Answer:
(a)In the attachment
(b)The road of length 35.79 km should be built such that it joins the highway at 19.52km from the perpendicular point P.
Step-by-step explanation:
(a)In the attachment
(b)The distance that enables the driver to reach the city in the shortest time is denoted by the Straight Line RM (from the Ranch to Point M)
First, let us determine length of line RM.
Using Pythagoras theorem

The Speed limit on the Road is 60 km/h and 110 km/h on the highway.
Time Taken = Distance/Time
Time taken on the road 
Time taken on the highway 
Total time taken to travel, T 
Minimum time taken occurs when the derivative of T equals 0.

Square both sides

The road should be built such that it joins the highway at 19.52km from the point P.
In fact,

Answer:
t = 139/490 + sqrt(75671)/490 or t = 139/490 - sqrt(75671)/490
Step-by-step explanation:
Solve for t:
4.9 t^2 - 2.78 t - 1.15 = 0
4.9 t^2 - 2.78 t - 1.15 = (49 t^2)/10 - (139 t)/50 - 23/20:
(49 t^2)/10 - (139 t)/50 - 23/20 = 0
Multiply both sides by 10/49:
t^2 - (139 t)/245 - 23/98 = 0
Add 23/98 to both sides:
t^2 - (139 t)/245 = 23/98
Add 19321/240100 to both sides:
t^2 - (139 t)/245 + 19321/240100 = 75671/240100
Write the left hand side as a square:
(t - 139/490)^2 = 75671/240100
Take the square root of both sides:
t - 139/490 = sqrt(75671)/490 or t - 139/490 = -sqrt(75671)/490
Add 139/490 to both sides:
t = 139/490 + sqrt(75671)/490 or t - 139/490 = -sqrt(75671)/490
Add 139/490 to both sides:
Answer: t = 139/490 + sqrt(75671)/490 or t = 139/490 - sqrt(75671)/490
The easiest way to solve this problem is by using the Pythagorean theorem :D
Pythagorean theorem: [-b +- sqrt(b^2 - 4ac)]/2a
5 = a
-1 = b
6 = c
therefore, by plugging these values in!
[-(-1) +- sqrt((-1)^2 - 4(5)(6))]/2(5)
[1 +- sqrt(1 - 120)]/10
Oh so we're using imaginary numbers, so in this case you'll need to know that i = sqrt(-1), so keep this in mind (I can see why this was so difficult)
[1 +- sqrt(119)*i]10
answers:
( 1 + i*sqrt(119) ) / 10
( 1 - i*sqrt(119) ) / 10
15. She bought:
3 jeans, $20 each
1 pants, $18
You find the total: 3(20) + 1(18) = $78 then you apply the discount
Since this is between 50-100, you use the 38% discount (decimal form: 0.38)
38% is being subtracted from your total, so you can do this:
100% - 38% = 62% This is your total after the discount
78(0.62) = 48.36 Your answer is C
16. D Income is the $ you make from your job
17. A Budgeting helps you organize and limit your spending and meet financial goals
18. C Liabilities impacts you negatively, since you have debts to pay off
19. A