Yes. It is a Geometric sequence!
Common ratio = a2/a1 = 12/6 = 2
In short, Your Answer would be Option A
Hope this helps!
Answer:
1, x/3, x^2/9, x^3/27, x^4/81
x^4 +3x^3+9x^2+27x+81
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81
Step-by-step explanation:
(x/3) ^i
i=0 (x/3)^0 = 1
i=1 (x/3)^1 = x/3
i=2 (x/3)^2 = x^2/3^2 = x^2/9
i=3 (x/3)^3 = x^3/3^3 =x^3/27
i=4 (x/3)^4 = x^4/3^4 =x^4/81
The sum is
1+(x/3) + x^2/9 + x^3/27 + x^4/81
We need a common denominator of 81
1*81/81 + x/3 *27/27 + x^2/9 *9/9 + x^3/27 *3/3 + x^4/81
81+27x + 9x^2 + 3x^3 +x^4
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81
Rewriting from largest power to smallest power
x^4 +3x^3+9x^2+27x+81
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81
Consider the bottom most rectangle.
Its length = 5 - 2 = 3 units
Its width = 3 - 1 = 2 units
So, its area = 3(2) = 6 square units.
Consider the middle rectangle.
Its length = 2 - 1 = 1 unit
Its width = 4 - 2 = 2 units
So, its area = 1(2) = 2 square units.
Finally, consider the top rectangle.
Its length = 6 - 1 = 5 units
Its width = 6 - 4 = 2 units
Its area = 5(2) = 10 square units.
Hence, total area = 6 + 2 + 10 = 18 square units. Correct option is C.
Step-by-step explanation:
We dimension of the smallest cube to be made from cuboids of sides 3 m , 4 m and 5 m will be the least common multiple of 3 m, 4 m and 5 m i.e. 60 m
12 cuboidal should be stacked along 5 m edge to 60 m, 15 cuboids should be stacked along 4 m edge and 20. cuboids should be stacked along edge to make a cube of 60 m edge, Hence number of cuboids are 12× 15 ×20=3600
hope it helps
