If the width of a rectangle is x + 2 and the length is twice as long, what is the area?
1 answer:
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Answer:
C: I and III only.
Step-by-step explanation:
We have two numbers x and y such that:
And we want to determine which of the following conditions must be true.
First, let's examine the third condition. We have:
To see if this is true, let's try a negative value for y. So, let's use -7. This will give:
Simplify:
This is saying we need a number less than -7 and greater than 7.
This is impossible. So, y <em>must be</em> greater than 0.
Therefore, Condition III must be true.
Next, let's see this compound inequality visually.
Picture the following number-line:
<----------(-y)----------0----------y---------->
Whatever y is, -y is just y on the negative side.
Now, since our condition is that -y<x<y, this means that our x can be anywhere between -y and y. Namely:
<----------(-y)----------0----------y---------->
To determine our correct condition, let's picture x anywhere on the bolded lines. I'm just going to put x here...
<----------(-y)-------(x)--0----------y---------->
Now, remember the alternative definition for absolute value. Namely, the absolute value of x is also the <em>distance</em> from 0 to x. And this distance is always positive.
Since our x is between -y and y, our distance from 0 to x will always be less than the distance from 0 to either y.
Therefore, Condition I is also true.
Let's try an example. Let's let y = 9. So, -y=-9. And let's let x be between 9 and -9, say, -2. So:
<---(-9)-----(-2)--0--------(9)------>
We can see that:
Also, this example counters Condition II, as our x can indeed be negative if we desire it.
Algebraically, if we have a negative x such that:
We can divide everything by -1 to obtain:
We can flip this to get:
Which is our original inequality. So, Condition II does not need to be true.
So, the two conditions that <em>must</em> be true is I and III.
So, our answer is C.