January: 32
February: 28
The percent change is a percent decrease.
Hey there!
Here is your answer:
<u><em>The proper answer to this question is "2×10^6".</em></u>
Reason:
<u><em>Translate these equations in standard from:</em></u>
<u><em>2×10^6=2,000,000 </em></u>
<u><em>9×10^5=900,000</em></u>
<u><em>2,000,000≥900,000</em></u>
<em>Therefore 2×</em><span><em>10^6 is greater!</em>
If you need anymore help feel free to ask me!
Hope this helps!
~Nonportrit</span>
Assuming that b>0 and c>0:
log(a)=3×log(b)-2×log(c)=
log(b³)-log(c²)=
log(b³/c²)
Hence a=b³/c²
Answer:
about 17 meters
Step-by-step explanation:
We can use the Pythagorean theorem to put an upper bound on the height of the bump in the rail. This assumes half the expanded rail length (d+e) is the hypotenuse of a right triangle whose legs are the bump height (b) and the 2500 meter distance (d) from the center of the rail to its end.
The Pythagorean theorem relates these distances this way:
b^2 + d^2 = (d+e)^2
Expanding the square on the right, we can simplify the expression to find b.
b^2 = (d^2 +2de +e^2) -d^2
b^2 = e(2d +e)
b = √(e(2d +e))
Using lengths in meters, we can fill this in to calculate b.
b = √(.06(2·2500 +.06)) = √300.0036
b ≈ 17.32 . . . . meters
_____
<em>Comment on this solution</em>
We don't expect rails to tear loose from the rail bed and rise up to a height matching that of a 3-story building. That is why there are typically expansion joints and shorter rail lengths used in the construction of railways.
The height is a little lower if we take physics into account and distribute the stress in the rail along its length. No doubt the final curve is somewhat more complicated than the triangle we have assumed.
If it were an ellipse, the height might only be 9.4 meters, with the steepest rise occurring near the ends of the rail. The math for this model is beyond the scope of this answer.
