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GMATPrep请大家帮忙

1. XY+Z=X(Y+Z)

Answer: X=1 or Z=0

But I chose X=1 and Y=1

不是也可以吗?

2.For a finite sequence of nonzero numbers, the number of variations in sign is defined as the number of pairs of consecutive terms of the two sequence for which the product of the two consecutive terms is negative. What is the number of variations is in sign for the sequence. 1, -3, 2, 5, -4, -6

Answer:3個

3. For which of the following function f is f(x)=f(1-x)

Answer: f(x)=x^2(1-x)^2


请大家帮帮忙!谢谢

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X=1 and Y=1是X=1 or Z=0的子集,所以后者.

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1. XY+Z=X(Y+Z)

Answer: X=1 or Z=0

But I chose X=1 and Y=1

不是也可以吗?

According to what you've provided, I think your answer should be right.

2.For a finite sequence of nonzero numbers, the number of variations in sign is defined as the number of pairs of consecutive terms of the two sequence for which the product of the two consecutive terms is negative. What is the number of variations is in sign for the sequence. 1, -3, 2, 5, -4, -6

Answer:3個

I'm not sure about the where does the 'highlighted' two come from.

My understanding is that, the question asks you how many pairs of consecutive terms in the sequence have a negative product.

In the sequence shown, there are three pairs.

Namely, 1&(-3), (-3)&2, 5&(-4)

3. For which of the following function f is f(x)=f(1-x)

Answer: f(x)=x^2(1-x)^2

请大家帮帮忙!谢谢

f(1-x)=(1-x)^2*[1-(1-x)]^2=(1-x)^2*x^2=f(x)

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1. XY+Z=X(Y+Z)

Answer: X=1 or Z=0

But I chose X=1 and Y=1

不是也可以吗?

楼主考虑的不全面.依楼主如果只考虑X为1的情况等式是可以成立的,但是如果仅考虑Y为1等式是不成立的,当Y=1时,等式为X+Z=X(1+Z),显然不成立,实际上还有一种情况Z为0则没有考虑

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