F ` ( x ) = ( x² )` · e^(5x) + x² · ( e^(5x) )` =
= 2 x · e^(5x) + 5 e^(5x) · x² =
= x e^(5x) ( 2 + 5 x )
f `` ( x ) = ( 2 x e^(5x) + 5 x² e^(5x) ) ` =
= ( 2 x ) ˙e^(5x) + 2 x ( e^(5x) )` + ( 5 x² ) ` · e^(5x) + ( e^(5x)) ` · 5 x² =
= 2 · e^(5x) + 10 x · e^(5x) + 10 x · e^(5x) + 25 x² · e^(5x) =
= e^(5x) · ( 2 + 20 x + 25 x² )
Answer:
Area of the resulting rectangle = 23.375 inch²
Step-by-step explanation:
After 1 fold width or length reduces to half.
Area = Length x width
So after 1 fold area will reduce to half.
So we have
Initial area = 8.5 x 11 = 93.5 inch²
Here the paper were folded completely in half, length is reduced to half and width reduced to half, that is
Number of folds = 2
Area after 2 folds = 23.375 inch²
Area of the resulting rectangle = 23.375 inch²
0.4(2x+1/2) = 3[0.2x-2]-4
0.8x+0.2 = 0.6x-6-4 Distribute it out
0.8x+0.2 = 0.6x-10 Combine like terms
0.2x = -10.2 Subtract 0.6x and 0.2 from both sides
x = -51 Multiply both sides by 5
The angle adjacent to angle 6 is the one we need to find first. To do this, add the measures of the intercepted arcs and divide by 2. 60 + 50 = 110, and half of that is 55. That means that both adjacent angles to the angle 6 are 55 (vertical angles are congruent). The measure of all the angles added together is 360 and angle 6 is vertical to the other "sideways" angle, so they are congruent as well. 360 - 55 - 55 = 250. Split that up between angle 6 and its vertical angle to get that each of those measure 125. Angle 6 measures 125, choice b from above.
