Answer:

Explanation: For this, it is often best to find the horizontal asymptote, and then take limits as x approaches the vertical asymptote and the end behaviours.
Well, we know there will be a horizontal asymptote at y = 0, because as x approaches infinite and negative infinite, the graph will shrink down closer and closer to 0, but never touch it. We call this a horizontal asymptote.
So we know that there is a restriction on the y-axis.
Now, since we know the end behaviours, let's find the asymptotic behaviours.
As x approaches the asymptote of 7⁻, then y would be diverging out to negative infinite.
As x approaches the asymptote at 7⁺, then y would be diverging out to negative infinite.
So, our range would be:
Since both trapezoids are similar, so LMNO is a scaled version of QRST by some magnitude
so we can easily use the concept of ratios to find x
in other words: x÷2=6÷5
so x=6×2÷5=12/5=2.4
x=2.4
Answer:
x>9 this is the correct answer
(-4/5) / 3 =
-4/5 * 1/3 =
-4/15 mile per minute
<span>2y-x=11</span>
<span>-x=11-2y
</span><span>x=-11+2y
</span><span><span>x=2y-11</span><span>
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