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Katen [24]
2 years ago
10

Brian drank 2 glasses of water and a half glass of orange juice. What percentage of

Mathematics
1 answer:
Pie2 years ago
8 0

15 is your percentage to the answer

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A fruit bowl contains apples, oranges, and bananas. The radius of one of the oranges is 1.3 inches. What is the approximate volu
shepuryov [24]

Answer:

Step-by-step explanation:

The volume of a sphere is as follows: (4/3) * (pi) * (radius cubed)

To solve this problem, we can substitute the given value for the radius and for pi.

Volume = (4/3) * (3.14) * (2.197)

Volume = 9.1981...

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2 years ago
Evaluate (2.4x10^4) (4.5x10^3)
goldfiish [28.3K]

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108000000x^2

Step-by-step explanation:

6 0
2 years ago
A chemist whishes to prepare 100 liters of 45% purity of sulphuric acid .He has two kinds of acid solutions in stock ,one is 55%
tester [92]

Answer:

the chemist should use 60 liters of 55% solution and 40  litres of 30% solution in order to prepare 100 liters of 45% purity of sulphuric acid.

Step-by-step explanation:

From the given information,

Let x be the litres of 55% pure solution

Let y be the litres of 30% pure solution

Also;

Given that our total volume of solution is  100 litres

x+y =100  ---- (1)

The total solution of pure by related by the sum of the individual pure concentrations to make up the concentration of final solution.

(0.55)(x)+(0.30)(y) = 0.45(100) ---- (2)

From equation (1)

Let ; y = 100 - x

Replacing the value for y = 100 - x into equation (2)

(0.55)(x)+(0.30)(100-x) = 0.45(100)

0.55x + 30 - 0.30x = 45

0.55x - 0.30x = 45 - 30

0.25x = 15

x = 15/0.25

x = 60 liters of 55% solution

From ; y = 100 - x

y = 100 - 60

y = 40  litres of 30% solution.

Therefore, the chemist should use 60 liters of 55% solution and 40  litres of 30% solution in order to prepare 100 liters of 45% purity of sulphuric acid.

7 0
3 years ago
a researcher in a personal submarine begins at the surface of thr ocean. The submarine descends 20.6 meters and the ascends 5 7/
Mumz [18]

Answer:

Depth = 14.9\ metres

Step-by-step explanation:

Given

Desc = 20.6\ metres

Asc = 5\frac{7}{10}\ metres

Required

Determine the depth

The depth is the distance between the two distances as follows'

Depth = |Desc - Asc|

Substitute values

Depth = |20.6 - 5\frac{7}{10}|

Convert fraction to decimal

Depth = |20.6 - 5.70|

Depth = |14.9|

Depth = 14.9\ metres

6 0
3 years ago
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