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Square root of .3 without a calculator?

I am doing a problem where I cannot use a calculator. The entire problem is x^2/.05 = (5.6 x 10^-10). So far, I did (5.6 x 10^-10) x .05, which I approximated to .3 x 10^-10

Now I don't know how to find an approxiation for the square root of .3 . I know that the square root of 3 is 1.8, but that doesn't help me. What shortcut can I use to do this?

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Keith Leannon
Keith LeannonLv2
19 Jan 2019

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I ALREADY KNOW X=0.189.


BUT, I AM STUMPED AS TO WHERE TO GO FROM HERE...PLEASE SHOW ME ASTEP BY STEP. PLEASE ASSUME I DON'T KNOW HOW TO DO THE NEXTSTEP.


Here was the original equation: 54.3=(2x)^2/(0.240-x)^2.

I understand you need to take the square root of both sides toget rid of the squares. The square root of 54.3=7.369.


I am drawing a blank! please help!


(square root symbol) 54.3= 2x/0.240-x


x=0.189.

BUUUUUUT......

How do I get x? Do I crossmultiply?

I am having trouble getting the image to appear apparently, I can see it when I submit the problem. If you can not see the image any suggestions would be appreciated to help me out.

Problems

Consider the discretized square plate in Figure 1.

Figure 1: Discretized Square Plate

Write a system of linear equations for this system and clearly identify the matrix A in the matrix equation Ax=b.

Compute the determinant of A.

Compute the classical adjoint matrix (i.e., the adjugate matrix) of A.

Using your answers from problems 2 and 3, solve for A-1 and x.

Use Cramer’s rule to check your answer by solving for just x1 and x4.

Show transcribed image text
Hello there. I just had one question dealing with binary operationsin abstract algebra. I need to see if this operation has anidentity element for all x in R. The operation is as follows:x*y=|x-y| This is what I did. I basically broke it up since thereis an absolute value. x*e=|x-e| = x-e if x> 0 or -(x-e) if xe,then x*e= x-e. We need to find e for x*e=x. So x-e=x and e=0. Nowmy first question is this: 1) If x>0, then why for thisequation, x-e=x, can x be values that are less than 0. It seemslike a contradiction to me. Is e=0 only valid for x>0? Thenafter that I tried out x*0= |x|= x if x>0 and -x if x<0.However, is it just a coincidence that it is x if x>0 and alsowe chose x*0 to be x-0 for x>0? I mean, it seems like either way( whether you use the |x| way or the (x-e) way, you get x as ananswer). My next question is that if we choose x<0 for the abovex*0=..., then doesn't e change? I am confused about this. I mean inone case we have |x|= -x if x<0, but if x<0, then e changes,doesn't it? Please just help me answer these questions and guide mestep by step through this problem. Thank you.

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