A town has a population of 5000 and grows at 4% every year. What will be the population after 7 years, to the nearest whole numb
1 answer:
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Answer:
x = 20.
Step-by-step explanation:
First, you should remember the relation:
Distance = Speed*Time.
First, we know that a man travels a distance of 20km at a speed of x km/h, in a time T.
We can write this as:
20km = (x km/h)*T
We know that the time is shortened by 12 minutes if the speed is increased by 5km/h
Rewriting these 12 minutes in hours (remember that 60min = 1 hour)
12 min = (12/60) hours = 0.2 hours
Then from this, he can travel the same distance of 20km in a time T minus 0.2 hours if the speed is increased by 5 km/h
We can write this as:
20km = (x + 5 km/h)*(T - 0.2 h)
Then we have a system of two equations, and we want to find the value of x:
20km = (x km/h)*T
20km = (x + 5 km/h)*(T - 0.2 h)
First, we should isolate the variable T in one of the equations, if we isolate it in the first one, we will get:
20km/(x km/h) = T
Replacing that in the other equation we get:
20km = (x + 5 km/h)*(T - 0.2 h)
20km = (x + 5 km/h)*( 20km/(x km/h) - 0.2 h)
Now we can solve this for x.
Removing the units (that we know that are correct) so the math is easier to read, we get:
20 = (x + 5)*(20/x - 0.2)
We only want to solve this for x.
20 = x*20/x - x*0.2 + 5*20/x - 5*0.2
20 = 20 - 0.2*x + 100/x - 1
subtracting 20 in both sides we get:
20 - 20 = 20 - 0.2*x + 100/x - 1 - 20
0 = -0.2*x + 100/x - 1
If we multiply both sides by x we get:
0 = -0.2*x^2 + 100 - x
-0.2*x^2 - x + 100 = 0
This is just a quadratic equation, we can solve it using the Bhaskara's equation, the solutions are:
Then the two solutions are:
x = (1 + 9)/-0.4 = -25
x = (1 - 9)/-0.4 = 20
As x is used to represent a speed, the negative solution does not make sense, so we should use the positive one.
x = 20
then the average speed initially is 20 km/h
<h3>Answer:</h3>
(x, y) ≈ (1.49021612010, 1.22074408461)
<h3>Explanation:</h3>This is best solved graphically or by some other machine method. The approximate solution (x=1.49, y=1.221) can be iterated by any of several approaches to refine the values to the ones given above. The values above were obtained using Newton's method iteration.
_____
Setting the y-values equal and squaring both sides of the equation gives ...
... √x = x² -1
... x = (x² -1)² = x⁴ -2x² +1 . . . . . square both sides
... x⁴ -2x² -x +1 = 0 . . . . . polynomial equation in standard form.
By Descarte's rule of signs, we know there are two positive real roots to this equation. From the graph, we know the other two roots are complex. The second positive real root is extraneous, corresponding to the negative branch of the square root function.