Given :
A boat is being pulled into a dock with a rope attached to the boat at water level. When the boat is 12 feet from the dock, the length of the rope form the boat to the dock is 3 feet longer than twice the height of the dock above the water.
To Find :
The height of the dock.
Solution :
This will make a right angle triangle as given in link below .
Now , applying Pythagoras theorem :
Now , h = 5 or h = -9 .
Now , height cannot be negative .
So , height of the dock is 5 ft .
Hence , this is the required solution .
Answer:
-2
Step-by-step explanation:
rise over run or, y2-y1/x2-x1
All are the exponential function.
<h2>Exponent</h2>
Exponential notation is the form of mathematical shorthand which allows us to write complicated expressions more succinctly. An exponent is a number or letter is called the base. It indicates that the base is to raise to a certain power. c is the base and x is the power.
<h3>Which functions represent exponential growth?</h3>
1. y = f(x) is a exponential function.
Because the range of the exponential function is from 0 to ∞ for all values of x. And it is 1 at x = 0.
2. y = h(x) is a exponential function.
Because the range of the exponential function is from 0 to ∞ for all values of x. And it is 1 at x = 0.
3. y = g(x) is a exponential function.
Because the range of the exponential function is from 0 to ∞ for all values of x. And it is 1 at x = 0.
4. y = k(x) is a exponential function.
Because the range of the exponential function is from 0 to ∞ for all values of x. And it is 1 at x = 0.
Thus, all are the exponential function.
More about the exponent link is given below.
brainly.com/question/5497425
Answer:
10-1/3x=4
Step-by-step explanation:
10 minus is 10-
1/3 of a number is 1/3x
is 4 means =4
10-1/3x=4
To solve this problem, we must assume that the man
undergoes constant acceleration as he goes down the river (therefore no other forces
must act on him except gravity). Therefore we can use the formula below to
calculate for the duration of his fall:
y = y0 + v0 t + 0.5 a t^2
where y is the distance and y0 = 0 since we set the
reference point at the bridge, v0 is the initial velocity and is also equal to
v0 = 0 since the man started from rest, therefore the equation becomes:
y = 0 + 0 t + 0.5 a t^2
y = 0.5 a t^2
Rewriting in terms of t:
t^2 = 2 y / a
t = sqrt (2y / a)
a is acceleration due to gravity = 9.8 m/s^2
t = sqrt [2 * 23 / 9.8]
t = 2.17 s
Therefore the jump last only about 2.17 seconds.
