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Maslowich
3 years ago
15

A 12-pint bucket of paint costs $51.60. What is the price per quart?

Mathematics
1 answer:
My name is Ann [436]3 years ago
6 0

Answer: $8.60

Step-by-step explanation:

1 pint = 0.5 quart

12 pint = 6

51.6/6 = 8.6

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Cones A and B both have volume 48pi cubic units, but have different dimensions. Cone A has radius 6 units and height 4 units. Fi
Andre45 [30]

Answer:

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7 0
2 years ago
Hii, does anyone know a example for 28 + 29? thanks
kaheart [24]

Answer:

57

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
I need help with number 12
fomenos
Ok so when the divisor get larger, the quotient gets smaller. 
For example, 15 ÷ 3 = 5. If we make the divisor larger, you can see the quotient get smaller. 15 ÷ 5 = 3.
The same thing goes for the other way around. If the divisor gets smaller, the quotient gets larger. Hope I helped!
7 0
3 years ago
8 is not a factor of 36.... Your answer doesn't work.
m_a_m_a [10]
Correct :)

36 is not divisible by 8.

The prime factorization of 36 is 2*2*3*3

To get 8 we need 2*2*2 but we don't have enough 2's.
7 0
3 years ago
Read 2 more answers
A metallic sphere is immersed in water in a cylindrical container causing a rise in the level of water by 7.5cm if the cylinder
SVETLANKA909090 [29]

Answer:

6.51 cm

Step-by-step explanation:

Since the sphere causes the water level in the cylindrical container to rise and thus increase by its own volume, the volume of the sphere is V = 4πr³/3 where r = radius of sphere. The volume rise of the container is thus    V' = πR²h where R = radius of base of cylinder = 7 cm and h = height of water level = 7.5 cm.

Since V = V',

4πr³/3 = πR²h

dividing through by π, we have

4r³/3 = R²h

multiplying both sides by 3/4, we have

r³ = 3R²h/4

taking cube-root of both sides, we have

r = ∛(3R²h/4)

Substituting the values of the variables into the equation, we have

r = ∛(3(7 cm)² × 7.5 cm/4)

r = ∛(3 × 49 cm² × 7.5 cm/4)

r = ∛(1102.5cm³/4)

r = ∛(275.625 cm³)

r = 6.508 cm

r ≅ 6.51 cm to 2 decimal places

3 0
3 years ago
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