Fidn volume
V=hpir^2
d/2=r
1.4/2=0.7=r
height=3.5
V=3.5*3.14*0.7^2
V=5.3851
1.5 times t=5.3851
divide both sides by 1.5
t=3.59
round
t=4
4 minutes
Answer:It is
3.49
⋅
10
4
Step-by-step explanation:
Let's use q to represent the number of quarts needed.
you would have the following:
2q= 107-19
2q=88
q=44
44 is the number of quarts in a in the barn.
Now you have to divide that number by 4 to get gallons because there is 4 quarts in a gallon.
There are 11 gallons of paint.
2 is the correct answer I'm almost definite
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; 