# Questions and Answers

# Your Questions About X1

Joseph asks…

## Can 2 different materials have exactly the same density and weight?

Is it possible for 2 different materials to have the exact same density and weight? I mean, 2 different materials, both cut 1′**x1**‘**x1**‘. If they both weighed the same and were just as dense, wouldn’t they be just as soft/firm too? Wouldn’t they have to have the same texture? Wouldn’t that make them the same material?

### Administrator answers:

Your supposition is that a material can have only one density and weight. Think about ice. A gallon of liquid water and a gallon of frozen water have different densities but are identical in every other way. By your rational, water and ice would qualify as different materials. Likewise, cold water is denser than hot water so even comparing seemingly identical materials you achieve different results.

That being said, I would have to say that elements will all have individual weights and densities, but compounds can be made up in any number of ways. Once you start changing the chemistry of a molecule you get into much more complex differentiations. Gas chromatography, where a small sample of a material is burned and the reflected light carries its own unique signature, is the standard for calculating the true differences between materials.

Michael asks…

## How to type axis descriptors in Word and Excel?

I’m taking a statistics class and must type a formula using mathematical character fonts such as **X1**, Y1, and R1. The numbers are smaller than the letter and sometimes are shifted below or above the letter. How do I type these characters in Word or Excel?

Thanks for helping!

### Administrator answers:

Goto Add/Remove in the control panel. Choose modify MS Office. Change. Make sure that Equation Editor is installed.

From the insert menu choose equation. You type A_1 for A sub 1 into the box.

Tricks to make this easier include saving a library of symbols and using menus or key-combinations (e.g., Alt-Shift A) to insert from the library.

You may want to consider a word processer especially designed for Math notation, Lyx. Http://www.lyx.org/

Ken asks…

## I need help creating an example of the slope formula in this format?

Use the slope formula: (y=y2-y1/x2-**x1**) to show why horizontal lines have a slope of 0, and why vertical lines have undefined slopes. (in other words, make an example with two points, and demonstrate what happens when you have two points from a horizontal line, and then two points from a vertical line.)

5 stars to best answer <3

### Administrator answers:

Your formula is wrong. See what you typed.

This is the correct one: m = (y2 – y1) / (x2 – x1)

Vertical lines have a common x-value. Thus, the slope is undefined.

(5, 2) (5, 10) <<< plot this and use the formula (the correct formula, not the one that you like)

Horizontal line have a common y-value. Thus, the slope is zero.

(2, -10) (5, -10) <<< plot this and use the formula

I suggest you see http://matrixlab-examples.com/equation-of-a-straight-line.html

.

Betty asks…

## How to find magnitude of instaneous velocity?

An object undergoes uniformly accelerated motion from point **x1** = 1 m at time t1 = 1 s to point x2 = 44 m at time t2 = 8 s.

(a) If the magnitude of the instantaneous velocity at t1 is v1 = 3 m/s, what is the instantaneous velocity v2 at time t2?

(b) Determine the magnitude of the instantaneous acceleration of the object at time t2.

### Administrator answers:

Uniform acceleration a = constant

v2 = v1 + a (t2 – t1)

v2 = 3 + a (8 – 1)

v2 = 3 + 7a

x2 = x1 + v1 (t2 – t1) + 1/2 a (t2 – t1)^2

44 = 1 + 3 (8 – 1) + 1/2 a (8 – 1)^2

44 = 22 + 49/2 * a

a = 44/49 = 0.9 m/s^2

v2 = 3 + 7 * 0.9 = 9.3 m/s

the instantaneous velocity v2 = 9.3 m/s

the instantaneous acceleration a = 0.9 m/s^2

Charles asks…

## How can I solve my two simultaneous equations using the matrix?

I can express my equations in the following format:

Y1=K1X1+K2X2 ………………….(1)

Y2=K3X1+K4X2 ………………….(2)

the values of Y1,Y2,K1,K2,K3 and K4 are known. I need to find the values of **X1** and X2 using the matrix.

thank you very much.

### Administrator answers:

You can express the system of

the two equations in terms of

matrix multiplication as:

[k1 k2] * [x1] = [y1]

[k3 k4] * [x2] = [y2] or briefly

[K]*[X]=[Y] where K,X,Y are

the matrices above.

Then if the determinant

D(K)=k1*k4-k3*k2 is non zero

the solution is given by the

matrix equation

[X]=[1/D(X)]*[[K}^-1]*[Y]

where K^-1 is the inverse

matrix of K

[k4 -k2]

[-k3 k1]

If D(K)=0 then the system has

either no solution or it has an

infinite number of solutions.

To find then what happens you

can follow other well known methods

Thomas asks…

## How do I capture the annual average yield on 10 year bond yield since 1970?

I need to get this data set for my (y) on a statistics project. Then another data set on the annual S&P 500 (**x1**) and then US dollar exchange rate against the Euro (x2). I know this information should be out there with relative ease. But I’m not familiar with the systems well enough to accomplish attaining these 3 data sets. Looking for help.

### Administrator answers:

The first thing you need is the name of the bond you are searching for…….provide that and we may be able to help you………

Linda asks…

## How do you find the tangent and normal lines of a general parabola WITHOUT using calculus and derivatives?

Assume the general equation of the parabola of the equation is a(x^2)+bx+c. How would you find both the equations of the tangent line and normal line at any given point say (**x1**,y1) on the parabola WITHOUT using calculus and derivatives? How would you find the slope?

### Administrator answers:

The tangent cuts the parabola at only ONE point.

Here’s an example: find the slope of the parabola x^2 + 3x + 1 at x = 5.

The equation of the parabola is y = x^2 + 3x + 1.

The point (5,41) is on the parabola, so the equation of a line through that point is y – 41 = m ( x – 5 ) with slope m.

Solve: x^2 + 3x + 1 = m(x-5) + 41.

Convert into a quadratic in x, with m as a parameter. This must have only ONE solution. What is the condition for a quadratic to have only one solution: B^2 = 4AC. The condition finally simplifies to m = 13.

Once you have the slope you have the equation of the tangent. The normal passes through (5,41) and the slope of the normal is related to the slope of the tangent, giving the equation of the normal.

Jenny asks…

## Why do so many professional wildlife photographers not use Full Frame sensors?

The crop factor of **x1**.6 or **x1**.5 is great for wildlife because it gets you closer. But are there any downsides to this?

I have been looking at professional photographers and it surprises me how many use DX sensors. Is FF only good for Landscape or am I missing something?

Should a serious wildlife photographer (who plans to sell his photographs) buy a dx or fx sensor?

### Administrator answers:

Perhaps cost is part of it. Although that is changing. Yeah, it was probably the crop factor that sold them on APS-C. It made their long lenses “seem” even longer. FF offers better noise control and acuity. Overall superior image with FF.

Now that FF cameras often have an APS-C mode I see no reason at all for getting an APS-C sensor camera.

“Should a serious wildlife photographer (who plans to sell his photographs) buy a DX or FX sensor?” IMHO, the bigger the better. I have a Canon 5D and love the image quality that the sensor is capable of. FF all the way.

Check out the comparison of the geese in this thread. It’s about Canon but it applies to Nikon as well: http://photo.net/nature-photography-forum/00UUhd

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