC = \sqrt{25}

AC = 5

Hence:

r = 5 --- radius

Calculate the circumference of the circle

C = 2\pi r

C = 2 * 3.14 * 5

C = 31.4yd

If 1 yard costs $30, 31.4 yards will cost:

Total = 31.4 * 30

Total = \$942

5 0
2 years ago
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