The information given in the question does not refer to an area but some sort of measure in 5 (or more)-dimensional hyperspace.
6*10^9 / 2*10^3 = (6/2) * 10^9 / 10^2 = 3*10^(9-3) = 3*10^6 or 3 million.
Total surface area = 2*(L*W + W*H + H*L) = 2*(10*6 + 6*3 + 3*10) = 2*(60 + 18 + 30) = 2*108 = 216 square units.
6 by 10 by 3 by 5 is not a measure of an area but of a four dimensional hypervolume.
(-6×10⁻¹) ÷ (-3×10⁻²) = (-6 ÷ -3)×(10⁻¹ ÷ 10⁻²) = 2×10⁽⁻¹⁾⁻⁽⁻²⁾ = 2×10¹ → -6×10⁻¹ is 2×10¹ = 20 times bigger (further away from zero) than -3×10⁻² (though -6×10⁻¹ is less than -3×10⁻²).
You need to calculate the surface area of the box: 2 faces at 5 x 3= 15 +15 2 faces at 3 x 2 = 6 + 6 2 faces at 5 x 2 = 10 +10 15 + 15 + 6 + 6 + 10 + 10 = 62 square feet
6*10^9 / 2*10^3 = (6/2) * 10^9 / 10^2 = 3*10^(9-3) = 3*10^6 or 3 million.
The LCM of 3 and 6 is 6. The LCM of 10 and 2 is 10. The LCM of 2, 3, 6 and 10 is 30.
Total surface area = 2*(L*W + W*H + H*L) = 2*(10*6 + 6*3 + 3*10) = 2*(60 + 18 + 30) = 2*108 = 216 square units.
6 by 10 by 3 by 5 is not a measure of an area but of a four dimensional hypervolume.
(-6×10⁻¹) ÷ (-3×10⁻²) = (-6 ÷ -3)×(10⁻¹ ÷ 10⁻²) = 2×10⁽⁻¹⁾⁻⁽⁻²⁾ = 2×10¹ → -6×10⁻¹ is 2×10¹ = 20 times bigger (further away from zero) than -3×10⁻² (though -6×10⁻¹ is less than -3×10⁻²).
You need to calculate the surface area of the box: 2 faces at 5 x 3= 15 +15 2 faces at 3 x 2 = 6 + 6 2 faces at 5 x 2 = 10 +10 15 + 15 + 6 + 6 + 10 + 10 = 62 square feet
Six and two-thirds, or 6 2/3 (2/3)*10= (20/3)= 6 (2/3)
(2 x 3) + 4 = 10 2 x 3 = 6 6 + 4 = 10
here are all the possible combinations 6(12)+8 6(10)+10(2) 6(9)+8(2)+10 6(8)+8(4) 6(7)+8+10(3) 6(6)+8(3)+10(2) 6(5)+8(5)+10 6(5)+10(5) 6(4)+8(2)+10(4) 6(4)+8(7) 6(3)+8(4)+10(3) 6(2)+8+10(6) 6(2)+8(6)+10(2) 6+8(3)+10(5) 6+8(8)+10 8(10) 8(5)+10(4) 10(8)
Seperate like this: 2 1 2 3 2 6 2 10 2 ? then categorize them two by two: 2 1 2 3 2 6 2 10 2 ? now this is the question : 1 3 6 10 ? and the answer is 15" 1 + 2 = 3 ---> + 3 = 6 ----> + 4 = 10 ----> + 5 =15
2/3 of 10 is 6 2/3
2, 3, 6, yes. 5, 10, no.