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LenKa [72]
3 years ago
14

An alternating current. I Amperes, is given by i = 10sin2nft.where f is the frequenct in Hertz and t the time in seconds.What is

the rate of change of current when t = 20ms given that f = 150 Hzπ​
Mathematics
1 answer:
defon3 years ago
8 0

Answer:

150

Step-by-step explanation:

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Scorpion4ik [409]
1. (C)
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7 0
3 years ago
Verify this cofunction identity by using the difference identity for cosine. Show your work! cos⁡(π/2-θ)=sinθ
taurus [48]
Cos(A-B) = cosAcosB + sinAsinB
<span>
cos(</span>π/2 - θ) = cos(π/2)cosθ + sin(π/2)sinθ

π/2 = 90°
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8 0
3 years ago
Find the volume of the solid generated when R​ (shaded region) is revolved about the given line. x=6−3sec y​, x=6​, y= π 3​, and
Dmitrij [34]

Answer:

V=9\pi\sqrt{3}

Step-by-step explanation:

In order to solve this problem we must start by graphing the given function and finding the differential area we will use to set our integral up. (See attached picture).

The formula we will use for this problem is the following:

V=\int\limits^b_a {\pi r^{2}} \, dy

where:

r=6-(6-3 sec(y))

r=3 sec(y)

a=0

b=\frac{\pi}{3}

so the volume becomes:

V=\int\limits^\frac{\pi}{3}_0 {\pi (3 sec(y))^{2}} \, dy

This can be simplified to:

V=\int\limits^\frac{\pi}{3}_0 {9\pi sec^{2}(y)} \, dy

and the integral can be rewritten like this:

V=9\pi\int\limits^\frac{\pi}{3}_0 {sec^{2}(y)} \, dy

which is a standard integral so we solve it to:

V=9\pi[tan y]\limits^\frac{\pi}{3}_0

so we get:

V=9\pi[tan \frac{\pi}{3} - tan 0]

which yields:

V=9\pi\sqrt{3}]

6 0
3 years ago
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RoseWind [281]
1440

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5 0
3 years ago
Read 2 more answers
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Nadusha1986 [10]

Answer: 1x+0

Step-by-step explanation:

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7 0
3 years ago
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