CPhill

Rosala.....look at my second way, again....it's just simple math.....

1st ..   Subtract 41-22 = 19 and add 1  = 20

2nd ..  Add 41 + 22 = 63

3rd ..   Multiply the "1st" times the "2nd"......(20)(63) = 1260

Now ..... divide that by 2 =  1260/2 = 630  !!

I know you can do that  ....  !!

Adding a long string of numbers is tedious.....and sometimes almost impossible.....this way is "easy-peezy"

What about 1,3,9,10,12 to -9?

Every number can only be used once and(unless you can find a different solution) all numbers are used.

So....you want to make -9 ???

(3 - 12) + (10 - 9) - 1 =

-9 + 1 - 1  =

-9 + 0  =

-9

75 =

25 *3 =

5 * 5 * 3 =

5² * 3

P.S. - you can do this on our calculator....just enter ...   factor(75)

The arc intercepted by the 71 degree angle is 2(71) = 142 degrees.

Since part of this arc is 104 degrees, the other part is the "C" degrees.

Therefore 142 - 104 = 38 degrees...and that is the measure of "C."

Likewise the arc intercepted by the 68 degee angle is 2(68) = 136 degrees.

Then,  the remaining minor arc - "B" - between the 71 degree angle and the 68 degree angle is just 360 - 136 - 104 = 120 degrees.  Therefore "A" = (1/2)(120 + 104) = (1/2)(224) = 112 degrees.

BTW......nice pic, Melody !!!

Melody, I believe he/she is saying that the arc intercepted by angle ABC (minor arc AC) is 110 degrees.

The sine is equal to 1/2 at 30 degrees and at 150 degrees on the interval 0 to 360 degrees, inclusive.

A more general form is:   30 +n360   and    150 + n360 ...   where n is an integer.

In radians, it would be pi/6 and 5*pi/6 with the general form being:

pi/6 + n(2pi)       and     5pi/6 + n(2pi)   ...   where n is an integer.

Let's get the derivative, first.....then, we can "plug in"

lim(Δx →0)  [1/(x +Δx) - 1/x]/Δx  =

lim(Δx →0)  [x/[x(x +Δx)] - (x +Δx)/[x(x +Δx)]]/Δx  =

lim(Δx →0)  [x - (x +Δx)]/(Δx[x(x +Δx)]) =

lim(Δx →0)  [-Δx]/(Δx[x(x +Δx)]) =

lim(Δx →0)  -1/[x(x +Δx)]=

lim(Δx →0)  -1/[x² +xΔx]  .....now, take the limit = -1/[x² + 0] =

-1/x²

And evaluating this at x = 2, we have

-1/2²  = -1/4

Note..you could go back and substitute "2" for "x"  in the limit stuff....you should arrive at the same answer without having to substitute at the end....!!!

I like the way you add the "check' to your answers, NinjaDevo......it really brings it home that your solutions are correct......3 points from me!!!

Thankee mightily, Morgan ...( or, if I can use thy formal appellation..Morgan^-1 )

Thou elixir has given me great insight into the possessor of  the object of mine quest...I should have suspected all along that it was the Great and Powerful Rom who holds the "Rom"an Numeral that I search for......I must must proceed carefully, however....It is known in the land that his "thirst" for power(s) is unrestrained by any bounds or limits.......mathematically speaking, of course

No thanks to Sisyphus, either....that lout and cad !!  I should have known all along that he was worthless....a real sypher, if you will......

I had made gift of flowers and chocolates for Lady Guinivere who celebrates her birthday this day. Apparently, they were unappreciated!! I think I will throw in Sisyphus as "lagniappe." ....(even though current records indicate that M'Lady was born on or about April First.....a noted US holiday..... fitting for one so regal..........)

Your Humble Servant,

Sir CPhill

Here's another way this could be done, rosala..it maybe isn't quite as nice and neat as Alan's, but it still "works"

The sum of the first n positive integers is just given by n*(n+1)/2

So, the sum of the first 41 integers is just 41*42/2

And the sum of the first 21 integers is just 21*22/2

So subtracting the second from the first, we get  41*42/2 - 21*22/2  .... or just......

(41*42 - 21*22)/2 =  1260/2 = 630

Note, that what we are really doing - in non-mathematical terms - is just taking the sum of the first 41 positive integers and "lopping off" the sum of its first 21 terms......so we have 22+23+24+25+.....+41 = 630

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Finally, rosala.....if you only want to work with the numbers themselves and not have to remember any "n's," etc., here's one more method.....(it's really Alan's in disguise!!)

(Subtract the two numbers and add 1)

(Add the two numbers)

Multiply the first thing times the second and divide by two

So we have   (41-22+1)(22+ 41) /2  =     20(63)/2 = 630.....no real "formulas" to remember....only a procedure.....whether this is "better" or "worse"...???.....I'll leave that up to you!!!

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P.S. -  Does the questioner want all sum of all the terms from 22 to 41 or between 22 and 41 ???.........well, no mind....we've given him/her a few different methods to achieve the second...