My Notes A street light is mounted at the top of a 15-ft-tall pole. A man 6 ft tall walks away from the pole with a speed of 4 f
t/s along a straight path. How fast is the tip of his shadow moving when he is 40 ft from the pole?
2 answers:
Answer:
6.667 ft/s
Step-by-step explanation:
Given
Height of street light = 15ft
Height of man = 6ft
Speed away from pole = 4ft/s
Let x represents the distance between the man and the pole, and Let y represents the distance between the tip of the man's shadow to the pole.
This forms a similar triangle (see attachment below).
From similar triangles, we have
(y - x)/6 = y/15 ----- Solve equation
15(y - x) = 6 * y
15y - 15x = 6y ---- Collect Like Terms
15y - 6y = 15x
9y = 15x --- divide through by 9
y = 15x/9
y = 5x/3 ---- differentiate with respect to time, t
dy/dt = 5/3 dx/dt
dx/dt represents rate of change of distance per time = 4ft/s
while dy/dt represents rate of movement of his shadow tips
dy/dt = 5/3 * 4
dy/dt = 20/3 = 6.667 ft/s
Answer: 6.67ft/s
Step-by-step explanation: PLEASE SEE PICTURE ATTACHED, IT IS A DIAGRAM THAT HELPS YOU TO ANALYSE THE QUESTION.
STEP 1: DEFINE ALL VARIABLES
The man's shadow distance from the pole (Y)
The man's height=6ft
The height of the pole= 15ft
The man's shadow length (X-Y)
The man's speed = 4ft/s
STEP 2: FIND THE MAN'S SHADOW DISTANCE FROM THE POLE (Y)
((X-Y)÷Y) = 6ft÷15ft
Open up bracket and cross multiply
15(X-Y)=6Y
15X-15Y=6Y
collecting like terms together
15X=9Y
Y= 15X/9
Y= 5X/3
STEP 3: FIND THE SPEED OF THE MAN'S SHADOW
Dy/dt = 4ft/s
Therefore
Dy/dx= (4ft/s) × (5X/3)
20ft/3s = 6.67ft/s
The speed of the shadow a cross the pole is 6.67ft/s
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The revenue function is a quadratic equation and the graph of the function
has the shape of a parabola that is concave downwards.
The correct responses are;
- (a) <u>R = -x² + 82·x</u>
- (b) <u>$1,645</u>
- (c) The graph of <em>R</em> has a maximum because the <u>leading coefficient </u>of the quadratic function for <em>R</em> is negative.
- (d) <u>R = -1·(x - 41)² + 1,681</u>
- (e) <u>41</u>
- (f) <u>$1,681</u>
Reasons:
The given function that gives the weekly revenue is; R = x·(82 - x)
Where;
R = The revenue in dollars
x = The number of lunches
(a) The revenue can be written in the form R = a·x² + b·x + c by expansion of the given function as follows;
R = x·(82 - x) = 82·x - x²
Which gives;
- <u>R = -x² + 82·x </u>
<em>Where, the constant term, c = 0</em>
(b) When 35 launches are sold, we have;
x = 35
Which by plugging in the value of x = 35, gives;
R = 35 × (82 - 35) = 1,645
- The revenue when 35 lunches are sold, <em>R</em> = <u>$1,645</u>
(c) The given function for <em>R</em> is R = x·(82 - x) = -x² + 82·x
Given that the leading coefficient is negative, the shape of graph of the
function <em>R</em> is concave downward, and therefore, the graph has only a
maximum point.
(d) The form a·(x - h)² + k is the vertex form of quadratic equation, where;
(h, k) = The vertex of the equation
a = The leading coefficient
The function, R = x·(82 - x), can be expressed in the form a·(x - h)² + k, as follows;
R = x·(82 - x) = -x² + 82·x
At the vertex, of the equation; f(x) = a·x² + b·x + c, we have;
Therefore, for the revenue function, the x-value of the vertex, is;
The revenue at the vertex is; = 41×(82 - 41) = 1,681
Which gives;
(h, k) = (41, 1,681)
a = -1 (The coefficient of x² in -x² + 82·x)
- The revenue equation in the form, a·(x - h)² + k is; <u>R = -1·(x - 41)² + 1,681</u>
(e) The number of lunches that must be sold to achieve the maximum revenue is given by the x-value at the vertex, which is; x = 41
Therefore;
- The number of lunches that must be sold for the maximum revenue to be achieved is<u> 41 lunches</u>
(f) The maximum revenue is given by the revenue at the vertex point where x = 41, which is; R = $1,681
- <u>The maximum revenue of the company is $1,681</u>
Learn more about the quadratic function here:
Answer:
C and F
Step-by-step explanation:
Given
(3x - 5)² = 19 ( take the square root of both sides )
3x - 5 = ± ← note plus or minus
Add 5 to both sides
3x = ± + 5 ( divide both sides by 3 )
x = ±
Separating the solutions
x = → C
x = → F