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Lynna [10]
3 years ago
10

An airplane was cruising at 23,000 feet altitude. Due to turbulence it dropped 4,300 feet.Once the air currents leveled off he w

ent back up 3,600 feet. How far was he from his original cruising altitude?
Mathematics
1 answer:
vladimir1956 [14]3 years ago
4 0

Answer:

-700 feet, aka 700 feet below/away from the original cruising altitude

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Before jumping into doing this problem, it is important to look at all the information's that have been made available in the question. Based on those given conditions the answer to the problem can be easily deduced.
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3 years ago
Let v be an eigenvector of a matrix A with a corresponding eigenvalue ?=2. Find one solution x of the system Ax=v.
Murljashka [212]

Answer with explanation:

For, a Matrix A , having eigenvector 'v' has eigenvalue =2

 The order of matrix is not given.

It has one eigenvalue it means it is of order , 1×1.

→A=[a]

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It is given that,

k=2

For, k=2, the matrix [a-2 I] will become singular,that is

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→I=[1]

→a=2

Let , v be the corresponding eigenvector of the given eigenvalue.

→[a-I] v=0

→[2-1] v=[0]

→[v]=[0]

→v=0

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3 years ago
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Gala2k [10]

Answer:

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a = 8

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3 years ago
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