Answer:
area = m²/2π
Step-by-step explanation:
∵ circumference of circle = 2πr = m
∴ r = m/2π
∵ area of circle = πr²
∴ area of circle = (m/2π)² × π = m²/2π² × π = m²/2π
hope this helps
Hello, what’s the question though
<span>3-2(Cosx)^2 - 3Sinx = 0.
Recall (Sinx)^2 + (Cosx)^2 = 1.
Therefore (Cosx)^2 = 1 - (Sinx)^2
Substitute this into the question above.
</span><span>3-2(Cosx)^2 - 3Sinx = 0
3 - 2(1 - (Sinx)^2) - 3Sinx = 0 Expand
3 - 2 + 2(Sinx)^2 - </span><span><span>3Sinx = 0</span>
1 + 2(</span><span>Sinx)^2 - 3Sinx = 0 Rearrange
2(Sinx)^2 </span><span><span>- 3Sinx + </span>1 = 0
Let p = Sinx
2p^2 - 3p + 1 = 0 Factorise the quadratic expression
2p^2 - p - 2p +1 = 0
p(2p -1) - 1(2p -1) = 0
(2p-1)(p -1) = 0
Therefore 2p-1=0 or (p-1) = 0
2p=0+1 or (p-1) = 0
2p=1 or p = 0 +1.
p=1/2 or p = 1 Recall p = Sinx
Therefore Sinx = 1/2 or 1.
For 0<u><</u>x<u><</u>360
Sinx =1/2, x = Sin inverse (1/2) , x = 30,
(180-30)- 2nd Quadrant = 150 deg
Sinx = 1, x = Sin inverse (1) , x = 90
Therefore x = 30,90 & 150 degrees.
Cheers.</span>
The answer:
<span>the two triangles are similar
in addition, the line UV is parallel to line BC, so we can use the theorem of Thales for proving the following ratios:
AV /AC = AU/ AB= UV/ BC
372/589=20x +80 / AB = 444 / 703
</span>so we get (372/589) AB - 80 = 20x, and x = ( (372/589) AB - 80 ) / 20
exact value of x depends on the value of AB
