Answer:

Step-by-step explanation:
Let L be the length of the ladder,
Given,
x = the distance from the base of the ladder to the wall, and t be time.
y = distance from the base of the ladder to the wall,
So, by the Pythagoras theorem,

,
Differentiating with respect to time (t),




Here,

Also,
( Ladder length = constant ),




Which is the required notation.
Trigonometry can be used to determine the height of a cell phone tower by using SOH CAH TOA or the Pythagorean theorem. If you look at it as a right triangle you can figure out how tall the tower is. If an angle is given (not a 90°angle) and the value of a side you can figure out all of the sides on the theoretical right triangle. Including the height of the tower.
Answer:
$14.23
Step-by-step explanation:
Multiply the original value by 1.035 to get your answer.
Answer:
<h2><u>
j = 38</u></h2>
Explanation:
ʲ⁄₋₂ + 7 = -12
Subtract 7 From Both Sides
ʲ⁄₋₂ + 7 - 7 = -12 - 7
Simplify
ʲ⁄₋₂ = -19
Multiply Both Sides By -2
(-2) ʲ⁄₋₂ = -19 (-2)
<h2><u>j = 38</u></h2>
Answer:
(x, y) = (0, -2248)
Step-by-step explanation:
<u>Substitution</u>
Substitute for y:
300x -2248 = 350x -2248
0 = 50x . . . . . . . . add 2248-300x to both sides
x = 0
y = 300·0 -2248
The solution is (x, y) = (0, -2248).
__
<u>Elimination</u>
Subtract the first equation from the second.
(y) -(y) = (350x -2248) -(300x -2248)
0 = 50x
x = 0
y = 300·0 -2248
The solution is (x, y) = (0, -2248).
__
<em>Additional comment</em>
If you think for a bit about what the graph looks like, you realize both lines cross at their y-intercept, (0, -2248). That point of intersection is the solution to the system of equations.