Solve for u . 3u+11=41
2 answers:
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Answer:
This is possible.
Step-by-step explanation:
We can say that m<E=m<E, because of the Reflexive Property
Then, we have angles JKL and ELJ, which are equal through the peripheral angle theorem.
With these two angles, we can say that triangles ELK and EJL are similar, by the Angle-Angle Postulate (AA).
Then we can create this ratio through the Corresponding Parts of Similar Triangles Theorem, (CPST), .
With this ratio, we can cross multiply to get the desired result
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Hope this helps with your RSM problem
Yup, i caught ya.
(a) ( 1, 0 ) is the eigen vector for '-1' and ( 0, 1 ) is the eigen vector for '1'.
(b) two eigen values of 'k' = 1, -1
for k = 1, eigen vector is
for k = -1 eigen vector is
See the figure for the graph:
(a) for any (x, y) ∈ R² the reflection of (x, y) over the y - axis is ( -x, y )
∴ x → -x hence '-1' is the eigen value.
∴ y → y hence '1' is the eigen value.
also, ( 1, 0 ) → -1 ( 1, 0 ) so ( 1, 0 ) is the eigen vector for '-1'.
( 0, 1 ) → 1 ( 0, 1 ) so ( 0, 1 ) is the eigen vector for '1'.
(b) ∵ T(x, y) = (-x, y)
T(x) = -x = (-1)(x) + 0(y)
T(y) = y = 0(x) + 1(y)
Matrix Representation of
now, eigen value of 'T'
after solving the determinant,
we get two eigen values of 'k' = 1, -1
for k = 1, eigen vector is
for k = -1 eigen vector is
Hence,
(a) ( 1, 0 ) is the eigen vector for '-1' and ( 0, 1 ) is the eigen vector for '1'.
(b) two eigen values of 'k' = 1, -1
for k = 1, eigen vector is
for k = -1 eigen vector is
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