Answer:
x + 7 ft and x + 15 ft as the dimensions of the outside border of the walkway.
Step-by-step explanation:
The dimensions of the fountain is x + 2 and x + 10
Adding the 5-ft walkway, we get x + 7 ft and x + 15 ft as the dimensions of the outside border of the walkway.
Answer:
a. The equation for the line representing Tim's account is y = 10·x + 40
b. The equation for the line representing Chloe's account is y = 20·x + 10
c. The solution to the system of equations is x = 3, y = 70
Step-by-step explanation:
a. The y-intercept of the graph for Tim's account = (0, 40)
A second point on the line graph of Tim's account = (3, 70)
The slope, m of a straight line equation, given two points on the line, (x₁, y₁), (x₂, y₂) is given as follows;
Therefore, the slope, m = (70 - 40)/(3 - 0) = 10
The equation for the line representing Tim's account in slope and intercept form is y = 10·x + 40
b. The y-intercept of the graph for Chloe's account = (0, 10)
A second point on the line graph of Chloe's account is also = (3, 70)
Therefore, the slope of the equation representing Chloe's account, m is given as follows;
m = (70 - 10)/(3 - 0) = 20
The equation for the line representing Chloe's account in slope and intercept form is y = 20·x + 10
c. The solution is found at the point where the two lines meet (are equal) as follows;
For Tim's account y = 10·x + 40
For Chloe's account y = 20·x + 10
Equating both equations gives;
y = y and 10·x + 40 = 20·x + 10
From which we have;
40 - 10 = 20·x - 10·x = 10·x
30 = 10·x
x = 30/10 = 3
x = 3
Therefore, haven found the x-coordinate value, the y-coordinate of the point which is the solution (where the two equations are equal or the two line graphs cross) of the two equations is given as follows;
y = 10·x + 40 where, x = 3, we have;
y = 10 × 3 + 40 = 70
y = 70
The point of the solution of the system of equations is (3, 70)
The solution to the system of equations is x = 3, y = 70.