No, this is not true every time. It depends on the conditions given.
For example-
Suppose we have a Box A with dimensions 10 x 10 x 1 and Box B with dimensions 5 x 5 x 5.
Box B has a surface area = 2(25+25+25) = 150 and a volume of 5*5*5= 125 cubic units
Box A has a larger surface area = 2(100+10+10) = 240 and a smaller volume = 10*10*1 = 100 cubic units
Similarly take an example of sphere.
Lets suppose the radius of the sphere is 2 cm
So, SA is 4πr² = 4*3.14*2*2 = 50.24 cm²
Volume of the sphere is = 4πr³ /3 = 33.50 cm³
Here also the SA is greater.
Answer:
339.24
Step-by-step explanation:
To find the surface area, you will need to the the area of the base of the cylinder. To do that, use the formula A=pi (r)^2
The radius is half the diameter do r=3
3.14(3)^2 should give you about 28.27.
This is the area of the top and bottom of the cylinder.
Now, do 28.27x10=282.7 to find the height surface area.
282.7+28.27+28.27=Surface area.
SA= about 339.24
Answer: The answer is 27
Given:
- Mr. Roberts will drive 56 miles
- He has 29 more miles to go
We will:
Subtract the total distance he plants to travel by the amount he has left.
<em> Equation: </em>
Our answer is 27. Best of Luck!
Let A( t , f( t ) ) be the point(s) at which the graph of the function has a horizontal tangent => f ' ( t ) = 0.
But, f ' ( x ) = [ ( x^2 ) ' * ( x - 1 ) - ( x^2 ) * ( x - 1 )' ] / ( x - 1 )^2 =>
f ' ( x ) = [ 2x( x - 1 ) - ( x^2 ) * 1 ] / ( x - 1 )^2 => f ' ( x ) = ( x^2 - 2x ) / ( x - 1 )^2;
f ' ( t ) = 0 <=> t^2 - 2t = 0 <=> t * ( t - 2 ) = 0 <=> t = 0 or t = 2 => f ( 0 ) = 0; f ( 2 ) = 4 => A 1 ( 0 , 0 ) and A 2 ( 2 , 4 ).