For these three questions, I will explain how I came to my conclusion. Please explain where I went wrong, and how I should have completed the problems. Thanks guys.
PROBLEM 1
For the first, my answer was 2^5. I have no idea how they got 2^8. My numerator was 2^6, my denominator was 2^1.
PROBLEM 2
For the second question, I translated as follows.
Statement 1: 2x-2y = 1
x-y = 1/2
x could equal 1, y could equal 1/2, and both numbers are positive.
x could equal 1/4, y could equal -1/4, and only one number is positive.
Stem 1 is insufficient
Statement 2: x/y > 1
X > Y
x could equal 1, y could equal 1/2, and both numbers are positive.
x could equal 1/4, y could equal -1/4, and only one number is positive.
Both statements are satisfied and we still dont know if both are positive or not. So my choice was E.
PROBLEM 3
The third problem, I don't understand at all. What is the question asking?
(2^2 + 2^2) = (2^2) + (2^2), is it not?
Can Someone Please Help EXPLAIN These Problems?? Thanks...
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Hey man, I got the first 2 but the third one has my number. Here they go:
Problem 1: The secret to this one is to look at it differently. First change the view so it looks like: 2^(4-1)^2 – 2^(3-2), from here since the denominator and numerator have similar bases you can just solve for
(4-1)^2 - (3-2) = 9-1 = 8. Answer 2^8
Problem 2: Start by looking at the statements separately:
Statement 1: 2x-2y = 1
In this case it helps to change the view to: 2(x-y) = 1. The main take away from this is that (x-y) has to be positive, because +2 times +(x-y) have to be +1 (+ * + = +). Now, there is no way to know whether x and y are both negative or positive, because they could be both positive: x = 5 & y = 4, which gives you +1, or they could be both negative x = -4 &
y = -5, which gives you +1. Insufficient.
Statement 2: x/y > 1
In this case I did the same thing you did to arrive at x > y, which doesn’t allows us to solve the problem by itself and if you keep it in the original form it still doesn’t clarify if they are both positive or both negative. Insufficient.
Together 1 & 2:
If you use x > y for the 2 examples I provided in 1, they both work because 5 > 4 and -4 > -5 ;however, if you look at statement 2 in the original form: x/y > 1 and plug my example numbers from statement 1 to it, you will get 1.25 and .80 respectively. Since we need the result to be over 1, X and Y are both positive.
Hope this helped and if someone has the answer to problem 3, please share with the forum.
[spoiler][/spoiler]
Problem 1: The secret to this one is to look at it differently. First change the view so it looks like: 2^(4-1)^2 – 2^(3-2), from here since the denominator and numerator have similar bases you can just solve for
(4-1)^2 - (3-2) = 9-1 = 8. Answer 2^8
Problem 2: Start by looking at the statements separately:
Statement 1: 2x-2y = 1
In this case it helps to change the view to: 2(x-y) = 1. The main take away from this is that (x-y) has to be positive, because +2 times +(x-y) have to be +1 (+ * + = +). Now, there is no way to know whether x and y are both negative or positive, because they could be both positive: x = 5 & y = 4, which gives you +1, or they could be both negative x = -4 &
y = -5, which gives you +1. Insufficient.
Statement 2: x/y > 1
In this case I did the same thing you did to arrive at x > y, which doesn’t allows us to solve the problem by itself and if you keep it in the original form it still doesn’t clarify if they are both positive or both negative. Insufficient.
Together 1 & 2:
If you use x > y for the 2 examples I provided in 1, they both work because 5 > 4 and -4 > -5 ;however, if you look at statement 2 in the original form: x/y > 1 and plug my example numbers from statement 1 to it, you will get 1.25 and .80 respectively. Since we need the result to be over 1, X and Y are both positive.
Hope this helped and if someone has the answer to problem 3, please share with the forum.
[spoiler][/spoiler]
- awesomeusername
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1) 2^(4-1)^2 = 2^(3^2) = 2^9
2^(3-2) = 2^1
(2^9)/(2^1) = 2^8
2^(3-2) = 2^1
(2^9)/(2^1) = 2^8
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Your analysis of Statement 2 isn't quite right. If x/y is greater than 1, then x/y is certainly positive. That means that x and y have the same sign: either both are positive, or both are negative. Now, if x and y are both positive, we can multiply both sides of the inequality by y to conclude that x > y > 0. On the other hand, if x and y are both negative, then when we multiply both sides of the inequality by y, we need to reverse the inequality; we get x < y < 0. So Statement 2 tells you that either:mkbigmoz wrote:
Statement 2: x/y > 1
X > Y
x could equal 1, y could equal 1/2, and both numbers are positive.
x could equal 1/4, y could equal -1/4, and only one number is positive.
Both statements are satisfied and we still dont know if both are positive or not. So my choice was E.
Case 1: 0 < y < x
or
Case 2: x < y < 0
So Statement 2 isn't sufficient on its own.
Notice that in Case 1, x - y is positive, while in Case 2, x - y is negative. So if Statement 1 is true, Case 2 is impossible, and x and y must both be positive.
It's asking you to evaluate f(a) + f(b) and f(a+b) for each function. Just looking at answer choice A, for example, ( f(x) = x^2 ) --mkbigmoz wrote: PROBLEM 3
The third problem, I don't understand at all. What is the question asking?
(2^2 + 2^2) = (2^2) + (2^2), is it not?
- to find f(a), plug a in for x into the function; f(a) = a^2.
- Similarly, f(b) = b^2
- So, f(a) + f(b) = a^2 + b^2.
- We need to decide if that's equal to f(a + b). Plug (a + b) into the function in place of x; you get (a + b)^2. Since (a+b)^2 = a^2 + 2ab + b^2, which is not equal to a^2 + b^2, we've just shown that f(a) + f(b) is not equal to f(a + b) for this function.
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