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Llana [10]
3 years ago
6

A) Find the next two numbers in the following sequences: i) 9, 18, 27, 45, 72, __________

Mathematics
2 answers:
postnew [5]3 years ago
8 0

Answer:

81 maybe not sure

Step-by-step explanation:

Mamont248 [21]3 years ago
7 0
99,135 you just multiply 9 by 1,2,3,5,8,11,15 or maybe i’m crazy
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I WILL GIVE BRAINLIEST FOR CORRECT ANSWER!!
Reil [10]

Answer:

2 1/2 divided by 5/2 is 1.

6 3/4 divided by 2 1/4 is 3

3 divided by 2/3 is 9/2.

1/3 divided by 3 is 1/9.

Step-by-step explanation:

hope I helped

7 0
3 years ago
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The marked price of a dozen of copies is Rs 600. If the shopkeeper allows 20% discount and then charges 10% VAT, how many copies
aalyn [17]

Answer:

sassy

Step-by-step explanation:

3 0
3 years ago
11x + 2y = 68 & 5x + y = 13
matrenka [14]

Answer:

i hope its correct

Step-by-step explanation:

11x + 2y=68 ------------1

5x+y=13 ------------2

y= 13-5x. ------------3

put 3 into 1

11x +2(13-5x) =68

11x +26 -10x =68

x=42

put x =42 into 2

5(42) +y =13

210+y=13

y=13-210

y=197

4 0
3 years ago
Help please ASAP
Nat2105 [25]

41. A

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6 0
3 years ago
Planners of an experiment are evaluating the design of a sphere of radius r that is to be filled with helium (0°c, 1 atm pressur
Vikentia [17]
The solution is:A mole of gas occupies 22.4 L A liter is 1000 cubic centimeters because 1 cubic centimeter = 1 mL
One Helium molecule (essentially one helium atom) has atomic mass 4 g/mol So for every 22400 cubic centimeters of volume, we have 4 grams of helium Density of helium = 4g / 22400 cm^3 = 1g / 5600 cm^3 
Volume of a sphere = (4/3)(pi)r^3 Volume of the outside sphere (the entire sphere) is (4/3)(pi)(R+T)^3 Volume of the inside sphere (the hollow region) is (4/3)(pi)R^3 
The difference for the volume of silver. (4/3)(pi)(R+T)^3 - (4/3)(pi)R^3 = (4/3)(pi)(3R^2T + 3RT^2 + T^3) 
The density of silver is 10.5g/cm^3 
So the mass of the silver is computed by:10.5*(4/3)(pi)(3R^2T + 3RT^2 + T^3) = (14*pi)(3R^2T + 3RT^2 + T^3) = (14pi)T(3R^2 + 3RT + T^2) 

Now for the mass of helium: volume x density = (4/3)(pi)R^3 (1/5600) = (pi/4200)R^3 

Set the two masses equal: (pi/4200)R^3 = (14pi)T(3R^2 + 3RT + T^2) R^3 = 58800*T(3R^2 + 3RT + T^2) R / T = 58800*(3R^2 + 3RT + T^2) / R^2 = 58800*( 3+T/R^2+(T/R)^2) 
then solve for xx = T / R 1/x = 58800*( 3+x/R+x^2) 1/58800 = x (3 + x/R + x^2) 1/58800 = 3x + x^3 x^3 + 3x - 1/58800= 0 x = ~ 5.66893x10^(-6)
6 0
3 years ago
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