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

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

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

Calculus1
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

What's stopping you from plugging t = 1 for the first problem?
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
Oh I see, you want the velocity, what you have there is the position, take the derivative and plug t = 1.

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

what would be the derivative?
Have you done calculus?
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.
\[s'(t) = -32t-20\] would be your velocity
First derivative of displacement is velocity, the second is acceleration.
where are you he 32 from o.o
Mhm, I just used the power rule
\[\frac{ d }{ dx } x^n = nx^{n-1}\]
more confused lol. ok show me how to solve this. o.o
Ok it seems you haven't got far at all in calculus yet lol
yup
Ok, so can you tell me what you know?
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?
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 :)
so what I am getting now for my first answer is -32. did I do that right?
No, the derivative is s'(t) = -32t-20, so you still have to plug in t=1 in this equation.
And for your second question average rate of change is \[\frac{ f(b)-f(a) }{ b-a }\]
ok I plugged t=1 in and got -52. not one of my choices.
How did you get that?
\[s(t) = -16t^2+20t+100,~~~s'(t) = -32t+20\] plug t = 1, into s'(t), so find s'(1).
Like I said s(t) is your position, but s'(t) is your velocity
And you're looking for velocity when t = 1
1=-32(1)+20 1= -32+20 1=-12 then what? t=-12?
Well you leave s'(1) alone, but yes, - 12.
I said -12 earlier lol
You said -32 and -52
But do you understand how that is?
You will need to read your book/ youtube videos I think, then you will really understand it.
yes. thanks. um on the second question a would be -2 and b is 5. right?
Yup
a = -2, b = 5
y=2+3? That's just a horizontal line
is there suppose to be an x there somewhere haha
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?
Yes, similarly
\[\frac{ f(5)-f(-2) }{ 5-(-2) }\]
making my answer for number 2 , --1? lol
Show your work
If it makes more sense, y = 3x+2 just means f(x) = 3x+2
You should not get -1
I got 3
That sounds good!
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
How did you get it
I went with this: y=mx+b formula looked at the slope of my equation which is -5. is my reasoning wrong?
No, all these problems require calculus
You need to use \[\lim_{h \rightarrow 0} \frac{ f(a+h)-f(a) }{ h }\] use this definition to find it
It's not a linear equation it's y = -5x^2+2, if it was y = -5x+2 we could say slope is -5.
how would is et it up though?
\[\lim_{h \rightarrow 0} \frac{ f(1+h)-f(1) }{ h }\]
Can you do the rest?
im assuming h is 0? making my anser =cannot be determined?
No, limit h -> 0 does not mean h = 0, you need to simplify the problem, before you start taking the limit.
Hey, why are they giving you these problems if you haven't learnt it yet..?
I don't know im in a pickle?. lol ok then is it -10?
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.
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
Use your equation you are given, plug it into the "formula" I gave you \[\lim_{h \rightarrow 0} \frac{ -5(1+h)^2+2-(-5(1)^2+2 )}{ h }\]
Now simplify
-3? I put that previously a
Oh, I don't know.
Well since you're out of time I guess I will show you how to do it, the answer I got was -5. @magy33
|dw:1435652296303:dw|
Note that we keep the limit till we actually take the limit itself.

Not the answer you are looking for?

Search for more explanations.

Ask your own question