Answer:
Required equation of tangent plane is
.
Step-by-step explanation:
Given surface function is,
To find tangent plane at the point (5,-1,1).
We know equation of tangent plane at the point $(x_0,y_0,z_0)[/tex] is,

So that,



Substitute all these values in (1) we get,


Which is the required euation of tangent plane.
Answer:
Step-by-step explanation:
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Recall: SOHCAHTOA
Sin M = Opp/Hyp
Reference angle = M
Opp = 3√/21
Hyp = 15
Sin M = (3-√21)/15
Sin M = √21/5
Cos M = Adj/Hyp
Reference angle = M
Adj = 6
Hyp = 15
Cos M = 6/15
Cos M = 2/5
Tan M = Opp/Adj
Reference angle = M
Opp = 3√21
Adj = 6
Tan M = (3√21)/6
Tan M = √21/2
:Therefore: Sin M = √21/5
Cos M = 2/5
Tan M = √21/2
Answer:
the solution is (2, -6)
Step-by-step explanation:
Substitute the second equation into the first, replacing y in the first:
2x - (-4x+2) = 10
Simplifying, we get:
2x + 4x - 2 = 10, or:
6x = 12, which yields x = 2.
Substituting 2 for x in the second equation yields y = -4(2) + 2 = 0, or y = -6
Then the solution is (2, -6).
90 + 20x + 10x = 180
30x = 90
x = 3