Answer:it forms a parallelogram where the green and the red conjoin........ i think that is correct but I am not positive on my particular response. a answer is better than no answer though.....
Step-by-step explanation:
Adding rational numbers involves adding decimals and repeating decimals like 4.6666666..... Whole numbers on the other hand involves adding 7+6 which arent exactly decimals
The equation you had:
6x + 5y = 45
Solve for Y:
5y = (-6x) + 45 divide by 5
y = (-6/5)x + 9
y = -1 1/5x +9
Short AnswerThere are two numbers
x1 = -0.25 + 0.9682i <<<<
answer 1x2 = - 0.25 - 0.9582i <<<<
answer 2 I take it there are two such numbers.
Let one number = x
Let one number = y
x + y = -0.5
y = - 0.5 - x (1)
xy = 1 (2)
Put equation 1 into equation 2
xy = 1
x(-0.5 - x) = 1
-0.5x - x^2 = 1 Subtract 1 from both sides.
-0.5x - x^2 - 1 = 0 Order these by powers
-x^2 - 0.5x -1 = 0 Multiply though by - 1
x^2 + 0.5x + 1 = 0 Use the quadratic formula to solve this.

a = 1
b = 0.5
c = 1

x = [-0.5 +/- sqrt(0.25 - 4)] / 2
x = [-0.5 +/- sqrt(-3.75)] / 2
x = [-0.25 +/- 0.9682i
x1 = -0.25 + 0.9682 i
x2 = -0.25 - 0.9682 i
These two are conjugates. They will add as x1 + x2 = -0.25 - 0.25 = - 0.50.
The complex parts cancel out. Getting them to multiply to 1 will be a little more difficult. I'll do that under the check.
Check(-0.25 - 0.9682i)(-0.25 + 0.9682i)
Use FOIL
F:-0.25 * -0.25 = 0.0625
O: -0.25*0.9682i
I: +0.25*0.9682i
L: -0.9682i*0.9682i = - 0.9375 i^2 = 0.9375
NoticeThe two middle terms (labled "O" and "I" ) cancel out. They are of opposite signs.
The final result is 0.9375 and 0.0625 add up to 1
Okay, so first you draw a picture and let x be the distance from point D to the rest stop. Then the distance from point to the rest stop is 8 - x
You know that the length of the new trail is y + z, where y is the distance from Ancaster to the rest stop and z is the distance from Dundas to the rest stop.
Now by the Pythagorean theorem, y^2 = 4^2 + x^2 and z^2 = 6^2 + (8 - x) ^2
So take square roots of these, add them, and minimize.
Note: I am assuming the path is perfectly straight, otherwise this approach fails.