Answer:
Step-by-step explanation:
Using synthetic division, the zeros goes into the box, which would be -1
-1 | 1 -3 -1 24
-1 4 -3
1 -4 3 21
The remainder is 21
First, let's establish a ratio between these two values. We'll use that as a starting point. I personally find it easiest to work with ratios as fractions, so we'll set that up:

To find the distance <em>per year</em>, we'll need to find the <em>unit rate</em> of this ratio in terms of years. The word <em>unit</em> refers to the number 1 (coming from the Latin root <em>uni-</em> ); a <em>unit rate</em> involves bringing the number we're interested in down to 1 while preserving the ratio. Since we're looking for the distance the fault line moves every one year, we'll have to bring that 175 down to one, which we can do by dividing it by 175. To preserve our ratio, we also have to divide the top by 175:

We have our answer: approximately
0.14 cm or
1.4 mm per year
Ratio 9:13:18......added = 40. This means that one person gets 9/40 of the total, one person gets 13/40 of the total and one person gets 18/40 of the total.
and if the total is 800...
9/40 * 800 = 7200/40 = 180 <==
13/40 * 800 = 10400/40 = 260 <==
18/40 * 800 = 14400/40 = 360 <==
Answer:
- train: 40 kph
- plane: 140 kph
Step-by-step explanation:
Let t represent the speed of the train in km/h. Then 4t-20 is the speed of the plane. Travel times are the same, so we can use the formula ...
time = distance/speed
and equate the travel times.
110/t = 385/(4t-20)
Cross multiplying gives ...
110(4t -20) = 385t
440t -2200 = 385t . . . . . eliminate parentheses
55t -2200 = 0 . . . . . . . . . subtract 385t
t -40 = 0 . . . . . . . . . divide by 55
t = 40 . . . . . . . . . . . add 40; train's speed is 40 kph
4t -20 = 140 . . . . . . find plane's speed; 140 kph
The train's speed is 40 km/h; the plane's speed is 140 km/h.
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<em>Check</em>
Train's travel time = 110 km/(40 km/h) = 2.75 h.
Plane's travel time = 385 km/(140 km/h) = 2.75 h.