We know that
surface area=area of the base+4*[area of one lateral triangle side]
area of the base=b²
b is the side length of the square
b=14 cm
area of the base=14²-----> 196 cm²
area of one lateral triangle side=b*h/2
b is the side length of the square
b=14 cm
h is the height of the lateral triangle side
h is equal to the slant height
h=15 cm
area of one lateral triangle side=14*15/2----> 105 cm²
surface area=196+4*[105]------> 616 cm²
the answer is
the surface area is 616 cm²
The answer is -4 + 7. Because its subtracting 4 to zero which makes it -4. And then its adding 7 so the answer is -4+7
The rate at which the water from the container is being drained is 24 inches per second.
Given radius of right circular cone 4 inches .height being 5 inches, height of water is 2 inches and rate at which surface area is falling is 2 inches per second.
Looking at the image we can use similar triangle propert to derive the relationship:
r/R=h/H
where dh/dt=2.
Thus r/5=2/5
r=2 inches
Now from r/R=h/H
we have to write with initial values of cone and differentiate:
r/5=h/5
5r=5h
differentiating with respect to t
5 dr/dt=5 dh/dt
dh/dt is given as 2
5 dr/dt=5*-2
dr/dt=-2
Volume of cone is 1/3 π
We can find the rate at which the water is to be drained by using partial differentiation on the volume equation.
Thus
dv/dt=1/3 π(2rh*dr/dt)+(
*dh/dt)
Putting the values which are given and calculated we get
dv/dt=1/3π(2*2*2*2)+(4*2)
=1/3*3.14*(16+8)
=3.14*24/3.14
=24 inches per second
Hence the rate at which the water is drained from the container is 24 inches per second.
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Answer:
8 11/100
Step-by-step explanation:
8.11 is the same thing as 8 whole numbers and since the .11 is in the hundreths place, it becomes 11/100.