How can 2182 be written as the sum of four consecutive whole numbers?
1 answer:
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1) (23^-2)^3 (5
3^2)^2 / (3^-2)(5
2)^2 = 2
2) (3^3) (4^0)^2 (32)^-3 (2^2) = 1/2
3) (3^74^7) (2
5)^-3 (5)^2 / (12^7) (5^-1) (2^-4) = 2
4) (2.3)^-1 (2^0) / (2.3)^-1 = 1
<u>Step-by-step explanation</u>:
Step 1 :
(23^-2)^3 (5
3^2)^2 / (3^-2)(5
2)^2
⇒ (2^3) (3^-6) (5^2) (3^4) / (3^-2) (10^2)
⇒ (2.2^2.5^2) (3^-6.3^4) / (3^-2) (10^2)
⇒ (2)(10^2) (3^-2) / (3^-2) (10^2)
⇒ 2
Step 2 :
(3^3) (4^0)^2 (32)^-3 (2^2)
Any number with power 'zero' is 1.
⇒ (3^3) (1^2) (3^-3) (2^-3) (2^2)
⇒ (2^(-3+2))
⇒ 2^-1 = 1/2
Step 3 :
(3^74^7) (2
5)^-3 (5)^2 / (12^7) (5^-1) (2^-4)
⇒ (12^7) (2^-3) (5^-3) (5^2) / (12^7) (5^-1) (2^-4)
⇒ (12^7) (2^-3) (5^-3) (5^2) / (12^7) (5^-1) (2^-4)
⇒ (2^-3) (5^(-3+2)) / (5^-1) (2^-4)
⇒ (2^-3) (5^-1) / (5^-1) (2^-4)
⇒ 1 / (2^-1)
⇒ 2
Step 4 :
(2.3)^-1 (2^0) / (2.3)^-1
⇒ (2^0)
⇒ Any number with power 'zero' is 1
⇒ 1