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Sagot :
The distance between the three places is an illustration of Pythagoras theorem.
- The unknown quantity is the distance from Bristol to Chester
- See attachment for sketch
- The equation about the variable is [tex]\mathbf{13^2 = x^2 + (17 - x)^2}[/tex]
- The value of x is 5 or 12
- The distance between Bristol and Chester is either 5 miles or 12 miles
Let:
A represents Amory
B represents Bristol
C represents Chester
(a) The unknown quantity
From the question, we understand that:
[tex]\mathbf{AC + CB = 17}[/tex] --- from Amory to Bristol, through Chester
[tex]\mathbf{AB = 17 - 4}[/tex] --- directly from Amory to Bristol (and you save 4 miles)
So, we have:
[tex]\mathbf{AB = 13}[/tex]
Let the distance from Bristol to Chester be x.
(b) The sketch
See attachment for the illustration
(c) An equation about the variable
Using Pythagoras theorem, we have:
[tex]\mathbf{AB^2 = BC^2 + AC^2}[/tex]
So, we have:
[tex]\mathbf{13^2 = x^2 + (17 - x)^2}[/tex]
(d) Solve the equation
In (c), we have:
[tex]\mathbf{13^2 = x^2 + (17 - x)^2}[/tex]
Evaluate the exponents
[tex]\mathbf{169 = x^2 + 289 - 34x + x^2}[/tex]
Collect like terms
[tex]\mathbf{x^2 + x^2 - 34x + 289 -169 = 0 }[/tex]
[tex]\mathbf{2x^2 - 34x + 120= 0 }[/tex]
Divide through by 2
[tex]\mathbf{x^2 - 17x + 60= 0 }[/tex]
Expand
[tex]\mathbf{x^2 - 12x - 5x + 60= 0 }[/tex]
Factorize
[tex]\mathbf{x(x - 12) - 5(x - 12)= 0 }[/tex]
Factor out x - 12
[tex]\mathbf{(x - 5)(x - 12)= 0 }[/tex]
Solve for x
[tex]\mathbf{x - 5 = 0\ or\ x - 12= 0 }[/tex]
[tex]\mathbf{x = 5\ or\ x = 12 }[/tex]
(e) The distance between Bristol and Chester
The above values of x means that:
The distance between Bristol and Chester is either 5 miles or 12 miles
Read more about Pythagoras equations at:
https://brainly.com/question/23936129
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