Answe: 3rd one
Step-by-step explanation:
The answer is 1/4 because all you have to really do is simplify
Answer:
Actually it's not polygon. it's a nonagon. With r=8.65mm″, the law of cosines gives us side a:
a=√{b²+c²−2bc×cos40°}
a=√{149.645−149.645cos40°}
Area Nonagon = (9/4)a²cos40°
=9/4[149.645−149.645cos40°]cot20°
=336.70125[1−cos(40°)]cot(20°)
Applying an identity for the cos(40°) does not get us very far…
= 336.70125[1−(cos2(20°)−1)]cot(20°)
= 336.70125[2−cos2(20°)]cot(20°)
= 336.70125[2−(1−sin2(20°))]cot(20°)
= 336.70125[1+sin2(20°)]cos(20°)sin(20°)
= 336.70125[cot(20°)+sin(20°)cos(20°)]mm²
Its easy you just have to do the pi formula
Answer:
We can use the sample about 42 days.
Step-by-step explanation:
Decay Equation:



Integrating both sides


When t=0, N=
= initial amount




.......(1)
.........(2)
Logarithm:
130 days is the half-life of the given radioactive element.
For half life,
,
days.
we plug all values in equation (1)






We need to find the time when the sample remains 80% of its original.







We can use the sample about 42 days.