No. If line WX would have been a tangent, the angle VWX should have been 90 degrees. If it would have been 90 degrees, using the pythagoras theorem, side VX should have been 5. But it's not 5, neither angle VWX is 90 degrees nor is the line WX a tangent to the circle
Answer:
Step-by-step explanation:
- sqrt(39) and square root 47
are the limits. The - square root of 39 is smaller than - 6. So the integer to use here is - 6
sqrt (47) = 6.686. Here the square root is larger than the closest integer.
The integer to use is 6
- 6 + 6 = 0
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:
2x-3
Explanation
12x-18
/2
6x-9
/3
2x-3
Answer:
The first table, because 1 has two outputs.
Step-by-step explanation:
The rest are fine because the numbers only have one counterpart.
The first table on the other hand, 1 is paired with both 10 and 15.
Hope this helps :)
