Cos(arccos(a))=a, a∈[-1;1]
a(t) = v'(t) = 4t³ - 4t
F(t) = m·a(t) = 2·(4t³ - 4t) = 8t³ - 8t.
F(3) = 8·3³ - 8·3 = 8·3·(3² - 1) = 192.
Відповідь: 192.
4sin2x + 10cos²x = 1
8sinxcosx + 10cos²x = sin²x + cos²x
sin²x - 8sinxcosx - 9cos²x = 0 |:cos²x
tg²x - 8tgx - 9 = 0
tg²x - 8tgx + 16 - 25 = 0
(tgx - 4)² - 5² = 0
(tgx - 4 - 1)(tgx - 4 + 5) = 0
(tgx - 5)(tgx + 1) = 0
tgx = 5 или tgx = -1
x = arctg5 + πn, n ∈ Z
x = -π/4 + πn, n ∈ Z
Ответ: x = arctg5 + πn, n ∈ Z; x = -π/4 + πn, n ∈ Z.
1. 1/5=2/10=0,2
3/4=75/100=0,75
7/8=875/1000=0,875
1/5=1:5=0,2
3/4=3:4=0,75
7/8=7:8=0,875
3/5=0,6
1/4=0,25
2/25=0,08
13/50=0,26
15/8=1,875
12/25=0,48
17/20=0,85
127/500=0,254
11/25=0,44
27/50=0,54
31/20=1,55
83/200=0,415
12/125=0,096
15/32=0,46875
7/16= 0,4375
5/64= 0,078125
Сейчас буду дальше писать.