Melody

I'm trying to do a calculation where 125+20*floor(wave/10) Where Floor is (Wave/10). X Would represent Wave, being the variable. I want to find the combined values of this equation where x is equal to 1-50. EXAMPLE If x = 1 then 125+20*.1 (127) + If x = 2 then 125+20*.2 (129) For a Total of 256. I wish to find the total for all the way up to x=50

\(f(x)=125+20* \frac{ x}{10} \\f(x)= 125+2x\\~\\ f(1)=125+2=127\\ f(2)=125+4=129\\ ....\\ f(50)=125+2*50=225\)

You want

127+129+131+       ......   +225      OR     Term 1 up to term 50

This is the sum of an AP and the formula is  

\(S_n=\frac{n}{2}(a+L)\\ S_{50}=\frac{50}{2}(127+225)\\ S_{50}=25*(352)\\ S_{50}=8800\ \)

3x4=12

That is assuming that you wear both those shoes at the same time. :)

Mmm

Wrt

With respect to....

I think it might help if you watch (and study) this video.     

You have made a little mistake EP

64-49=15 not 16

Otherwise i think your method would work so long as you were very careful about the negative signs.

There is another way to do this using the actual number plane but you we have laready given avdb33  have 2 methods :))

Let θ be an angle in quadrant III such that sinθ=−7/8

. 
Find the exact values of secθ and cotθ

In the third quadrant cos is negative so sec is negative too    (sec=1/cos)

and

In the third quadrant tan is positive so cot is positive too    (cot=1/tan)

Draw a right angled triangle and put theta as one of the acute angles.

sin theta =7/8     you can ignor the negative sign for this.

so

put 7 on the opposite side to theta and 8 on the hypotenuse.

Use pythagoras's theorum to work out the adjacent side

\(opp=7\\ hyp=8\\ adj=\sqrt{64-49}=\sqrt{15}\)

\(sec(\theta)= -\frac{ hyp}{adj} =-\frac{8}{\sqrt{15}}=\frac{-8\sqrt{15}}{15}\\ cot(\theta)= \frac{adj}{ opp} =+\frac{\sqrt{15}}{7}\\\)

I have not checked my answer and there is a lot of room for careless error but this is what I did.

Steps.

1)   I used polynomial division to get the remainder - which I got as     (5k-2)/4

2)   then I looked at

      \(\frac{5k-2}{4}=\frac{15-i}{32}\\ ...\\ 40k=31-i\\ k=\frac{31}{40}-\frac{i}{40}\)

Hi avdb33,

I think something is missing from your question.

There sould be something on both sides of the equal sign.  :)

also, do you mean

sin(theta)-78

or

sin(theta-78)         ( I assume the angles are in degrees.)

What are you talking about ?????

Thanks Heureka