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∙ 7y ago23 divided by 135 is 0.17037
3 is the remainder when 23 is divided by 10. 2 x 10 = 20, and 23 is 3 more than 20.
112
-53/4- 67/20= -23/4 - 127/20 = -115/20 - 127/20 = -242/20 = =121/10 = -121/10-53/4- 67/20= -23/4 - 127/20 = -115/20 - 127/20 = -242/20 = =121/10 = -121/10-53/4- 67/20= -23/4 - 127/20 = -115/20 - 127/20 = -242/20 = =121/10 = -121/10-53/4- 67/20= -23/4 - 127/20 = -115/20 - 127/20 = -242/20 = =121/10 = -121/10
20
23 divided by 135 is 0.17037
20 20 - 1978 2009-10-23 was released on: USA: 23 October 2009
range = biggest number - smallest number in this case range = 135 - 8 = 127
3 is the remainder when 23 is divided by 10. 2 x 10 = 20, and 23 is 3 more than 20.
112
45*2 = 90 so 45*20 = 900 45*3 = 135 So 45*23 = 45*20 + 45*3 = 900 + 135 = 1035
-53/4- 67/20= -23/4 - 127/20 = -115/20 - 127/20 = -242/20 = =121/10 = -121/10-53/4- 67/20= -23/4 - 127/20 = -115/20 - 127/20 = -242/20 = =121/10 = -121/10-53/4- 67/20= -23/4 - 127/20 = -115/20 - 127/20 = -242/20 = =121/10 = -121/10-53/4- 67/20= -23/4 - 127/20 = -115/20 - 127/20 = -242/20 = =121/10 = -121/10
20
To calculate the mass of 3.011 x 10^23 nitrogen atoms, first find the molar mass of nitrogen which is 14.01 g/mol. Then, divide the number of atoms by Avogadro's number (6.022 x 10^23) to get the number of moles. Finally, multiply the number of moles by the molar mass to find the mass of the atoms. The calculation would be 3.011 x 10^23 / 6.022 x 10^23 = 0.5 moles, then 0.5 moles x 14.01 g/mol = 7.005 grams.
There are 17.037037... (repeating) lots of 135 cm in 23 metres.
10 18 15 23 20
There are 5 halves in 10 whole ones.