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RideAnS [48]
3 years ago
6

An hourglass, composed of two identical cones, is 12 cm tall. The radius of each cone is 3 cm. If you want to fill the bottom ha

lf of the hourglass full of salt, how much salt will you need? Explain the method you would use to find the amount of salt.
Mathematics
1 answer:
EleoNora [17]3 years ago
6 0
The volume of a cone is V= (1/3)pi r²h
it was given 
h=12cm
radius=3cm 
the volume of the first cone is V1= 1/3)pi r²h1
where h1= h/2=6cm 
so its volume is V1=1/3)xpix3²x6=56.52cm^3

and since the two cones are identical so V1=V2=<span>56.52cm^3,
we need </span><span>56.52cm^3 or salt </span>
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The first time that Arnold ran for mayor in a small town, he received 5% of the votes. This year, when Arnold ran for mayor agai
sammy [17]

Answer:

20 votes

Step-by-step explanation:

You start with the percents: You always start with 100%

100-5=95

Then you multiply

95 x 4

4 is how many votes he received the 2nd time

95 x 4= 380

Then you subtract

380-360= 20

Your Welcome :)

6 0
2 years ago
Which of the following expressions is equal to 3x^2 + 27
Kisachek [45]
Factor: 
3x^2 + 27
= 3(x^2  + 9)
Answer is 3(x^2 + 9), when factored.


A) (3x + 9i)(x + 3i)
= (3x + 9i)(x + 3i)
= (3x)(x) + (3x)(3i) + (9i)(x) + (9i)(3i)
= 3x^2 + 9ix + 9ix + 27i^2
= 27i^2 + 18ix + 3x^2



B) (3x - 9i)(x + 3i)
= (3x +  - 9i)(x + 3i)
= (3x)(x) + (3x)(3i) + ( - 9i)(x) + (- 9i)(3i)
= 3x^2 + 9ix - 9ix - 27i^2
= 27i^2 + 3x^2


C) (3x - 6i)(x + 21i)
= (3x +  - 6i)(x + 21i)
= (3x)(x) + (3x)(21i) + (- 6i)(x) + ( -6i)(21i)
= 3x^2 + 63ix - 6ix - 126i^2
=  - 126i^2 + 57ix + 3x^2






D) (3x - 9i)(x - 3i)
=  (3x +   - 9)(x +  - 3)
= (3x)(x) + (3x)( - 3i) + (- 9)(x) + ( - 9)( - 3i)
= 3x^2 - 9ix - 9x + 27i
= 9ix + 3x^2 + 27i - 9x









Hope that helps!!!

8 0
3 years ago
Read 2 more answers
If AB=20, then AB+CD=?​
Leona [35]

Answer:

20*20+CD=400+CD

Step-by-step explanation:

Multiply  

20

by  

20

6 0
3 years ago
6. A bus station has 8 buses that hold a total
lesantik [10]

Answer:

1040 passengers

Step-by-step explanation:

416÷8=52

52×20=1040

5 0
2 years ago
Read 2 more answers
Particle 1 of charge q1 �� ��5.00q and particle 2 of charge q2 �� ��2.00q are fixed to an x axis. (a) as a multiple of distance
Naya [18.7K]

<span>Assuming that the particle is the 3rd particle, we know that it’s location must be beyond q2; it cannot be between q1 and q2 since both fields point the similar way in the between region (due to attraction). Choosing an arbitrary value of 1 for L, we get </span>

<span>
k q1 / d^2 = - k q2 / (d-1)^2 </span>

Rearranging to calculate for d:

<span> (d-1)^2/d^2 = -q2/q1 = 0.4 </span><span>
<span> d^2-2d+1 = 0.4d^2 </span>
0.6d^2-2d+1 = 0  
d = 2.72075922005613 
d = 0.612574113277207 </span>

<span>
We pick the value that is > q2 hence,</span>

d = 2.72075922005613*L

<span>d = 2.72*L</span>

3 0
2 years ago
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