Answer:
Step-by-step explanation:
The directional derivative of a function in a particular direction u is given as the dot product of the unit vector in the direction of u and the gradient of the function
g(x,y) = sin(π(x−5y)
∇g = [(∂/∂x)î + (∂/∂y)j + (∂/∂z)ķ] [sin(π(x−5y))
(∂/∂x) g = (∂/∂x) sin (πx−5πy) = π [cos(π(x−5y))]
(∂/∂y) g = (∂/∂y) sin (πx−5πy) = - 5π [cos (π(x−5y))]
∇g = π [cos(π(x−5y))] î - 5π [cos (π(x−5y))] j
∇g = π [cos (π(x−5y))] [î - 5j]
So, the question requires a direction vector and a point to fully evaluate this directional derivative now.
400 - 3x = 163
Add 3x to both sides.
400 = 163 + 3x
Subtract 163 to both sides.
237 = 3x
237/3 = 79
Answer:
m<NQS = 32°
Step-by-step explanation:
Given:
m<BQS = 80°
m<BQN = 48°
Required:
m<NQS
SOLUTION:
Angle BQN and angle NQS are adjacent angles having a common line, QN, and a common corner point, Q.
Therefore:
m<BQN + m<NQS = m<BQS (angle addition postulate)
48° + m<NQS = 80° (substitution)
m<NQS = 80 - 48° (Subtraction of 48 from each side)
m<NQS = 32°
Answer:
38km
Step-by-step explanation:
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