It would be sometimes because it can be 82.2% but if you round it the closest one would be 80% if it was 84.7% its closer to 85 so it would be 85% I hope I helped you out rate the answer if I got it right thanks!
Answer:
<h2>
AC = 36.01</h2>
Step-by-step explanation:
Given ΔABC and ΔADB, since both triangles are right angled triangles then the following are true.
From ΔADB, AB² = AD²+BD²
Given AB = 24 and AD = 16
BD² = AB² - AD²
BD² = 24²-16²
BD² = 576-256
BD² = 320
BD = 
BD = 17.9
from ΔABC, AC² = AB²+BC²
SInce AC = AD+DC and BC² = BD² + DC² (from ΔBDC )we will have;
(AD+DC)² = AB²+ (BD² + DC²)
Given AD = 16, AB = 24 and BD = 17.9, on substituting
(16+DC)² = 24²+17.9²+ DC²
256+32DC+DC² = 24²+17.9²+ DC²
256+32DC = 24²+17.9²
32DC = 24²+17.9² - 256
32DC = 640.41
DC = 
DC = 20.01
Remember that AC = AD+DC
AC = 16+20.01
AC = 36.01
1 m = 100 cm, so 5 m = 500 cm plus the 32 cm, so she has a rope that is 532 cm long.
532 - 249 = 283 cm
So the other piece of rope is 283 cm long.
(a) x = 4
First, let's calculate the area of the path as a function of x. You have two paths, one of them is 8 ft long by x ft wide, the other is 16 ft long by x ft wide. Let's express that as an equation to start with.
A = 8x + 16x
A = 24x
But the two paths overlap, so the actual area covered will smaller. The area of overlap is a square that's x ft by x ft. And the above equation counts that area twice. So let's modify the equation by subtracting x^2. So:
A = 24x - x^2
Now since we want to cover 80 square feet, let's set A to 80. 80 = 24x - x^2
Finally, let's make this into a regular quadratic equation and find the roots.
80 = 24x - x^2
0 = 24x - x^2 - 80
-x^2 + 24x - 80 = 0
Using the quadratic formula, you can easily determine the roots to be x = 4, or x = 20.
Of those two possible solutions, only the x=4 value is reasonable for the desired objective.
(b) There were 2 possible roots, being 4 and 20. Both of those values, when substituted into the formula 24x - x^2, return a value of 80. But the idea of a path being 20 feet wide is rather silly given the constraints of the plot of land being only 8 ft by 16 ft. So the width of the path has to be less than 8 ft (the length of the smallest dimension of the plot of land). Therefore the value of 4 is the most appropriate.
