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14 Dec 2016, 01:30
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Which of the following is the approximation of \((4^4+8^6)/(4^8+16^8\))?
A. \(2^-^1^2\)
B. \(2^-^1^4\)
C. \(2^-^1^6\)
D. \(2^1^2\)
E. \(2^1^4\)
==> Because of the word “approximation”, you get \((4^4+8^6)/(4^8+16^8)= 8^6/16^8\).
Also, from \(8^6/16^8 =(2^3)^6/(2^4)^8=2^1^8/2^3^2=2^1^8^-^3^2=2^-^1^4\), the answer is B.
Answer: B
A. \(2^-^1^2\)
B. \(2^-^1^4\)
C. \(2^-^1^6\)
D. \(2^1^2\)
E. \(2^1^4\)
==> Because of the word “approximation”, you get \((4^4+8^6)/(4^8+16^8)= 8^6/16^8\).
Also, from \(8^6/16^8 =(2^3)^6/(2^4)^8=2^1^8/2^3^2=2^1^8^-^3^2=2^-^1^4\), the answer is B.
Answer: B
15 Dec 2016, 01:16
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In the x-y coordinate plane, the distance between (m, n) and (3, 4) is\(√26\). If m and n are integers, how many possible cases are there?
A. 5 B. 6 C. 7 D. 8 E. 9
==> From \((m-3)^2+(n-4)^2\)=(√26)2, you get (m-3, n-4)=(±1, ±5) or (±5, ±1), so there are total of 8 numbers. Therefore, the answer is D.
Answer: D
A. 5 B. 6 C. 7 D. 8 E. 9
==> From \((m-3)^2+(n-4)^2\)=(√26)2, you get (m-3, n-4)=(±1, ±5) or (±5, ±1), so there are total of 8 numbers. Therefore, the answer is D.
Answer: D
16 Dec 2016, 00:35
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If (n+2)!/n!=156, n=?
A. 2/131 B. 9 C. 10 D. 11 E. 12
==> From (n+2)!/n!=156 and (n+2)(n+1)n!/n!=(n+2)(n+1)=156=13*12, you get n+2=13, n=11.
Therefore, the answer is D
A. 2/131 B. 9 C. 10 D. 11 E. 12
==> From (n+2)!/n!=156 and (n+2)(n+1)n!/n!=(n+2)(n+1)=156=13*12, you get n+2=13, n=11.
Therefore, the answer is D
