PEMDAS my friend. Start with the squared term in the denominator, evaluate the rest of the fraction. Then remember to multiplyOriginally Posted byboondockthenadd.

Jameson

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- November 4th 2005, 05:54 PM #1

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- November 4th 2005, 05:59 PM #2

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- November 4th 2005, 07:03 PM #3

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- November 4th 2005, 11:43 PM #4

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- November 5th 2005, 03:11 AM #5

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how do you solve this

Lv = {[Av *1.84 *10^8] / [Vb *1728 *(Fb/0.159)^2]} -{0.823*sqrt(Av)}

if av=63 vb=6.0 fb=45

what are the steps you use

Lv = {[63 *1.84 *10^8] / [6.0 *1728 *(45/0.159)^2]} -{0.823*sqrt(63)}

Lv = {[115.92 *10^8] / [10,368*(283.019)^2]} -{0.823 *7.937}

Lv = {[115.92 *10^8] / [10,368*80,100]} -{6.532}

Lv = {[115.92 *10^8] / [830,476,800]} -{6.532}

Lv = {[115.92 *10^8] / [8.305 *10^8]} -{6.532}

Lv = {(115.92 / 8.305) *10^(8-8)} -6.532

Lv = {13.958 *10^(0)} -6.532

Lv = {13.958 *1} -6.532

Lv = 13.958 -6.532

Lv = 7.426 -----------------answer.

- November 5th 2005, 09:15 AM #6

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- November 5th 2005, 09:56 AM #7

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- November 5th 2005, 01:09 PM #8

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how did you get Lv = {[115.92 *10^8] / [8.305 *10^8]} -{6.532}

where did this come from [8.305 *10^8]

--------------------------

how did you get Lv = {[115.92 *10^8] / [8.305 *10^8]} -{6.532}

Lv = {[Av *1.84 *10^8] / [Vb *1728 *(Fb/0.159)^2]} -{0.823*sqrt(Av)}

Lv = {[63 *1.84 *10^8] / [6.0 *1728 *(45/0.159)^2]} -{0.823*sqrt(63)}

Lv = {[115.92 *10^8] / [10,368*(283.019)^2]} -{0.823 *7.937}

Lv = {[115.92 *10^8] / [10,368*80,100]} -{6.532}

Lv = {[115.92 *10^8] / [830,476,800]} -{6.532}

Lv = {[115.92 *10^8] / [8.305 *10^8]} -{6.532}

Lv = {(115.92 / 8.305) *10^(8-8)} -6.532

Lv = {13.958 *10^(0)} -6.532

Lv = {13.958 *1} -6.532

Lv = 13.958 -6.532

Lv = 7.426 -----------------answer.

You are asking me how I got the 6th line.

If I say I got it by doing the first 5 lines above it, would you believe me?

-----------------------

where did this come from [8.305 *10^8]

Lv = {[115.92 *10^8] / [830,476,800]} -{6.532}

There is the 830,476,800.

It is 830Millions, +476Thousands, +800.

The decimal point is at the end of the 800, or at the end of the 830,476,800.

If you want to express the 830,476,800 in engineering mode, or in a small number times 10^(something), then you place the decimal point all the way to the left, after the first significant number, after the the 8 of the 830Millions.

In so doing, how many decimal places, or how many numbers did you pass over? Count them. From the decimal point being at the end of 830,476,800 to the decimal point being moved to between 8 and 3 of the 830Millions?

Did you count 8 decimal places? If yes, then you are correct.

So, now the 830,476,800. is transformed into

8.30476800 *10^8

or, 8.304768 *10^8

Since in your original equation you used only 3 decimal places for the numbers less than 1, [1.84, 0.159, 0.823], then I followed your lead. I reduced/expressed the 8.304768 to 8.305 only.

Do I have to ask you why I used "5" only for the whole "4768"?

You have not been taught in school about "rounding up" yet?

-----------

May I add this:

830,476,800

is the same as 8.305 *10^8,

the same as 83.048 *10^7,

the same as 8,304.768 *10^5,

the same as 0.00083 *10^12

- November 5th 2005, 03:14 PM #9

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- November 5th 2005, 03:19 PM #10

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- November 5th 2005, 05:02 PM #11

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- November 5th 2005, 06:20 PM #12

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- November 5th 2005, 08:00 PM #13