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Nonamiya [84]
3 years ago
14

I need the answer. Please help me . It’s due today

Mathematics
1 answer:
Phoenix [80]3 years ago
4 0

9514 1404 393

Answer:

  graph: lower right

  π units to the right of the graph of f(x)

Step-by-step explanation:

The transformation of f(x) to f(x-h) is a translation to the right by h units.

Here, the function f(x) = 1/2cos(x) is transformed to f(x-π) = 1/2cos(x-π), so is a translation of f(x) to the right by π units.

The graph of the translated function is shown below.

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Answer:

<h3>2</h3>

Step-by-step explanation:

<h3><em>For</em><em> </em><em>every</em><em> </em><em>action</em><em>,</em><em> </em><em>there</em><em> </em><em>is </em><em>a</em><em>n</em><em> </em><em>equal</em><em> </em><em>and</em><em> opposite</em><em> </em><em>reaction</em><em>.</em></h3>
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Help me with the picture please
Sergio [31]
<h3>Answer:  115</h3>

=====================================================

Explanation:

Refer to the diagram below. I've drawn diagonal that slopes upward. This diagonal cuts the quadrilateral into two triangles: One is equilateral and the other is isosceles.

The equilateral triangle marked in blue has all three angles 60 degrees each.

Note that the 60 and y angles combined to form 130, so,

60+y = 130

y = 130-60

y = 70

Then focus on the isosceles triangle (angles y, w and w). These three interior angles must add to 180

y+w+w = 180

70+2w = 180

2w = 180-70

2w = 110

w = 110/2

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What expression is equivalent to 4-(-7)
Inessa05 [86]
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Help with numer 5 please. thank you​
Alex17521 [72]

Answer:

See Below.

Step-by-step explanation:

We are given that:

\displaystyle I = I_0 e^{-kt}

Where <em>I₀</em> and <em>k</em> are constants.

And we want to prove that:

\displaystyle \frac{dI}{dt}+kI=0

From the original equation, take the derivative of both sides with respect to <em>t</em>. Hence:

\displaystyle \frac{d}{dt}\left[I\right] = \frac{d}{dt}\left[I_0e^{-kt}\right]

Differentiate. Since <em>I₀ </em>is a constant:

\displaystyle \frac{dI}{dt} = I_0\left(\frac{d}{dt}\left[ e^{-kt}\right]\right)

Using the chain rule:

\displaystyle \frac{dI}{dt} = I_0\left(-ke^{-kt}\right)  = -kI_0e^{-kt}

We have:

\displaystyle \frac{dI}{dt}+kI=0

Substitute:

\displaystyle \left(-kI_0e^{-kt}\right) + k\left(I_0e^{-kt}\right) = 0

Distribute and simplify:

\displaystyle -kI_0e^{-kt} + kI_0e^{-kt} = 0 \stackrel{\checkmark}{=}0

Hence proven.

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