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 ).
Answer:
A
Step-by-step explanation:
If you translate left one unit, then reflect across the x axis, you get the ending figure
To find out how much is needed for 1 you will need to divide each given value by 4:
12 marshmallows / 4 = 3
8 graham crackers / 4 = 2
4 smores / 4 = 1
you then need to multiply each value by 3 :
3 marshmallows x 3 = 9
2 graham crackers x 3 = 6
1 smore x 3 = 3
therefore the answer would be :
9 marshmallows
6 graham crackers
hope this helps you
5/6 times 3/1= 15/6 so that will be your answer
Answer:
The approximate circumference is 62.8 mm
Step-by-step explanation:
Circumference is basically the perimeter of the circle
Circumference is given by
Here the diameter is 20 mm, so the radius will be half of diameter = 10 mm
Substituting the given values, in above equation, we get -
Circumference =
mm
The approximate circumference is 62.8 mm
