This is my first time teaching 8th grade and I'm learning.

I mean I'm learning to teach

I'm confident about my math :)

This is tricky, would it throw people off if you tried to put a hold on this stuff, return to just standard-fare rearranging, and then go back and try to pick this up again?

Yeah, @randomgirl jumping around in the textbook is not a good idea at that age :-)

/would it make you lose too much time?

Ah, yeah teaching for the first time is a big challenge

Hell, the first time I tried to give a one-off lecture to peers, it was a total disaster

I think I'm going to insert my own lesson between the book's lessons that seem to make too big of jump.

the position at time t =>0, of a particle moving at along a coordinate line is x = 10cos(t+pi/4)

Try going back and covering the prerequisites.

What is the particles starting point? could anyone help me with this problem

I taught college algebra for a few years so I'm kinda used to a little higher thinking. I will adjust. I know I will.

Thank you guys.

8th grade needs patience.

@MATHASKER what does t equal at the starting point

Have the other 8th grade teachers had similar issues?

And yeah good luck @randomgirl!

well i'm the only math teacher at my senior/jr high school

small school

very small

Wow!

0 @orbit-stabilizer

so I just plug in zero for t

@MATHASKER good, so do you know the answer now?

Yes.

yea i think so thanks

np

Try returning to first principles @randomgirl

@Daminark so a semidirect product is basically where we have two groups, whose interesection is trivial and one is trivial

one is normal*?

You are probably right. I would rather them get this now then me reteach it to them in 9th grade or some other grade.

but what if we don't know what the two groups are embedded in

That's the internal semidirect product

@randomgirl teach them Gaussian elimination /s

Okay, then for the external, we don't require either of them to be normal?

The external one is of two abstract groups

Building a strong foundation always works with algebra @randomgirl

If im asked to find the average rate of change of a given function given the x and y coordinates, how would I solve it? I have the function $(4x+1)^(1/2)$ over the point $[0,3/4]$ should I take the derivaitive of the function and plug in the x value, what about the y value when do i plug that in?

Over the point or over the interval?

over the given interval

@orbit-stabilizer

You have $f(x) = \frac{4x+1}{2}$ right?

no they give me f(x) = √(4x+1)

Okay, so you want the average rate of change. So you know you that you should take the derivative

Since you want the average, there are many ways to do it, but a common way that is taught is to plug in the end points

plug in the end point?

I know I should take the derivative but end points?

Have you covered integration?

yea

little bit

The average value of a function is defined in terms of integrals - so we should use that instead

Rather than my end points thing

tutorial.math.lamar.edu/Classes/CalcI/AvgFcnValue.aspx

the question is asking for like the rate of change issn't that the slope?

Slope = derivative evaluated at a point, yes

so should'nt there be a way to do it with derivatives but not integrals>

Are you okay with the fact that the derivative of a function gives another function that gives you the slope of the original function at each point?

Well, you don't need to take an integral if you realize what $f$ is.

yea I think thats how I was taught but i forgot how to do it lol

should i use the h method for this one

or is it the c one

@GFauxPas halp

Is it true that the limit of L^p norms as p->q is the L^q norm? Assuming f is in all the applicable L^p

average rate of change for the function $f(x)=(4x+1)^{\frac{1}{2}}$ over the given interval would just be $\frac{f(\frac{3}{4})-f(0))}{\frac{3}{4}-0}$

oops (f(3/4)-f(0))/(3/4-0)

Oh ok thanks so no need to take derivative>

unless it asked for rate of change aka instantaneous rate of change at a particular value

but yeah no need for derivative

no need for derivative because of FTC

oh ok thanks for helping!