Melody

Find the angle of least positive measure that is coterrminal with -3061  (degrees I will assume)

3060/60 = 51

so this is the same as -1 degree

-1 is equavalent to  359 degrees

This is coterminal with +1 degree

If a ball is thrown vertically upward with a velocity of 128ft/s, then its height after t seconds is s=128t-16t^2.

a) what is the velocity of the ball when it is 240 ft above the ground on its way up? (consider pos direction)

b) what is the velocity of the ball when it is 240ft above the ground on its way down?

- I calculated maximum height reached by the ball which is 256 feet.   :)  You didn't need to do this.

To start with it is symmetrical so whatever the velocity is going up it will be the same only minus going down.

To avoid confusion I am going to replace the s which is a height with the letter y.

\(y=128t - 16t^2\\ \dot y=128-32t\\ \text {you are being asked to find } \dot y \text{ when y=240 }\\ \text{But to do this you need to find t when y=240 first}\\ 240=128t-16t^2\\ 16t^2-128t+240=0\\ t^2-8t+15=0\\ (t-3)(t-5)=0\\ t=3\;\;or\;\;t=5\\ when\;\;t=3\\ \dot y=128-32*3\\ \dot y=128-96\\ \dot y=32\;feet/sec\\ \text{The velocity on the way up is } 32f/sec\\ \text{The velocity on the way down is } -32f/sec\\ \)

Hi guest,

This is a good question, if you appreciate your answer and you will use this site again, how about becoming a members.  Members often get priority when we are answering question - especially polite members that are trying hard to learn from our answers. :)    

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-1500*[(1-(1+.07)^-N)/(.07)]+40396.2*(1+.07)^-N-10000=0

\(-1500*\frac{1-(1+.07)^{-N}}{0.07}+40396.2*(1+.07)^{-N}-10000=0\\ -1500*\frac{1-(1.07)^{-N}}{0.07}+\frac{40396.2}{(1.07)^{N}}-10000=0\\ \frac{-150000[1-(1.07)^{-N}]}{7}+\frac{40396.2}{(1.07)^{N}}-10000=0\\ \frac{-150000[1-(1.07)^{-N}]}{1}+\frac{7*40396.2}{(1.07)^{N}}-70000=0\\ -220000+150000(1.07)^{-N}+\frac{7*40396.2}{(1.07)^{N}}=0\\ -220000(1.07)^{N}+150000(1.07)^{-N}(1.07)^{N}+7*40396.2=0\\ -220000(1.07)^{N}+150000+282773.4=0\\ -220000(1.07)^{N}+432773.4=0\\ 220000(1.07)^{N}=432773.4\\ (1.07)^{N}=\frac{432773.4}{220000}\\ (1.07)^{N}\approx 1.967151818\\ log(1.07)^{N}\approx log(1.967151818)\\ Nlog(1.07)\approx log(1.967151818)\\ N\approx \frac{ log(1.967151818}{log(1.07)}\\ \)

log(1.967151818)/log(1.07) = 10.0000034615204508

N = 10      approximately.

check:

-1500*((1-(1+.07)^-10.0000034615204508)/(.07))+40396.2*(1+.07)^(-10.0000034615204508)-10000 = 0.0000029048699091459732438786634643045252961939322074922

Well, that is pretty close to zero :)))

Err....  could it be 210  ???

2- -1=  2 + 1 = 3

New birthday post for Rosala   2016  (and 2014)

2016

Mmm maybe you are right ElecticPavlov.  I never was great at English.

BUT

If I was talking about shoe that belonged to Dennis I would call it Dennis's shoe.  Maybe that is wrong - I don't know.

Anyway, in this question the singular word is product. The plural word is products.

The apostrophy indicates ownership so the data on safety and effectiveness belongs to more than one of the companies products.  It may only refer to one type of product at a time, like a specific machine but the company makes more than one of these machines so it is still plural.

That is my 2 cents worth but I am a mathematician - not a grammar expert :))

Find the nth derivative of each function by calculating the first few derivatives ....

\(\begin{align}a)\quad f(x)&=x^n\\ f'^{(1)}(x)&=nx^{n-1}\\ f'^{(2)}(x)&=n(n-1)x^{n-2}\\ ...\\ f'^{(n)}(x)&=n(n-1)......1x^{n-n}\\ f'^{(n)}(x)&=n!\\ \end{align}\)

\(\begin{array} \\ b)\\\; f(x)&=&x^{-1}\\ f'^{(1)}(x)&=&-1x^{-2}\\ f'^{(2)}(x)&=&(-1*-2)x^{-3}\\ f'^{(3)}(x)&=&(-1*-2*-3)x^{-4}\\ ...\\ f'^{(n)}(x)&=&(-1*-2*-3*.....*-n)x^{-(n+1)}\\ f'^{(n)}(x)&=&(-1)^n n!\;x^{-(n+1)}\\ f'^{(n)}(x)&=&\frac{(-1)^n n!\;}{x^{(n+1)}}\\ \end{array}\)

 You had best be very careful to check through my working and my logic but this answer seems ok to me.

WHAT ARE YOU TALKING ABOUT?

What's the difference between (-x)^y and -x^y?

THEY ARE NOT THE SAME :)

\((-x)^y = (-1)^y\times x^y\\~\\ -x^y=-1\times x^y\\~\\ \text{these two expressions will be the same only if }(-1)^y=-1\\ \text{when will this be true?}\\~\\ \text{Well, it will be true if y is odd, it will be false if y is even}\\ \text{so they are definitely not the same thing.}\)