What does <em>x </em>stand for?
The length of the unknown sides of the triangles are as follows:
CD = 10√2
AC = 10√2
BC = 10
AB = 10
<h3>Triangle ACD</h3>
ΔACD is a right angle triangle. Therefore, Pythagoras theorem can be used to find the sides of the triangle.
where
c = hypotenuse side = AD = 20
a and b are the other 2 legs
lets use trigonometric ratio to find CD,
cos 45 = adjacent / hypotenuse
cos 45 = CD / 20
CD = 1 / √2 × 20
CD = 20 / √2 = 20√2 / 2 = 10√2
20² - (10√2)² = AC²
400 - 100(2) = AC²
AC² = 200
AC = √200 = 10√2
<h3>
Triangle ABC</h3>
ΔABC is a right angle triangle too. Therefore,
Using trigonometric ratio,
cos 45 = BC / 10√2
BC = 10√2 × cos 45
BC = 10√2 × 1 / √2
BC = 10√2 / √2 = 10
(10√2)² - 10² = AB²
200 - 100 = AB²
AB² = 100
AB = 10
learn more on triangles here: brainly.com/question/24304623?referrer=searchResults
Let's say the length of the ladder is "d". Since the distance from the bottom of the ladder to the building is 16 feet less than the length, we can write this as d - 16.
The distance from the ground to the top of the ladder is 2 feet less than the length of the ladder, so we can denote this as
d - 2.
The shape made by the ground, ladder, and side of the building is a right triangle, and we just found its legs. We can use the Pythagorean Theorem to solve for "d" and find how far up the side of the building the ladder is:
(d - 16)^2 + (d - 2)^2 = d^2
(d^2 - 32d + 256) + (d^2 - 4d + 4) = d^2
2d^2 - 36d + 260 = d^2
d^2 - 36d + 260 = 0
(d - 26) * (d - 10) = 0
d = 26 or d = 10
The answer d = 10 does not fit in this case because it isn't possible to have 16 feet less of 10 feet; it would be negative!
So, d = 26. The top of the ladder to the ground is 2 less than d, so we do 26 - 2 = 24 feet.
The answer is 24 feet.
<u>Answer:</u>
<u>Step-by-step explanation:</u>
Let's take two points on a line. (2,0) and (0,2).
Now, determine the rise and run.
Slope = Rise/Run
=> 2/2
<u>Conclusion:</u>
Therefore, the slope is 1.
Hoped this helped.
13500-1500= 12000/5=2400
12000-2400= 9600
$9600 most expensive painting
