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rewona [7]
3 years ago
10

Two cities are building bicycle paths. Charles City has built 5km of bicycle paths by the end of the first month, and the total

length of the paths doubles each month. Tinsel Town has built 21km of bicycle paths by the end of the first month, and the total length of the paths increases by 5km per month. At the end of which month does the total length of the bicycle paths in Charles City first exceed the length in Tinsel Town?
Mathematics
1 answer:
sergey [27]3 years ago
6 0

Answer:

Month 4

Step-by-step explanation:

To solve this problem we must propose an equation that models the number of kilometers of bicycle trails made monthly in Charles City and in Tinsel Town

<em>For Charles City </em>we know that the number of kilometers built in the first month is 5, and it doubles every month. Then we have an exponential equation, in base 2, whose initial value is 5.

This equation has the following form:

y = C_1 (2^{x-1})

Where:

C_1 is the number of kilometers built in month 1

x is the number of months: {1, 2, 3, 4, 5 ... x}

x = 1 represents month 1.

So:

Charles\ City\ (km) = 5(2^{x-1})

<em>For Tinsel Town</em> we know that the number of kilometers built in the first month is 21 and increases at a fixed rate of 5 kilometers per month. This can be modeled by a linear equation.

Tinsel\ Town\ (km) = 21 + 5(x-1)

Where x is the number of months. x: {1, 2, 3, 4, 5 ...}

We want to know at the end of what month the total length of the cycle lanes in Charles City first exceeds the length in Tinsel Town

Then we equal both equations and clear x.

5 (2^{x-1}) = 21 + 5(x-1)\\\\5 (2^{x-1}) - (21 + 5(x-1)) = 0

Clearing x from this equation is very difficult, so to find x, we iterate until we get the value of x that causes the equation to approach 0.

For x = 3

5(2^{3-1}) - (21 +5(3-1)) = 0\\\\20- 31= -11

For x = 4

5(2^{4-1}) - (21 +5(4-1)) = 0\\\\40 - 36 = 4

<em>The value must be between x = 3 and x = 4</em>

For x = 3.8

5 (2^{3.8-1}) - (21 +5(3.8-1)) = 0\\\\34.82 - 35 = 0.18

Then x ≈ 3.8 months.

Finally we have that By the end of the fourth month the total length of the cycle lanes in Charles City exceeds the length in Tinsel Town

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Solve:<br> 2/3x+15=17<br><br> 3x+8-x=7<br><br> 4(2x-6)=2
Volgvan

Answer:

x = 3

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x = 13/4

Step-by-step explanation:

Solve for x:

(2 x)/3 + 15 = 17

Put each term in (2 x)/3 + 15 over the common denominator 3: (2 x)/3 + 15 = (2 x)/3 + 45/3:

(2 x)/3 + 45/3 = 17

(2 x)/3 + 45/3 = (2 x + 45)/3:

1/3 (2 x + 45) = 17

Multiply both sides of (2 x + 45)/3 = 17 by 3:

(3 (2 x + 45))/3 = 3×17

(3 (2 x + 45))/3 = 3/3×(2 x + 45) = 2 x + 45:

2 x + 45 = 3×17

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2 x + 45 = 51

Subtract 45 from both sides:

2 x + (45 - 45) = 51 - 45

45 - 45 = 0:

2 x = 51 - 45

51 - 45 = 6:

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Divide both sides of 2 x = 6 by 2:

(2 x)/2 = 6/2

2/2 = 1:

x = 6/2

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Answer:  x = 3

______________________________________________________

Solve for x:

3 x - x + 8 = 7

Grouping like terms, 3 x - x + 8 = (3 x - x) + 8:

(3 x - x) + 8 = 7

3 x - x = 2 x:

2 x + 8 = 7

Subtract 8 from both sides:

2 x + (8 - 8) = 7 - 8

8 - 8 = 0:

2 x = 7 - 8

7 - 8 = -1:

2 x = -1

Divide both sides of 2 x = -1 by 2:

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Solve for x:

4 (2 x - 6) = 2

Divide both sides of 4 (2 x - 6) = 2 by 4:

(4 (2 x - 6))/4 = 2/4

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The gcd of 2 and 4 is 2, so 2/4 = (2×1)/(2×2) = 2/2×1/2 = 1/2:

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Add 6 to both sides:

2 x + (6 - 6) = 1/2 + 6

6 - 6 = 0:

2 x = 1/2 + 6

Put 1/2 + 6 over the common denominator 2. 1/2 + 6 = 1/2 + (2×6)/2:

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Divide both sides by 2:

x = (13/2)/2

2×2 = 4:

Answer: x = 13/4

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