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4 years ago

14

BEDE

1 answer:

fenix001 [56]4 years ago

8 0

Answer:

$\large\boxed{y=\dfrac{1}{4}x+4}$

Step-by-step explanation:

$Let:\\\\k:y=m_1x+b_1\\\\l:y=m_2x+b_2\\\\l\ \perp\ k\iff m_1m_2=-1\to m_2=-\dfrac{1}{m_1}\\\\l\ \parallel\ k\iff m_1=m_2\\============================\\\\\text{We have the equation of the line:}\ y=-4x+9\to m_1=-4.\\\\\text{Therefore}\ m_2=-\dfrac{1}{-4}=\dfrac{1}{4}.\\\\\text{Put the value of the slope and the coodrdinates of the given point (4, 5)}\\\text{to the equation of a line}\ y=mx+b:\\\\5=\dfrac{1}{4}(4)+b\\\\5=1+b\qquad\text{subtract 1 from both sides}\\\\4=b\to b=4\\\\\text{Finally:}\\\\y=\dfrac{1}{4}x+4$

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Use a graphing utility to graph the function and the damping factor of the function in the same viewing window.

Bas_tet [7]

Answer:

The function touches the damping factor

at x= $\frac{(4n-3)\pi}{2}$ and x= $\frac{(4n-1)\pi}{2}$

The x-intercept of f(x) is

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Step-by-step explanation:

Given function is f(x)= $e^{-3x} sin(x)$ and damping factor as y= $e^{-3x}$ and y= $(-1)e^{-3x}$

To find when function touches the damping factor:

For f(x)= $e^{-3x} sin(x)$ and y= $e^{-3x}$

Equating the both the equation,

$e^{-3x} sin(x)=e^{-3x}$

$sin(x)=1$

x= $\frac{(4n-3)\pi}{2}$

For f(x)= $e^{-3x} sin(x)$ and y= $(-1)e^{-3x}$

Equating the both the equation,

$e^{-3x} sin(x)=(-1)e^{-3x}$

$sin(x)=(-1)$

x= $\frac{(4n-1)\pi}{2}$

Therefore, The function touches the damping factor x= $\frac{(4n-3)\pi}{2}$ and x= $\frac{(4n-1)\pi}{2}$

To find x-intercept of f(x):

For x-intercept, y=0

f(x)= $e^{-3x} sin(x)$

y= $e^{-3x} sin(x)$

$e^{-3x} sin(x)=0$

Hence, $e^{-3x}$ is always greater than zero.

Therefore, $sin(x)=0$

x= $n\pi$

Thus,

The x-intercept of f(x) is at x= $n\pi$

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nirvana33 [79]

Answer:

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ss7ja [257]

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Answer:

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