Hello :
by idetity : 1+cos(2a) = 2cos²(a)
let : a = <span>θ/2
1+cos(2×</span>θ/2)=2cos²(θ/2)
1+cos(θ) = 2cos²(θ/2)
solve : 2 cos^2 (θ/2) - 3 cos(θ) = 0 ......(1)
(1) : 1+cos(θ)-3 cos(θ) =0
2cos(θ) = 1
cos(θ) = 1/2
in : [0, 2π<span>[
</span>( θ = π/3) or (θ = 5π/3)
X² + 2x = 24
Organizing:
x² + 2x - 24 = 0
<<<< (<em>Quadratic </em><span><span><em>equation</em>)
</span>
</span><span>Delta:
</span>Δ = b²<span> - 4.a.c </span>
Δ = 2²<span> - 4 * 1 * -24 </span>
<span>Δ = 4 - 4 * 1 * -24 </span>
Δ = 100
<span>Bhaskara:
</span><span>x = (-b +- √Δ)/2*a
</span>
x' = (-2 + √100)/2*1
x' = 8 / 2
x' = 4
x'' = (-2 - √100)/2*1
x'' = -12 / 2
x'' = -6
Result: x = 4 or<span> x = - 6 .
</span>
Good studies! :)
1. B
2. A
3. B
4. A
Is your answer
The second answer i think is the best one
Answer: 1/6
Step-by-step explanation:
Because there's only one number that's greater than 5 on a numbered cube and that's '6'
