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i downvoted all the the dumbo's posts in here.

no worries guys

No problem, we are here to help!

Thanks You All I really apretiate it!!!!

you must be having a good laugh with your retarded SMILEY FACES, ARENT YOU!?!

vietas is also easy to prove.

lets say you have x^2+bx+c

a=1 for simplicity, it will work for any other case too

so you have (x+j)(x+k).

x^2+(j+k)x+jk.

j and k are the opposites of the roots.

so then -b/a is the sum.

and two negations still have the same product, so it is still c/a.

hope this proof helped!

Vieta's formulas are formulas that are like shortcuts.

two of them are where you have ax^2+bx+c.

and the sum of roots is -b/a

product is c/a.

asdf is correct! Good job! 722.

ok so number two.

1. calculate the first 2*1

2*1 is 4-2+1 or 3.

2. now you have 2*(1*(2*(1*3))).

calculate 1*3.

1*3 is 3-1+3 or 5.

3. now you have 2*(1*(2*5)).

calculate 2*5.

2*5 is 20-2+5 or 23.

4. now you have 2*(1*23).

calculate 1*23.

1*23 is 23-1+23 or 45.

5. just calculate 2*45.

this is 180-2+45 or 225-2 or 223.

HOPE THIS HELPED!

How many times do you want us to read this goddamn question? 

Speak your piece, or forever STFU!

GA

ok thanx

explanation for i)

4/64 simplifies to 1/16

16 only has factors of 2 (5 would not change either)

therefore, it is terminating.

You are correct. I wrote down the wrong number from WolframAlpha!

i gave you a hint in the same question here: https://web2.0calc.com/questions/help_57731

and number 2: keep your work organized and solve

isnt 808 the sum of the proper divisors of 432?

i thought you had to do it twice.

so you would have 2^3*101

then using formulas, you get 15*102 or 1530

then take away 808 to get 722.

for it to be terminating, it must only have factors of 2 and 5 because 17 is prime.

therefore 17/40 or 17/(2^3*5) is terminating.

70 has a factor of 7, so it will be recurring

HOPE THIS HELPED!

1) a = 4, b = 9, You should be able to work out c and d easily from this.

3) 722.  I just used WolframAlpha to answer this. (Edited to correct answer - see asdf below).

i) 4/64 = 0.625 and it ends there so therefore it is TERMINATING

ii) We are given, 
             3/70 = 0.04285714285...
= 3/10*7 = 3/2*5*7

Clearly, Denominator has one more factor other than 2 and 5,
Therefore it has RECURRING DECIMALS

Like so

Thanks Alan :)

Here's a derivation of the relation:

NB For the person who posted the question:  Please note that this is not a proof by induction!

Does that mean that after I learn how to factor large numbers, I have to give up eating bananas and climbing trees? 

I think it’s better to factor large numbers while sitting in the treetops eating bananas.  Plus, it’s easier to throw banana peels at you ground-dwellers while you speak techno-babble. 

I suspect this rude, careless dumb-dumb intended this equation:

\(\LARGE x^2 - ax + b = 0\\ \)

Geometric solutions for this quadratic form are found by plotting A(0, 1) and B(a, b) and using the midpoint and distance formulas for finding (y) intercepts in circles. This seems consistent with the question.  Examples of these are sometimes included in pre-calculus texts.