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Let us suppose that there exists  integers a, b such that

√2  =   a / b      where a/b   is reduced to lowest terms  [a and b have no factors in  common ]

So

b√2 = a           square both  sides

2b^2  = a^2

But a^2  must be even......so  ......"a" must be even......

Then   a =  2m    ....so a^2  = 4m...so......

2b^2 = (2m)^2

2b^2 =  4m^2        divide both sides by 2

b^2 = 2m^2      then b must also be even

But....we assumed that a, b had no factors in common.....so.....they can't both be even

So....our aqssumption that

√2 = a / b  must be false, since it leads to a false conclusion

100-2+1=99

But Chris you have still assumed that a person is heavier than all the others, just as EP assumed the opposite.

We do not know if the odd person is too heavy or too light. :/

I am sure some of the other answers are excellent but i have not wrapped my head around them yet.

I still like my answer though.  

The graph of y=ax^2+bx+x is given below, where a,b , and c are integers. Find a+b+c.

I assume that is meant to be

The graph of y=ax^2+bx+c is given below, where a,b , and c are integers. Find a+b+c.

parabola

axis of symmetry is x=1

so

\(\frac{-b}{2a}=1\\ b=-2a\)

so we have

\(y=ax^2-2ax+c\)

When x=0 y=1 so   c=1

\(y=ax^2-2ax+1\)

When x=1, y=3

so

\(y=ax^2-2ax+1\\ 3=a-2a+1\\ 2=-a\\ a=-2 \)

\(\boxed{y=-2x^2+4x+1}\)

ok thank you can u please explain where 100-2+1 come from please

Put 6 people on one side of the see-saw and 6 on the other

Obviously....the heavy person will be on the lower side

Divide this side into two groups of 3......again...the heavy person will be on the lower side

Choose any 2 of these 3 people to weigh

We have two possibilities for the third weighing :

1) They balance.....the heavy person is the one not weighed

2) The don't balance....again....the heavy person is on the lower side

Note, YEEEEEET.....we are summing  99 terms   [ 100 - 2] + 1  = 99

So....

(1/6) (99) ( 2*(99)^2 + 9*(99) + 13)    =    338,349

Check this against    sum (r + 1)^2  from r = 1 to r = 99  = 338,349

i got -18, 18

by sqrting both values

Modulo 13 inverses of the first 12 positive integers are:
1, 7, 9, 10, 8, 11, 2, 5, 3, 4, 6, 12 =78 - Their total sum.
78 mod 13 =0 - Remainder.

Two distinct possible values for B.

5 x 6^5  + 3 x6^4  + 1 x 6^3 + 3 x 6^2 + 4 x 6 + 0 = 43116       43116/210 = 205 R 66

16  36  are the only two I can conjure,,,,,   so two :    1 and 3

S(n) = \(\dfrac{Sn + 1}{(n+1)!}\).

We can take a look at the constant term, 18. 18 factors into 2*3*3, so the two values of x must be 6 and 3. (They must be in a ratio of 2 to 1)

We can write the factored form of x^2 + bx + 18 = 0 as (x - 3)(x - 6) = 0. If we plug in x = 3 or x = 6, the equation is true, so 3 and 6 are solutions. 

If we factor out (x-3)(x-6), we have x^2 - 9x + 18, so \(\boxed{b = 9}\).

Hope this helps,

- PM

S=1/1+1/2+1/3+...+1/n

S(n) ?

15/30 does not equal 25/30.

1/2 does not equal 5/6.

We need to know what the probability of you making throwing the ball into the bowl when there is a certain volume of water inside the cone. We also need to know if the ball will go in or out of the bowl if the ball is thrown straight onto the rim. 

I think I need more information to solve this problem, but I can provide some help.

The area of a cone is \(\dfrac{1}{3}\pi r^2 h\).  The radius of this cone is 0.9 m, and the height is 1m. The volume of the cone is \(0.27\pi\), which is around 0.85 m^3. Since 1 cm^3 = 1 mL^3, and you pour 10mL each time you make the ball into the bowl, you will need to make the ball into the bowl 9 times for the cone to fill up with water. 

If the probaility that you make the ball into the bowl each time is P, and X is the number of times you need to throw the ball before the cone is full of water, your equation would be P*X = 9. 

EP, the odd one out might be the heavy one. 

Thx who answer my question

I answered your question in your previous post.

If you want to check Devin's solution, we can use the same equation of y = 1.5x + 0.2.

x = 2.2 and y = 1.7. 

We have 1.7 = 1.5(2.2) + 0.2 --> 1.7 = 3.3 + 0.2 --> 1.7 = 3.5. This is not true, so Devin's solution is not correct. Adam's is correct.

I do not think we need a graph for this. We can just solve the system of equations to find out who is right. 

      y = -0.5x + 3

-     y = 1.5x + 0.2

_______________

      0 = -2x + 2.8 --> 2x = 2.8 --> x = 1.4

Inserting x = 1.4 into the equation y = 1.5x + 0.2, we have y = 2.1 + 0.2 --> y = 2.3. The solution is (1.4, 2.3)

Adam's solution is correct.

- PM

6x^2-47x+15 can be positive or negative or 0. We can factor \(6x^2-47x+15\) into \((3x-1)(2x-15)\). For (3x-1)(2x-15) to be prime, either (3x-1) or (2x-15) must be equal to 1 or -1. There are 4 scenarios. 

1) 3x - 1 = 1. Simplifying, we have x = 2/3. Putting this into (2x-15), we have 4/3 - 15, which is not an integer or prime. 

2) 3x - 1 = -1. Simplifying, we have x = 0. Putting this into (2x-15), we have -15. Absolute value makes it 15, which is not a prime.

3) 2x - 15 = 1. Simplyfying, we have x = 8. Putting this into (3x-1), we have 24-1 = 23. 23 is a prime.

4) 2x - 15 = -1. Simplifying, we have x = 7. Putting this into (3x-1), we have 21-1 = 20. 20 is not a prime. 

The largest integer x such that |6x^2-47x+15| is prime is \(\boxed{8}\).

- PM