Вот в картинке примерное решение уравнения:
Решение
<span>a</span>₂ <span>= a</span>₁ <span>+ d
</span><span>a₉ </span><span>=a₁ </span><span>+ </span><span>8d
</span><span>a</span>₂ <span>= 3a</span>₉
<span>3(a</span>₁ <span>+ 8d) = a</span>₁ <span>+ d
</span><span>3a</span>₁ <span>+ 24d = a</span>₁ <span>+ d
</span><span>2a</span>₁ <span>+ 23d = 0
</span><span>2a</span>₁ = - 23d
<span>a</span>₁ = - 11,5d
<span>Sn = [2a</span>₁ + d*(n - 1)*n]/2
S₂₀ = [(a₁ + a₁ + 19d)*20]/2 = (a₁ + a₂₀)*10
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(8,1х-3х)-(5х+4,9)= 8,1х-3х-5х-4,9= 8.1х-3х-5х-4.9=0.1х-4.9
Ответ будет 6:1
Тк 6 раз бросают и 6 сторон)
2(1-sin^2 2x)+5sin2x-4=0
2-2sin^2(2x)+5sin2x-4=0
2sin^2(2x)-5sin2x+2=0
sin2x=t
2t^2-5t+2=0
|t|<=1
t=1/2
sin2x=1/2
x=П/12+Пk k=1 13П/12
x=5П/12+Пk k=1 17П/12
5cos2x+7cos(x-3п/2)+1=0,при cos< или=0
5(cos^2x-sin^2x)+7cos(3П/2-x)+1=0
5-10sin^2x-7sinx+1=0
10sin^x+7sinx-6=0
sinx=1/2
x=5п/6+2Пk