michaelcai

In order to find the points where the parabola intersects, we set the equations to equal to each other in order to find a common x term. We get:

-x^2-x+1=2x^2-1 

Simplfying, we get 3x^2+x-2=0, and using the quadratic formula, we get that x can be equal to 2/3 or -1.

These are the x-values of the coordinates of the intersection points, or a and c. SInce c is greater than a, we get that c=2/3, and a=-1. 2/3-(-1) is equal to 1 2/3, or 5/3, which is our answer.

Ans=5/3

Conversation between friend and me: 

So you can write f(x) = (x-p)(x-q)(x-r)(x-s)+5, which will satisfy the thing and then minimize it since everything is integer

oh yeah

but how to minimize it]

isnt the minimumn 5?

f(x)>5 though

Plug in x=t

ok

Basically minimize positive value of four different integers (t-whatever)

Value of product of four

 f(t) = (t-p)(t-q)(t-r)(t-s)+5>5, so   (t-p)(t-q)(t-r)(t-s)>0

?

 (t-p)(t-q)(t-r)(t-s) would be 1*2*3*4?

at minimum

Yeah so the product of the four t's need to be positive and distinct since p,q,r,s are different

so the answer wold be 24?

No

Wait

The product has to be positive but not each term

oh

And add 5

-1, 1, -2, 2 product=4, plus 5 is 9?

so it would be 9?

And the answer is 9

The answer is -2, 0, 2 and four. 

n^2+n+1 is equal to (n-1)(n+2)+3.

dividing this by n-1 gives n+1-3/n-1. from there, we find all integers where 3/n-1 is an integer, thus we get n=-2, 0, 2, and 4

I really need help on these questions. Any help is appreciated!

 (-inf, 1] is the answer

c=-3

The answer was 3.

Solved it: : Noticing that left-hand sides are symmetric in $a$, $b$, and $c$ (in that any relabelling of the variables in one of the left-hand sides gives one of the others), we multiply all three equations to get $(a^2b^2c^2)^2=(28)(27)(21)$, which implies $a^2b^2c^2=\sqrt{(28)(27)(21)}=\sqrt{(4\cdot 7)(9\cdot 3)(3\cdot 7.2\cdot 3\dot3\cdot 7=126.$ Dividing this equation by each of the three original equations gives $c^2 = 9/2$, $a^2 = 6$, and $b^2=14/3$, respectively. Since $b

This is correct. Thanks!

The correct answer is actually 36sqrt2.

k=4

a=5, 50, ....

r=2, slowest going up

\(\begin{align*} S &= \frac{a(1-r^{n})}{1-r}= \frac{2}{3} \cdot \frac{1-\left(\frac{2}{3}\right)^{10}}{1-\frac{2}{3}}\\ & = \frac{2}{3}\cdot\frac{1-\frac{1024}{59049}}{\frac{1}{3}}=\frac{2}{3}\cdot\frac{3}{1}\cdot\frac{58025}{59049}=\frac{2\cdot58025}{59049}\\ & = \boxed{\frac{116050}{59049}}. \end{align*}\)

I figured it out

Edit: 63 was correct

yeah, I got 160 as well, but my answer checking thing says its incorrect...

1000 more than 298,370 is equal to 298,370+1000 which is equal to 299370

Nvm, i solved it. answer is (3,10)