You have :
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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:
Morgan, Dakota and Taylor showed their work solving the equation 7x + 5x +1= 73. Identify the student with the invalid first step and justify your reasoning. Morgan 7x + 5x + 1 = 73 12x + 1 = 73 Dakota 7x + 5x + 1 = 73 7x + 6x = 73 Taylor 7x + 5x + 1 = 73 7x + 5x = 72
Step-by-step explanation:
Answer:
P(working product) = .99*.99*.96*.96 = .0.903
Step-by-step explanation:
For the product to work, all four probabilities must come to pass, so that
P(Part-1)*P(Part-2)*P(Part-3)*P(Part-4)
where
P(Part-1) = 0.96
P(Part-2) = 0.96
P(Part-3) = 0.99
P(Part-4) = 0.99
As all parts are independent, so the formula is P(A∩B) = P(A)*P(B)
P (Working Product) = P(Part-1)*P(Part-2)*P(Part-3)*P(Part-4)
P (Working Product) = 0.96*0.96*0.96*0.99*0.99
P(Working Product) = 0.903
Hello,
I am going to remember:
y'+3y=0==>y=C*e^(-3t)
y'=C'*e^(-3t)-3C*e^(-3t)
y'+3y=C'*e^(-3t)-3Ce^(-3t)+3C*e^(-3t)=C'*e^(-3t) = t+e^(-2t)
==>C'=(t+e^(-2t))/e^(-3t)=t*e^(3t)+e^t
==>C=e^t+t*e^(3t) /3-e^(3t)/9
==>y= (e^t+t*e^(3t)/3-e^(3t)/9)*e^(-3t)+D
==>y=e^(-2t)+t/3-1/9+D
==>y=e^(-2t)+t/3+k
