To answer your question, this could be the possible answer and i hope you understand and interpret it correctly:
<span>[Integrate [0, 1/2] xcos(pi*x
let u=x so that du=dx
and v=intgral cos (xpi)dx
v=(1/pi)sin(pi*x)
integration by parts
uv-itgral[0,1/2]vdu just plug ins
(1/pi)sinpi*x]-(1/pi)itgrlsin(pi*x)dx from 0 to 1/2
(1/pi)x sinpi*x - (1/pi)[-(1/pi) cos pi*x] from 0 to 1/2
=(1/2pi)+(1/pi^2)[cos pi*x/2-cos 0]
=1/2pi - 1/2pi^2
=(pi-2)/2pi^2 ans</span>
Answer:
Parallelograms I, II, and IV
Step-by-step:
Area of parallelograms:
I. A=3*5=15 units squared
II. A=5*3=15 units squared
III. A=4*4=16 units squared
IV. A=5*3=15 units squared
So, parallelograms I, II, and IV have the same area of 15 units squared.
A = 30
B = 41
C = 70
A < B + C
30 < 41+70 ==> 30 < 111 (TRUE)
B < A + C ==>
41 < 30 + 70 ==> 41 < 100 (TRUE)
C < A+B
70 < 30 + 41 ==> 70 < 71 (TRUE)
Yes, can!
Answer:
m and n are parallel because if you add 55 and 125 together you get 180
Step-by-step explanation:
