BMWX1

Questions and Answers

Your Questions About X1

April 7, 2013

Charles asks…

Is HCL ME tablet x1 is a worth one as it is cheap?

Many consumers complaint about hcl me x1 tablet. But it is a cheaper having android 4.0.4 . Is it a worth one or not?

Administrator answers:

Well yeah for such a low price it is fine, but not at gaming or handling heavy apps. If you want the most out of your buy then buy a micromax tablet at the same range it might give you the satisfaction as hcl me tab is too buggy . Go for the micromax tab
Hope this helps safe shopping.

James asks…

What are coordinates of the center of mass for the uniform plate shown in the figure below?

What are coordinates of the center of mass for the uniform plate shown in the figure below? (x1 = 21 cm, y1 = 15 cm, and y2 = 4 cm. The figure is not shown to scale.)

http://www.webassign.net/hrw/W0176-Nalt.jpg

Find the x coordinate and the y coordinate.

Administrator answers:

Xcm = ?[? (20 + y1)x dx (-10 ? X ? 0) + ? (y2 + 10)x dx (0 ? X ? 10) + ? (y2)x dx (10 ? X ? X1)] / {?[(20 + y1)(10) + (y2 + 10)(10) + (y2)(x1 - 10)]}

xcm = [? 35x dx (-10 ? X ? 0) + ? 14x dx (0 ? X ? 10) + ? 4x dx (10 ? X ? 21)] / [35(10) + 14(10) + 4(11)]

xcm = (1/2)[35(-100) + 14(100) + 4(341)] / 534

xcm = -736 / 1068

xcm = -184/267 cm ? -0.69 cm

ycm = ?[? 20y dy (-20 ? Y ? -10) + ? 10y dy (-10 ? Y ? Y1 - y2) + ? (10 + x1)y dy (y1 - y2 ? Y ? Y1)] / {?[20(10) + 10(y1 - y2 + 10) + (10 + x1)(y2)]}

ycm = [? 20y dy (-20 ? Y ? -10) + ? 10y dy (-10 ? Y ? 11) + ? 31y dy (11 ? Y ? 15)] / [20(10) + 10(21) + (31)(4)]

ycm = (1/2)[20(-300) + 10(21) + 31(104)] / 534

ycm = -2566 / 1068

ycm = -1283/534 cm ? -2.40 cm

Mandy asks…

How do you write inches and feet with quotation marks?

I am looking at a college to bring list and they have 2′x3′x1‘ and I’m wondering if that means 2 feet by 3 feet by 1 foot or is that inches?

Administrator answers:

Feet is written with one quotation mark.Inches is written with two.
So 2′x3′x1′ is 2 feet by 3 feet by 1 foot.

Donald asks…

How to solve for coefficient in exponential decay?

Sorry, it’s been a long time (10 years) since I’ve had a math class and I haven’t had to use this in some time. If I have an equation like e^(-ax1) = e^(-ax2) + e^(-ax3) and I have values for x1, x2, and x3, how do I solve for a?

I tried some algebra and got to 1=e^(a*{x1-x2}) + e^(a*{x1-x3}) but that really just leaves me in the same place I was before. I suppose I could transform it to

ln[1] = ln[e^(a*{x1-x2}) + e^(a*{x1-x3})]
0 = ln[e^(a*{x1-x2}) + e^(a*{x1-x3})]

but I don’t see how that gets me further to my answer.

Thanks for your help in advance!

Administrator answers:

A possible way, would be to, assuming x3>x2, operate as follows:

E^(-a*x2) – E^(-a*x3)=E^(-a*x3)*(1 + E^((x3 – x2)*a))

And log the result:

Log[E^(-a*x3)*(1 + E^((x3 - x2)*a))]
Log[E^(-a*x3)]+Log[(1 + E^((x3 - x2)*a))]
(-a*x3)+Log[(1 + E^((x3 - x2)*a))]

If you also log the left hand side of the equation, you get the equality again:

(-a*x1)=(-a*x3)+Log[(1 + E^((x3 - x2)*a))]

Differentiate on both sides with respect to a:

x1 = x3+ 1/(-x2 + x3) + E^(-a (-x2 + x3))/(-x2 + x3)

You can then isolate a:

x1 – x3 – 1/(-x2 + x3) == E^(-a (-x2 + x3))/(-x2 + x3)
(x1 – x3)*(-x2 + x3) – 1 = E^(-a (-x2 + x3))
-a (-x2 + x3) = Log[(x1 - x3)*(-x2 + x3) - 1]
a = -Log[(x1 - x3)*(-x2 + x3) - 1]/(-x2 + x3)

I didn’t double check the algebra… But the idea is factorize the right hand side. Then log both sides. Differentiate both sides and then solve for a. And Log is here the natural logarithm.

Donna asks…

How to replace an equality equation by 2 inequalities ?

Have come accross a question where the equality X1 + X2 = 2 was replaced with X1 + X2 <= 1 and X1 + X2 >= 1

Could not understand the concept form this trivial problem … could you please give a more concrete example with the concept ?

Administrator answers:

X1 + X2 = 2 equal to X1 + X2 <= 1
2 equal to 1 is not true
there is some thing wrond in your description.

Carol asks…

How would I wire this switch up?

Details: I have a normally open switch, 120v Illuminated LED on off switch. The on-off part of the switch I have no problem with, I run the line through the switch to my motor. When I turn the switch ON (or close the switch) the motor turns on, no problem. But the switch is supposed to illuminate. There are two terminals for the illumination marked X1 and X2 . How would I wire this to have the switch illuminate as it should?

Administrator answers:

If you want it to light when on:
t2 to x1
n to x2

If you want it on all the time then
t1 to x1
n to x2

Jenny asks…

Whats the difference between the Box Motorbike helmets BX 2R and the BX 1?

Hubby has tried on the B X1 (Scope) in XXL, and it fit perfect, but he likes the design of the BX 2R ( Target )and the BX 2 (Wired) , does anyone know what the difference is or are they the same fit just a different code due to the design?

Administrator answers:

Any decent shop should be happy to order the xxl in either of the helmets he likes so that he can try them on. In theory they should fit the same, but only one way to find out for certain!

David asks…

does the ear force x1 headset by turtle beach work with hdmi?

does the ear force x1 headset by turtle beach for the xbox 360 work with hdmi or do you have to use the av cables supplied with xbox 360 to use the headset?

Administrator answers:

Just buy headphones that are “microsoft certified” on them or the actual green and white or black x-box 360 headphones………

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