Answer: The field F has a continuous partial derivative on R.
Step-by-step explanation:
For the field F has a continuous partial derivative on R, fxy must be equal to fyx and since our field F is ∇f,
∇f = fxi + fyj + fzk.
Comparing the field F to ∇f since they at equal, P = fx, Q = fy and R = fz
Since P is fx therefore;
∂P ∂y = ∂ ∂y( ∂f ∂x) = ∂2f ∂y∂x
Similarly,
Since Q is fy therefore;
∂Q ∂x = ∂ ∂x( ∂f ∂y) = ∂2f ∂x∂y
Which shows that ∂P ∂y = ∂Q ∂x
The same is also true for the remaining conditions given
Answer:
120in.
Step-by-step explanation:
The pic attached
Answer:
-45
Step-by-step explanation:
-34+15=-19
-29-(-3)= -29+3= -26
-19 + -26 = -45
Let events
A=Nathan has allergy
~A=Nathan does not have allergy
T=Nathan tests positive
~T=Nathan does not test positive
We are given
P(A)=0.75 [ probability that Nathan is allergic ]
P(T|A)=0.98 [probability of testing positive given Nathan is allergic to Penicillin]
We want to calculate probability that Nathan is allergic AND tests positive
P(T n A)
From definition of conditional probability,
P(T|A)=P(T n A)/P(A)
substitute known values,
0.98 = P(T n A) / 0.75
solving for P(T n A)
P(T n A) = 0.75*0.98 = 0.735
Hope this helps!!
the answer for this is 20.8558