help! fan and medal!
a ball is thrown vertically upward from the top of a 100-foot tower, with an initial velocity of 20ft/sec. its position function is s(t)=-16t^2+20t+100. what is its velocity in ft/sec when t=1 second?
a. -12
b.-44
c.100
d.-32
what is the average rate of change of y with respect to x over the interval [-2,5] for the function y=3+2?
a. -9
b.3
c.1/3 ( wrong answer )
d.-1

- anonymous

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- Astrophysics

What's stopping you from plugging t = 1 for the first problem?

- anonymous

im getting a choice that is not presented. I don't know what im doing wrong
see:
-16(1)^2+20(1)+100
I get 104

- Astrophysics

Oh I see, you want the velocity, what you have there is the position, take the derivative and plug t = 1.

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## More answers

- anonymous

what would be the derivative?

- Astrophysics

Have you done calculus?

- anonymous

I am taking the class. lol I am stuck on this lesson. what would be the answer to my questions ?would you be able to work the problems out and show me how to do it. im confused.

- Astrophysics

\[s'(t) = -32t-20\] would be your velocity

- Astrophysics

First derivative of displacement is velocity, the second is acceleration.

- anonymous

where are you he 32 from o.o

- Astrophysics

Mhm, I just used the power rule

- Astrophysics

\[\frac{ d }{ dx } x^n = nx^{n-1}\]

- anonymous

more confused lol. ok show me how to solve this. o.o

- Astrophysics

Ok it seems you haven't got far at all in calculus yet lol

- anonymous

yup

- Astrophysics

Ok, so can you tell me what you know?

- anonymous

pretty much what I first posted while trying to find an answer to my questions lol.
I have limited time so im fussing over what I am doing wrong lol. o.)
here is my guess:
1- -44
2- -9
am I right?

- Astrophysics

You can use other methods such as the kinematic equations I suppose, but this question requires you to understand calculus, I gave you the derivative already, so you should be able to figure it out :)

- anonymous

so what I am getting now for my first answer is -32. did I do that right?

- Astrophysics

No, the derivative is s'(t) = -32t-20, so you still have to plug in t=1 in this equation.

- Astrophysics

And for your second question average rate of change is \[\frac{ f(b)-f(a) }{ b-a }\]

- anonymous

ok I plugged t=1 in and got -52. not one of my choices.

- Astrophysics

How did you get that?

- Astrophysics

\[s(t) = -16t^2+20t+100,~~~s'(t) = -32t+20\] plug t = 1, into s'(t), so find s'(1).

- Astrophysics

Like I said s(t) is your position, but s'(t) is your velocity

- Astrophysics

And you're looking for velocity when t = 1

- anonymous

1=-32(1)+20
1= -32+20
1=-12
then what?
t=-12?

- Astrophysics

Well you leave s'(1) alone, but yes, - 12.

- anonymous

I said -12 earlier lol

- Astrophysics

You said -32 and -52

- Astrophysics

But do you understand how that is?

- Astrophysics

You will need to read your book/ youtube videos I think, then you will really understand it.

- anonymous

yes. thanks. um on the second question a would be -2 and b is 5. right?

- Astrophysics

Yup

- Astrophysics

a = -2, b = 5

- Astrophysics

y=2+3? That's just a horizontal line

- Astrophysics

is there suppose to be an x there somewhere haha

- anonymous

y=3x+2 im supposed to look for the average rate of y with respect to x over the interval [-2,5] for the function y=3x+2. so this is basically like my slope formula? m=y2-y1/x2-x1?

- Astrophysics

Yes, similarly

- Astrophysics

\[\frac{ f(5)-f(-2) }{ 5-(-2) }\]

- anonymous

making my answer for number 2 , --1? lol

- Astrophysics

Show your work

- Astrophysics

If it makes more sense, y = 3x+2 just means f(x) = 3x+2

- Astrophysics

You should not get -1

- anonymous

I got 3

- Astrophysics

That sounds good!

- anonymous

quicl question. lol
would this answer of mine be correct:
what is the slope for the function y=-5x^2+2 at the point x=1?
a- -5 ( my choice)
b- -10
c--3 ( not the answer)
d-slope no determined

- Astrophysics

How did you get it

- anonymous

I went with this: y=mx+b formula looked at the slope of my equation which is -5. is my reasoning wrong?

- Astrophysics

No, all these problems require calculus

- Astrophysics

You need to use \[\lim_{h \rightarrow 0} \frac{ f(a+h)-f(a) }{ h }\] use this definition to find it

- Astrophysics

It's not a linear equation it's y = -5x^2+2, if it was y = -5x+2 we could say slope is -5.

- anonymous

how would is et it up though?

- Astrophysics

\[\lim_{h \rightarrow 0} \frac{ f(1+h)-f(1) }{ h }\]

- Astrophysics

Can you do the rest?

- anonymous

im assuming h is 0? making my anser =cannot be determined?

- Astrophysics

No, limit h -> 0 does not mean h = 0, you need to simplify the problem, before you start taking the limit.

- Astrophysics

Hey, why are they giving you these problems if you haven't learnt it yet..?

- anonymous

I don't know im in a pickle?. lol ok then is it -10?

- Astrophysics

If you show your work I will tell you whether you're right or not, use the draw tool if you want to, because otherwise to me it seems you're just guessing.

- anonymous

in this question I am lol if h does not equal 0 its just approaching 0? how would I be able to find h before solving o.o

- Astrophysics

Use your equation you are given, plug it into the "formula" I gave you \[\lim_{h \rightarrow 0} \frac{ -5(1+h)^2+2-(-5(1)^2+2 )}{ h }\]

- Astrophysics

Now simplify

- anonymous

-3? I put that previously a

- Astrophysics

Oh, I don't know.

- Astrophysics

Well since you're out of time I guess I will show you how to do it, the answer I got was -5. @magy33

- Astrophysics

|dw:1435652296303:dw|

- Astrophysics

Note that we keep the limit till we actually take the limit itself.

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