Line is divided into 4 equal parts.
we have to find a point which is closest to point A.
So that means required point P(x,y) is at 1 unit away from A(-1,1) and 3 unit away from B(8,4)
Now we just need to use section formula to get the coordinate of required point using m1=1 and m2=3
So the final answer is 
.
Answer:
Angle = 45
Supplement = 135
Complement = 45
Step-by-step explanation:
You have :
--------------
DE arc = ( pi ) ( AD ) ( 2.36 radians / 2 pi radians ) = ( 2/3 ) ( AB ) ( 2.36 radians / 2 )
DE arc = ( 2/3 ( AB ) ( 1.18 radians )
BC arc = ( pi ) ( AB ) ( 1.18 radians / 2 pi radians )
BC arc = ( AB ) ( 0.59 radians )
BC arc / DE arc = ( AB ) ( 0.59 radians ) / ( 2/3 ) ( AB ) ( 1.18 radians )
BC arc / DE arc = ( AB ) ( 0.59 rad ) / ( 2/3 ) ( AB ) ( 1.18 rad )
BC arc / DE arc = ( 3/2 ) ( .59 rad / 1.18 rad ) = 3/4 <-------
Answer:
$95.78
Step-by-step explanation:
f(t) = 300t / (2t² + 8)
t = 0 corresponds to the beginning of August. t = 1 corresponds to the end of August. t = 2 corresponds to the end of September. So on and so forth. So the second semester is from t = 5 to t = 10.
$T₂ = ∫₅¹⁰ 300t / (2t² + 8) dt
$T₂ = ∫₅¹⁰ 150t / (t² + 4) dt
$T₂ = 75 ∫₅¹⁰ 2t / (t² + 4) dt
$T₂ = 75 ln(t² + 4) |₅¹⁰
$T₂ = 75 ln(104) − 75 ln(29)
$T₂ ≈ 95.78
A dilation by a scale factor of 6 will cause the radius to increase by a factor of 6. So
. If we plug this value of r into the formula for the circumference of a circle we get
So basically N = 12r. You haven't given me any original radius, so I can't give you a constant for N, but if you do have that original radius you can just plug that into 12r.
Based on your comment, N = 12 * 9 = 108 inches
