Answer:
Step-by-step explanation:
The resulting course vector (magnitude 300, angle = 160° with the positive x axis (= 290° bearing)) with components (300cos160,300sin160) is the sum of the vector of the actual flight vector and the wind vector.
The wind vector has magnitude 18 and direction 46° with the positive x-axis ( it blows from - 134° with the positive x axis (= 224° bearing) into direction 46° with the positive x-axis). So its components are (18cos46,18sin46).
The course that the plane has to steer is :
(300cos160,300sin160)-(18cos46,18sin46)
=(300cos160 - 18cos46, 300sin160 - 18sin46)
=(-294.41, 89.6579)
The magnitude of this steering vector is = 307.749 knots and its direction will be:
with the negative x-axis (286.94° bearing).
Thus, the drift angle will be=290° - 286.94° = 3.06°.
Let me see the whole thing, theres no enough info
Answer:
C. x-5=0 or x+3=0
Step-by-step explanation:
Given,
( By multiplication )
( By operating like terms )
( Subtracting 12 on both sides )
( By middle term splitting )
By the zero product property,
Hence, option C is correct.
It’s 80 because it is the same shape just flipped which means it has the same angles
Answer:
Step-by-step explanation:
18^y = 4/6 Take the log of both sides
log(18^y) = log(4/6)
log(18^y) = y * log(18)
log (4/6) = - 0.17609
y * log(18) = y * 1.25527
So
y = - 0.17609 / 1.25527
y = - 0.14056
Check
18^-0.14056 = 0.6661 which is close enough to 0.666666666
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z^4 = 8/5 Take the log of both sides.
log(z^4) = log(8/5) Simlplify the left. Take the log of the right.
4 * log(z) = 0.20416 Divide by 4
log(z) = 0.20416 /4
log(z) = 0.051029 Take the antilog of both sides.
z = 10^(0.051029)
z = 1.12467
Check
1.12467 ^ 4 = 1.6 which is what it should be.
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I can't really make out the third one.