# Questions and Answers

# Your Questions About X1

Ruth asks…

## Find intersection coordinates given the distance of two points?

I have the map coordinates of point A and B.

I have to find the coordinate of point C knowing that this point is **X1** meters distant from point A and X2 meters distant from point B.

I am sure this can only be done using UTM coordinates, but how?

Thanks for any help.

- No drawing with compass

### Administrator answers:

Us a compass to draw a circle that is x1 meters in radius around point A, and another that is x2 meters in radius around point B. They’ll intersect at 2 points, so you’ll have to pick one based on other info (if they gave you any!)

Donna asks…

## What is the equilibrium concentration of hydronium ion in a solution?

HX is a weak acid that reacts with water according to the following equation:

HX(aq) + H2O(l) H3O1+(aq) + **X1**-(aq)Ka = 1.58e-05

What is the equilibrium concentration of hydronium ion in a solution that is 0.0944 M in HX and 0.136 M in **X1**- ion?

[H3O1+] = ____M

### Administrator answers:

Set up your equation

Ka = Products / Reactants

Ka = [H3O+][X-] / [HX]

Note: (H20 can be ignored)

Plug in your values:

1.58×10^-5 = [H3O+][.136M] / [.0944M]

1.49×10^-6 = [H3O+][.136M]

[H3O+]eq = 1.10×10^-5M

Mark asks…

## How do you find the slope and y-intercept in this equation: x = -2 and graph it?

Also, there is a question “What does **x1** stand for?” and “What does y1 stand for?”

Only the 1′s are little 1′s that are a little below the number. It’s what they use in equations where you substitute in a value for x or y, but what would it ‘stand’ for?

### Administrator answers:

X = -2 is a vertical line. It is a line going up and down through -2 on the x-axis.

There is no y-intercept b/c it will not cross the y-axis

For a vertical line, there is no slope. Your answer would be: “no slope”

(Do not say you have zero slope b/c that would be a horizontal line with y = (any number) like y = 4)

As for the x1 and y1 question…there is a formula for finding slope and it is:

y2 – y1

———

x2 – x1

You use this formula if you are given two points on a line:

Example: (4, 2)(5,

y2 = 8

y1 = 2

x2 = 5

x1 = 4

You are numbering the coordinates really to work out the formula.

( 4 , 2 )…..(5 , (x1,y1)……(x2,y2)

(8 – 2)

——–

(5 – 4)

so my slope would be 6 / 1 or 6

You label them so you know how to plug them into the formula to get your slope.

Hope this helps!

Betty asks…

## What is the coefficient of friction between the block and the table?

A block weighing 2 lb is forced against a horizontal spring of negligible mass, compressing the spring an amount **x1** = 6 inches. Upon releasing the block, it moves on a horizontal table top a distance x2 = 2ft before coming to rest. The spring constant k is 8 lb/ft.

### Administrator answers:

The potential energy of the spring must equal the work done to slide the block.

The potential energy of the spring E is ½ k x², where k = 8 lb/ft and x = 0.5 ft.

The work done to move the block is W = F x, where F is the frictional force and x is the distance the block moved, 2 ft. The friction force F is the normal force N multiplied by the coefficient of friction ?. Here N = 2 lb.

Equating Energy to Work,

½ (8 lb/ft) (0.5 ft )² = ? (2 lb) (2 ft)

? = 0.25

Jenny asks…

## How do I find the eigenvectors of this matrix?

[2 3 0

3 2 0

0 0 1]

I can find the eigenvalues (5,-1,1) easily but I seem to be doing something wrong to find the vectors. (3-u)**x1**+2y1=0 and I should be able to find my vectors by plugging in the eigenvalues but! when I check my work (plug in the eigenvalues into another equation) I get an extra eigenvector.

Help!

### Administrator answers:

Subtract your matrix from xI and get

x-2 -3 0

-3 x-2 0

0 0 x-1

Now, find the determinant:

[(x-2)^2 - 9](x – 1) = (x – 5)((x + 1)(x – 1)

5, -1 and 1 are eigen values.

Now multiply your matrix by a vecorr (x1 x2 x3)

For the eigenvalue 1:

2×1+3×2 = x1

3×1+2×2 = x2

x3 = x3

or

x1 + 3×2 = 0

3×1 + x2 = 0

You don’t have to work too hard to find that x1=x2=0, and x3 is any real number.

For the eigenvalue 5

2×1+3×2 = 5×1

3×1+2×2 = 5×2

x3 = 5×3

Well, x3 = 0

Let’s find the dependency between x1 and x2.

-3×1+3×2=0

-2×1+2×2=0

Well, x1=x2

Take (1,1,0) as an eigenvector of 5

Robert asks…

## How do I answer this question about midpoint?

The question is,

A line segment has (**x1**, y1) as one endpoint and (xm, ym) as its midpoint. Find the other endpoint (x2,y2) of the line segment in terms of **x1**, y1, xm, and ym.

I’ve used mid point before but I don’t know what to do when I’m not given the values of x and y.

### Administrator answers:

Try here..

Http://www.visharadsoftware.com/forum/

you can ask calculus, physics, and math there…

There’s a genius person there who answers all the questions asked..

She answered correctly all of my physics questions..

Try it for urseLf…

Goodluck!

^_^

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