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Black_prince [1.1K]
3 years ago
11

In a water treatment plant, water passes through a cone-shaped filter with a height of 4 m and a diameter of 9 m. The water flow

s from the filter into a cylinder below it. One full cone-shaped filter fills one-fifth of the cylinder.
What are the dimensions of the cylinder? Use 3.14 to approximate pi.

A.
h = 3 m; r = 3 m

B.
h = 4.23 m; r = 5 m

C.
h = 15 m; r = 6 m

D.
h = 33.75 m; r = 2 m
Mathematics
2 answers:
QveST [7]3 years ago
7 0
\frac{9/2*9/2*4* \pi }{3}
= \frac{81*3.14 }{3}
=27*3.14 (the cone's v)
so multiply 5 to the v = 135*3.14 (cylinder's v)
A:3*3*3*3.14=27*3.14 X
B:5*5*4.23*3.14=105.75*3.14 X
C:6*6*15*3.14=540*3.14 X
D:2*2*33.75*3.14=135*3.14 O
Answer: is .....D
:)
Maslowich3 years ago
5 0
<span>so volume of a cone=1/3 times (area of base) times height

area of base=circle so
1/3 times pi times radius^2 times height

height=4
diameter=2radius
diameter/2=radius
9/2=4.5=radius

height=4
radius=4.5
aprox pi at 3.14
1/3 times (3.14 times 4.5^2) times 4=84.78
that's the volume of the cone

volume of cone=1/5 of cylinder

84.78=1/5cylinder
multiply both sdies  by 5
423.9=volume of cylinder

volume of cylinder=height times area of base
area of base=3.14 times r^2
v=h3.14r^2
423.9=3.14hr^2
divide both sdies by 3.14
135=hr^2
since we are not given the specific ratio of height to radius or such to the cone to the cylinder, we must use trial and error to elimninate wrong choices


135=hr^2
let's try A

A
135=3(3^3)
135=3(9)
135=27
false
A is wrong

B
4.23=h  5=r
135=4.23 times 5^2
135=4.23 times 25
135=105.75
false
B is wrong


C
h=15 r=6
135=15 times 6^2
135=15 times 36
135=540
false
C is wrong


D
33.75=h 2=r
135=33.75 times 2^2
135=33.75 times 4
135=135

the answer is D since it fitst the dimentions</span>
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Answer:

x=3−2d,5,−2(1+d),5,−27−2d,5,−2(2+d),5,2(y−d),5

Step-by-step explanation:Solving for x. Want to solve for y or solve for d instead?

1 Simplify  0-20−2  to  -2−2.

3,-2,-27,-2-2,02y=1,5x+2d3,−2,−27,−2−2,02y=1,5x+2d

2 Simplify  -2-2−2−2  to  -4−4.

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3 Subtract 2d2d from both sides.

3-2d,-2-2d,-27-2d,-4-2d,02y-2d=1,5x3−2d,−2−2d,−27−2d,−4−2d,02y−2d=1,5x

4 Divide both sides by 1,51,5.

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​−2(1+d)

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​−27−2d

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​−2(2+d)

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3-2d,5,\frac{-2(1+d)}{1},5,\frac{-27-2d}{1},5,\frac{-2(2+d)}{1},5,\frac{2(y-d)}{1},5=x3−2d,5,

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​−27−2d

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​

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