Answer:
Length of arc AB = 1.57 yds
Step-by-step explanation:
Given:
Central angle (θ) = 30°
Radius (r) = 3 yds
Required:
Length of arc AB approximated to the nearest hundredths
Solution:
Length of arc = θ/360 * 2πr
Length of arc AB = 30/360*2*π*3
Length of arc AB = 1.57079633
Length of arc AB = 1.57 yds (nearest hundredths)
It would be in the 40,000,00 range
Hope this helps
Answer:8
Step-by-step explanation:
8N is the equation
And a coefficient is the number in front of a letter
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:
{2, 8}
Step-by-step explanation:
We want to find x for ...
15 = 4|x -5| +3
12 = 4|x -5| . . . . subtract 3
3 = |x -5| . . . . . . divide by 4
±3 = x -5 . . . . . . show the meaning of absolute value
5 ±3 = x = {2, 8} . . . . . add 5
The values of x for which f(x) = 15 are 2 and 8.
