Answer:
the rate of change of the water depth when the water depth is 10 ft is; 
Step-by-step explanation:
Given that:
the inverted conical water tank with a height of 20 ft and a radius of 8 ft is drained through a hole in the vertex (bottom) at a rate of 4 ft^3/sec.
We are meant to find the rate of change of the water depth when the water depth is 10 ft.
The diagrammatic expression below clearly interprets the question.
From the image below, assuming h = the depth of the tank at a time t and r = radius of the cone shaped at a time t
Then the similar triangles ΔOCD and ΔOAB is as follows:
( similar triangle property)


h = 2.5r

The volume of the water in the tank is represented by the equation:



The rate of change of the water depth is :

Since the water is drained through a hole in the vertex (bottom) at a rate of 4 ft^3/sec
Then,

Therefore,

the rate of change of the water at depth h = 10 ft is:




Thus, the rate of change of the water depth when the water depth is 10 ft is; 
The answer is 44.85. Hope this helped! :)
Answer:
<h2><u><em>
3.5 Km</em></u></h2>
Step-by-step explanation:
the length of the hypotenuse is 4 Km, it is probable that you are looking for the value of the cathetus a.
it is a right triangle and we use Pythagoras
a² = 4² - 2²
a² = 16 - 4
a² = 12
a = √12
a = 3.46 (round 3.5)
The complete question in the attached figure
we know that
length side AB=8 units
length side DE=4 units
[ABC]=[DEF]*[scale factor]
then
[scale factor ]=[ABC]/[DEF]---------> 8/4--------> 2
the answer is
the scale factor for a dilation image of DEF to obtain ABC is 2
Step-by-step explanation:
x=30° { ins angle inscribed angle Standing on arc are equal }
hope it helps
<h2>stay safe healthy and happy...</h2>