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Because if that shift wasn't there, it would be possible to obtain negative values. But by vertically shifting the function up by the same magnitude as the amplitude (0.5), we remove the negative numbers.

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Because if that shift wasn't there, it would be possible to obtain negative values. But by vertically shifting the function up by the same magnitude as the amplitude (0.5), we remove the negative numbers.

OH OK. I've been wanting to know this the whole time. haha thanks!

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OH OK. I've been wanting to know this the whole time. haha thanks!

Haha no worries. Math 12?

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yup you are

i understand why its 0.5 cos[π(t − 8)/6] but i dont understand why its 0.5 cos[π(t − 8)/6]+0.5. Why vertical shift by .5?

Also, the vertical displacement for sine and cosine is always found by using (max + min)/2. It's the average value of the function.

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given that f (x)= square root [1+(lnx)^2] for x greater or equal to 1, find the inverse function

how do you do this?

x=root 1+ (lny)^2

x^2=1+(lny)^2

x^2-1=(lny)^2

root (x^2-1)=lny

e^(root(x^2-1) = y

what am i doing wrong?

The bolded should be the correction.

Edited by babych

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A block of mass m = 0.570 kg rests on top of a block of mass M = 2.30 kg. A string attached to the block of mass M is pulled so that its tension is T = 6.20 N at a 20o angle to the horizontal as shown. The blocks move together. The coefficient of static friction at the surface between the blocks is μs = 0.43; there is no friction at the surface between block M and the floor.

The tension T is now increased - what is the maximum tension, Tmax, with which the string can be pulled such that the blocks continue to move together (i.e. that the block of mass m does not start to slide on top of the block of mass M)

so i found Ff

mewFN=(0.43)(9.8)(0.570

=2.4

then

F=ma

Ftx-2.4=(2.3+0.57)(2.03)

Ftx=8.228M

Ftx/cos20=tmax

tmax=8.76N

Is this right?

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Anyone here inolved with electrical engineering or knows anything about series parallel circuits?

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A 15.0 g bullet with a speed of v = 235 m/s is fired into a block of wood that is initially at rest on a frictionless surface. The block of wood is attached to a horizontal massless spring with a spring constant k = 195 N/m that is initially uncompressed. The other end of the spring is attached to a wall as shown in the figure below. The bullet becomes imbedded in the block of wood which then compresses the spring a distance of 28.0 cm before momentarily coming to a stop.

1) determine the mass of the block

i know you use conservation of momentum to do this but how do you find final velocity which is needed to solve the problem?

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Figured this would be the right place to ask this:

Does anyone know what day Dan Russell's birthday is?

We need to do a biography on someone and need a birthdate but I can't find it. Probably shouldn't be this hard.

Edited by Peaches

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Amateur, need help, can someone help me with this:

B=7.5 A=15.3 h=?

h2 = a2 + b2

h2 = (15.3)2 + (7.5)2

h = √ (15.3)2 + (7.5)2

h = √ 234.09 + 56.25

h = 23.2

Okay, I'm super amateur, I'm guessing the square root of the h gets transferred into √.... how does this work?

I don't know how it works so what I did was, 234.09+56.25=290.34 which has the square root of 17.03 (Wrong answer)

+'s to anyone who can teach me in laymen terms lol

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APA format. Help a brotha out.

I know this is late, but owlpurdue is great for this.

I'm also getting quite familiar with it due to my psych research class.

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Scrolling through all these math equations is starting to make my head hurt. Just checking out the thread, mostly math eh? I recognize some of the pre-calc questions, barely passed

pre-calc 11 :)

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Scrolling through all these math equations is starting to make my head hurt. Just checking out the thread, mostly math eh? I recognize some of the pre-calc questions, barely passed

pre-calc 11 :)

You think you can help me with mine?

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