Question

Lesson 23 godine by og

Answer 1

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 $\mathbf{13^2 = x^2 + (17 - x)^2}$
• 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:

$\mathbf{AC + CB = 17}$ --- from Amory to Bristol, through Chester

$\mathbf{AB = 17 - 4}$ --- directly from Amory to Bristol (and you save 4 miles)

So, we have:

$\mathbf{AB = 13}$

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:

$\mathbf{AB^2 = BC^2 + AC^2}$

So, we have:

$\mathbf{13^2 = x^2 + (17 - x)^2}$

(d) Solve the equation

In (c), we have:

$\mathbf{13^2 = x^2 + (17 - x)^2}$

Evaluate the exponents

$\mathbf{169 = x^2 + 289 - 34x + x^2}$

Collect like terms

$\mathbf{x^2 + x^2 - 34x + 289 -169 = 0 }$

$\mathbf{2x^2 - 34x + 120= 0 }$

Divide through by 2

$\mathbf{x^2 - 17x + 60= 0 }$

Expand

$\mathbf{x^2 - 12x - 5x + 60= 0 }$

Factorize

$\mathbf{x(x - 12) - 5(x - 12)= 0 }$

Factor out x - 12

$\mathbf{(x - 5)(x - 12)= 0 }$

Solve for x

$\mathbf{x - 5 = 0\ or\ x - 12= 0 }$

$\mathbf{x = 5\ or\ x = 12 }$

(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:

brainly.com/question/23936129