jfan17

This is the exact same question that has already been posted, and its formatting is just as horrendous(not trying to be rude). 

Hey Jenny! See you got some conics to deal with!

Let's start with the general equation of a circle, the one that most people are taught:

(x - a)² + (y-b)² = r², where (a,b) is the center of the circle, and r is the radius of the given circle.

Because the question gives us 3 points, we know they all must satisfy this equation.

Let's pick the pair of points (7,-4) and (-3,-4) to plug in(there's a good reason for picking this pair. Do you see why?)

We can first look at (7,-4). This gives us:

(7-a)² + (-4-b)² = r²

Let's then look at (-3,-4). It'll start to become clear why I picked these two points. We get:

(-3-a)² + (-4-b)² = r²

We realize that we can then set these two equations equal to each other, to get:

(7-a)² + (-4-b)² = (-3-a)² + (-4-b)²

Now here's where our selection of points comes in handy. Realize we can just subtract 

(-4-b)² on both sides, which gives us only a, allowing us to solve for a without getting a nasty(ier) expression than if we chose another pair.

We get:

(7-a)² = (-3-a)² 

Expanding, we get:

49-14a + a² = 9+6a + a^2

Combining like terms and subtracting appropriately, we get:

40 = 20a

a = 2

To get b, let's create another statement with two different points, say (7, -4) and (6,-7)

We get:

(7-2)² + (-4 -b)² = (6-2)² + (-7-b)²

This gives us:

25 + (-4 -b)² = 16 + (-7-b)²

9 + 16 + 8b + b² = 16 + 49 + 14b + b²

Combining like terms and subtracting where is necessary, we obtain the equation:

6b = -40

b = -20/3

Our circle's equation is then:

(x - 2)² + (y+20/3)² = r²

I'll leave finding the radius to you(it should be fairly simple, just plug in one of the points to find it, and then take the square root of the result. 

Also just wanted to put this out there. If you have lots of time to do this problem, I'd actually recommend just bashing out all of the possibilities divisible by 11. Because there are only ~ 80 - 81(do the math?) divisors of 11 from 100 - 999, I would recommend listing all of them out and then testing on a case by case basis(of course, this heavily relies on you having lots of time). 

Kudos to you to!

Sorry I'm not CU, but I'll try this question. I've done a variant of it before I think.

Let's call N = 100a + 10b + c

Then we know:

100a + 10b + c = 11m for some arbitrary integer m

m = a^2 + b^2 + c^2 

100a + 10b + c = 11a^2 + 11b^2 + 11c^2.

Using divisibility rule of 11, we know:

Case 1:

a-b+c = 0

a+c = b

or Case 2: 

a-b+c = 11

b = a+c-11

this splits the problem into two cases for us:

1. b = a +c where the first case for divisibility of 11 holds true.

100a + c + 10a + 10c = 11a^2 + 11c^2 + 11(a+c)^2

10a + c = 2a^2 + 2ac+ 2c^2

10a + c = 2a^2 + 2ac + 2c^2

Here, we're kind of stuck. Realize that we can factor out 2 from the right hand side, meaning that c must be even since a is also even.

we can write c = 2z, with z being either 0, 1, 2, 3, or 4, and substitute it in

we get:

10 a + 2z = 2a^2 + 4az + 8q^2

Dividing by 2 on both sides, this gives us:

5a + z = a^2 + 2az + 4z^2

We just bash out cases now, and eventually we find:

when we have z = 0, the case works out, with a being 5, and c = 0  b = 5. From this case, we get that N = 550. All other cases don't work(2,4,6,8)

I'll leave the rest of the problem to you; there's only one other case!

2. ?????

also edit: sorry for the bad "typography". Had to do this in a hurry

FORMAT PLEASE!!!! I can't emphasize this enough, but recently people have just been straight pasting problems from other sites, so it becomes unreadable. I'd rather just not attempt a problem that I can't clearly interpret(probably most other people too), so please make sure you're not just c + p ing !

Name the quadratic you try to find as :

\(x^2+ax+b\)

Next we define the roots as:

x1 and x2

By vietas formulas:

x1 + x2 = -a

x1 * x2 = b

Since we know that the roots are one more than the last two coefficients respectively, we have:

x1 = a + 1

x2 = b+ 1

Substituting back in, we get:

a + 1 + (b+1) = -a

a+b + 2 = -a

2a = -b-2

a = -b/2 - 1

(a+1) * (b+1) = b

(-b/2 -1 + 1) * (b+1) = b

(-b/2) * (b+1) = b

(-b2/2 - b/2) = b

b2/2 + b/2 = -b

Dividing by b on both sides, we get:

b/2 + 1/2 = -1

b/2 = -3/2

b = -3

plugging back into our first equation, we get:

(a + 1) + (-3 + 1) = -a

a + 1 -2 = -a

a -1 = -a

2a = 1

a = (1/2)

our two roots are 1/2 and -3

This is unreadable. Please repost your question with proper formatting so it's readable

What is the value of  if \( x\cdot(3x+1)  is what appears on my screen. Please reformat it for readability!

Why must I imagine this scenario? Aren't we only looking at what's given in the problem, not anything outside of it? Imagine if I asked you to solve for the volume of a pyramid, and then saying "wait, in real life, a pyramid like that would never be constructed!" That's not the point here. Even if you're rounding to the nearest hundredth, that's not very hard to do; what's important is the process to get to the answer, not so much the answer itself.