<span>o,t,t,f,f,s,S,E,N,T,E,T </span>
<span>It's the first letter of all the numbers starting with one, two, three, four,... </span>
<span>The second pattern I might guess would be move the O back two spaces then move one space back, then repeat. </span>
<span>JKLMNO, JKLOMN, JKOLMN, OJKLMN, JKLMON, JKOLMN, ... </span>
<span>The second one I'm not sure about because there are only three starting terms to work with. It's hard to see the real pattern with that little to work with. </span>
<span>Hope this helps. </span>
Answer:
m<MON = 100°
Explanation:
Given:
Area of shaded sector LOM = 2π cm²
NL = 6 cm
Required:
m<MON
Solution:
m<MON = 180° - m<LOM (angles in a straight line)
We don't know m<LOM. Therefore, let's find m<LOM.
Area of a sector = θ/360 × πr²
Area of sector LOM = 2π cm²
r = 3 cm
θ = m<LOM = ?
Plug in the values
2π = m<LOM/360 × π × 3²
2π = m<LOM/360 × 9π
2π = m<LOM × 9π/360
2π = m<LOM × π/40
Multiply both sides by 40
2π × 40 = m<LOM × π
80π = m<LOM × π
Divide both sides by π
80π/π = m<LOM
80 = m<LOM
m<LOM = 80°
✔️m<MON = 180° - m<LOM (angles in a straight line)
Substitute
m<MON = 180° - 80°
m<MON = 100°
