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Katen [24]
2 years ago
9

Could someone help me out?

Mathematics
1 answer:
worty [1.4K]2 years ago
8 0

Since the grade of the numerator and the denominator is the same, then the limit exists and is distinct from 0. The limit of the expression is 4/7.

<h3>How to determine the limit of a rational expression when x tends to infinite</h3>

In this problem we must apply some algebraic handling and some known limits to determine whether the limit exists or not. The limit exists if and only if the result exists.

\lim_{x \to \infty} \frac{4\cdot x - 1}{7\cdot x + 3}

\lim_{x \to \infty} \frac{4\cdot x - 1}{7\cdot x + 3} \cdot \frac{x}{x}

\lim_{x \to \infty} \frac{4 - \frac{1}{x} }{7 + \frac{3}{x} }

\lim_{x \to \infty} \frac{4}{7}

4/7

Since the grade of the numerator and the denominator is the same, then the limit exists and is distinct from 0. The limit of the expression is 4/7.

To learn more on limits: brainly.com/question/12207558

#SPJ1

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let t : r2 →r2 be the linear transformation that reflects vectors over the y−axis. a) geometrically (that is without computing a
tangare [24]

(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 \left[\begin{array}{c}0\\1\end{array}\right]

for k = -1 eigen vector is \left[\begin{array}{c}1\\0\end{array}\right]

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 T = \left[\begin{array}{cc}-1&0\\0&1\end{array}\right]

now, eigen value of 'T'

T - kI =  \left[\begin{array}{cc}-1-k&0\\0&1-k\end{array}\right]

after solving the determinant,

we get two eigen values of 'k' = 1, -1

for k = 1, eigen vector is \left[\begin{array}{c}0\\1\end{array}\right]

for k = -1 eigen vector is \left[\begin{array}{c}1\\0\end{array}\right]

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 \left[\begin{array}{c}0\\1\end{array}\right]

for k = -1 eigen vector is \left[\begin{array}{c}1\\0\end{array}\right]

Learn more about " Matrix and Eigen Values, Vector " from here: brainly.com/question/13050052

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6 0
1 year ago
  Simplify............
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