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ليس لديها ما يكفي من المال.

They are asking you for a graph of height (of the bottom of the spring) with respect to time.

Time in seconds will be the horizontal axis, and height is the vertical axis.

The midline is the bottom of the weight when the spring is at rest. That is 1.6metres above the ground

The mass is pulled down by 1.1 metres - so that is the amplitude.

the wavelength is 1 second (1/2 +1/2)

The way I am looking at it, the spring starts at the bottom of its cycle so this is a reflection of the usual cos graph about the line y=1.6

so the equation is

y=-1.1cos[(2pi/wavelength)x]+1.6

y = -1.1cos(2pi*x)+1.6

I didn't use the decimals either.  I found out quite quickly that decimal would make my answer less accurate, and an exact answer was needed.

Admittedly I did start using  0.75 and 0.25 but I changed them back to fractions when I actually used them.  

Thank you so much. I wasn't using decimals so that might have made the problem a bit more tricky to follow.

this is a geometric progression problem  

a1 = 320; r = 1/4, Sn = 6825/16       find n

For a GP

\(S_n=\frac{a(1-r^n)}{1-r}\\ a=320, \quad r=0.25\quad S_n=6825/16 = 426.5625\\~\\ \frac{6825}{16}=\frac{320(1-0.25^n)}{1-0.25}\\ \frac{6825}{16}=\frac{320(1-0.25^n)}{0.75}\\ \frac{6825}{16}*\frac{3}{4}*\frac{1}{320}=1-0.25^n\\ \frac{1365}{16}*\frac{3}{4}*\frac{1}{64}=1-0.25^n\\ \frac{4095}{4096}=1-0.25^n\\ 0.25^n=1-\frac{4095}{4096}\\ 0.25^n=\frac{1}{4096}\\ \frac{1}{4^n}=\frac{1}{4096}\\ 4^n=4096\\ 4^n=4^6\\ n=6\)

LaTex:

S_n=\frac{a(1-r^n)}{1-r}\\
a=320, \quad r=0.25\quad S_n=6825/16 = 426.5625\\~\\

\frac{6825}{16}=\frac{320(1-0.25^n)}{1-0.25}\\
\frac{6825}{16}=\frac{320(1-0.25^n)}{0.75}\\
\frac{6825}{16}*\frac{3}{4}*\frac{1}{320}=1-0.25^n\\
\frac{1365}{16}*\frac{3}{4}*\frac{1}{64}=1-0.25^n\\
\frac{4095}{4096}=1-0.25^n\\
0.25^n=1-\frac{4095}{4096}\\
0.25^n=\frac{1}{4096}\\
\frac{1}{4^n}=\frac{1}{4096}\\
4^n=4096\\
4^n=4^6\\
n=6

It says that you are composing an answer, Melody. Are you actually there? This has happened a few times and nothing happened. I do need help. Can I repost the question so more people see it?

Thx Guys!

I figured that is what you meant.   

Also, are you saying I'm doing an AP problem in a normal algebra class? That would make me feel better about not being able to solve it.

I've already tried. I got to 455/256 = 1/4^n but when I converted it into log and inputted it into my calculator, it said it couldn't do the problem. I also got -181999 = 1/4^n but that also wouldn't work when I tried getting the final answer through converting it into log. I hope this makes sense. I was using log to simplify for n in both instances. That's the only way I know how to do it. I'm not even sure if 455/256 and -181999 are the right end numbers. I'm confused and I need help! I hope I've shown you that I tried. Along the way, I also tried different ways of simplifying and canceling out too. I'm stuck.

Oops, I meant 34 not 37. 

13+34 = 47. 

Thank you for noticing. 

=^._.^=

13 + 37 = 47  ???

I just did a similar one to this, only with an AP.  Have a go by yourself.

Show us what you try, even if you don't get far.

It has nothing to do with intelligence,  most likely you have not learned it yet.

It is a topic called sequences and series.

Let me see.

AP

a=9

Tn=a+(n-1)d = 72

so

9+(n-1)d=72

Sn=n/2(a+Tn) = 405

n/2 * (9+72) =405

n/2 * 81 = 405

n=10

OK. Thanks!

Looking at this problem, the first thing that I notice is that 13 + 37 = 47. 

34 = -13 (mod 47)

13^(-1) = 29 (mod 47)

-13^(-1) = -29 (mod 47)

-29 = 18 (mod 47)

=^._.^=

I'm a little confused on the question, or it just may be I'm not smart enough to understand..

It was not removed by a moderator.

It was automatically censored, probably because you used the word 'death'

12 players, how many pairs of people can we get?

12*11/2 = 66

=^._.^=

We answered this very recenty. Go look for it.

OK.......glad to  help   !!!!

Thank you so much! 

I understand it!

I guess this  is supposed  to  be

sqrt ( 7 - 3x)  +  3  =  x                

sqrt ( 7 - 3x)   =  x - 3         (1)              square  both sides

7 - 3x  =   (x - 3)^2

7 - 3x   =  x^2  - 6x   +  9

x^2  -6x  + 9  - 7  + 3x   =  0

x^2  - 3x  +  2   =   0          factor

(x - 2)  ( x - 1)  =  0

Setting   each  factor  to  0 and solving for  x    gives us

x = 2      or    x   =  1

However, looking at (1)......either  value of  x produces  a negative  value  on  the right.....and  we  can  never get a negative  value   from the positive root on the left

So.....we  have no real solutions

MY BAD! I was doing it all wrong. It was the exponent of 1/5 because I was multiplying it by the exponent of 5. Thank you! I figured it out.

It has already got an answer.  If you don't like it then make a post on the original thread saying why.