June 8, 2013

## How do I find the slope of a line passing through two points?

The points are: (a-b,0) and (a,b)

I need the slope of the line as an integer or a fraction. I understand that the slope formula is m=y2-y1 over x2-x1 but the a-b has thrown me off.

M = (b – 0)/(a – (a – b)

m = b/(a – a + b)

m = b/b

m = 1

## How do you solve a power regression problem for two variables?

By power regression, I mean in the format of y = ax^b. I know it has something to do with logarithms, but I just can’t figure it out. I know you need two x and y points. So if you have (x1,y1) and (x2,y2), how do you solve for two variables?

By logs, solve for two variables…

Good luck!

## Any tips to help sleep changes because of the return to standard time?

My 4-month old little one used to wake up between 6-7 am. With the change back to standard time, she’s now waking up at 5am and nothing I do seems to help. I’ve pushed nap times and bedtime forward, but she’s waking earlier in the night to nurse (x1) and this 5 am wake up time isn’t shifting at all. Any tips??!!

Just be patient. She’ll adjust. My son is on a tight schedule (self-inflicted)…he’s asleep by 8, and up at 6. This past week has been rough, he’s out by 7:15 and up at 5:15! Everyday it gets better, by next week, she’ll be back on track.

## How to convert a ‘general’ form of line to a ‘normal’ form?

General:
Ax+By+C=0
Normal:
y=m(x-x1)+y1

Now understand the normal form and how it works,
but I have no idea how the general form should work?
What’s the slope, where are the origin coordinates?
I’m confused

Can somebody help me to better understand how this General Form of Line works?
Thanks!

First, you have to realize the “Normal” form of the line is essentially the point-slope form where the y coordinate is moved to the side with the x variable.

As lines consist of infinitely many points, then there are an infinite number of “Normal” equations for any given General equation, just as there are an infinite number of General equations that can be found by multiplying by a constant.

Just solve for y.

Ax + By + C = 0

By = -Ax – C

y = (-Ax – C)/B {Now you need to put it in a similar form as above.

Y = -A/B x – C/B

Notice this is the slope-intercept form of the line, where y1 = 0

The slope is always -A/B, or the coefficient on x when you have y = mx+b.

If you want the Normal form, you need to pick a point on the line.
If you are given one, great.
If not, just go straight to slope intercept. That will inherently give you a point (0, -C/B)

## Does a tattoo on the back of your neck hurt?

I want to get a 2 x1 inch tattoo on the back of my neck, right below my hairline. How badly did it hurt if at all? Especially for a girl. Thanks
I’m just getting a little paintbrush with the word passion along the handle of it. Just a contour with a bit of shading, so it is rather minimal.

I know someone who has one on the back of her neck, and she said it didn’t really hurt. And she’s really skinny and bony, so i think you’ll be okay.

## What is the value of d^2 where d is the distance between X1 and X2?

Let ABCD be a rectangle such that AB=5 and BC=12. There exist two distinct points X1 and X2 on BC such that ?AX1D=?AX2D=90?. Suppose that d is the distance from X1 to X2. What is d^2?

X = AX?
Y = X?D
z = BX?

Let X? Be closer to B than D. Then x < y

Area(?AX?D) = 1/2 * 12 * 5 = 30
Area(?AX?D) = 1/2 * x * y = xy/2
xy/2 = 30
xy = 60
y = 60/x

x² + y² = 12²
x² + 3600/x² = 144
x? + 3600 = 144x²
x? ? 144x² = ?3600
x? ? 144x² + 72² = ?3600 + 72²
(x² ? 72)² = 1584
x² ? 72 ± 12?11
x² = 72 ± 12?11

x² = 72 ? 12?11
y² = 72 + 12?11

AB² + BX?² = AX?²
5² + z² = x²
25 + z² = 72 ? 12?11
z² = 47 ? 12?11
z² = 36 ? 12?11 + 11
z² = (6 ? ?11)²
z = 6 ? ?11

BX? = 6 ? ?11
Similarly, CX? = 6 ? ?11

d = X?X? = 12 ? 2(6??11) = 2?11
d² = 44

## How can I show a function presents increasing returns to scale?

Consider the production function f(x1, x2) = x1^(1/2) + x2^(3/4). Show that it
presents increasing returns to scale.

What you described is what’s called a Cobb-Douglas production function. If the sum of the exponents on L and K are > 1, you have increasing returns to scale. When they’re = 1, you have constant returns to scale, and when they’re 1. Now the original production function = f(L,K) = L^a * K^b becomes:

=> f(Z*L,Z*K) = (Z*L)^a * (Z*K)^b
=> f(Z*L,Z*K) = Z^(a+b) * L^a * K^b
=> f(Z*L,Z*K) = Z^(a+b) * q

(The part L^a * K^b is the original production function and becomes the letter q for elegance purposes.)

Now we see that we multiply the original production function (q) by a number > 1 that has an exponent bigger than 1, which means the percentage extra production (you can think of this as Z^(a+b)) will be bigger than the percentage extra inputs (you can think of this as Z).

Hope this helped!