Randy and Jason have $5.00 each after they pay for their tickets. The shared a big popcorn for $2.50 and a jumbo soda for $3.50.
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Answer:
<em>A=3 and B=6</em>
Step-by-step explanation:
<u>Increasing and Decreasing Intervals of Functions</u>
Given f(x) as a real function and f'(x) its first derivative.
If f'(a)>0 the function is increasing in x=a
If f'(a)<0 the function is decreasing in x=a
If f'(a)=0 the function has a critical point in x=a
As we can see, the critical points may define open intervals where the function has different behaviors.
We have
Computing the first derivative:
We find the critical points equating f'(x) to zero
Simplifying by -6
We get the critical points
They define the following intervals
Thus A=3 and B=6
(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
Learn more about " Matrix and Eigen Values, Vector " from here: brainly.com/question/13050052
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