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iragen [17]
3 years ago
14

What would the length be?

Mathematics
1 answer:
Natalka [10]3 years ago
4 0

Answer:

EF = ~3.8

Step-by-step explanation:

Apply the sine theorem:

EF/sinD = DE/sinF

=> EF = DE*(sinD/sinF)

         = 3*(sin75/sin50)

         =3.78 = ~3.8

Hope this helps!

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Wendell is looking over some data regarding the strength, measured in Pascals (Pa), of some building materials and how the stren
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The logarithmic model for the length when the strength is of 8 Pascals is given by:

  • f^{-1}(8) = \log_{2}{8} = \log_2{2^3} = 3
  • That is, the length is of 3 units.

<h3>What is the function?</h3>

The strength in Pascals for a building of length x is given by:

f(x) = 2^x

To find the length given the strength, we apply the inverse function, that is:

2^y = x

\log_{2}{2^y} = \log_2{x}

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Hence, when the strength is of 8 Pascals, x = 8, and the length is given by:

f^{-1}(8) = \log_{2}{8} = \log_2{2^3} = 3

You can learn more about logarithmic functions at brainly.com/question/25537936

6 0
2 years ago
let t : r2 →r2 be the linear transformation that reflects vectors over the y−axis. a) geometrically (that is without computing a
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(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 )

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∴ 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

#SPJ4

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Answer:

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