On a test victor asked to find the square root of 0.83. Wich of the following is closest to the square root of 0.83?

I do not know what choices you have, but 0.9110433579144299 is the square root of 0.83

Step-by-step explanation:

Step 1:

Divide the number (0.83) by 2 to get the first guess for the square root.

First guess = 0.83/2 = 0.415.

Step 2:

Divide 0.83 by the previous result. d = 0.83/0.415 = 2.

Average this value (d) with that of step 1: (2 + 0.415) / 2 = 1.2075 (new guess).

Error = new guess - previous value = 0.415 - 1.2075 = 0.7925.

0.7925 > 0.001. As error > accuracy, we repeat this step again.

Step 3:

Divide 0.83 by the previous result. d = 0.83/1.2075 = 0.6873706004.

Average this value (d) with that of step 2: (0.6873706004 + 1.2075) / 2 = 0.9474353002 (new guess).

Error = new guess - previous value = 1.2075 - 0.9474353002 = 0.2600646998.

0.2600646998 > 0.001. As error > accuracy, we repeat this step again.

Step 4:

Divide 0.83 by the previous result. d = 0.83/0.9474353002 = 0.8760492667.

Average this value (d) with that of step 3: (0.8760492667 + 0.9474353002) / 2 = 0.9117422835 (new guess).

Error = new guess - previous value = 0.9474353002 - 0.9117422835 = 0.0356930167.

0.0356930167 > 0.001. As error > accuracy, we repeat this step again.

Step 5:

Divide 0.83 by the previous result. d = 0.83/0.9117422835 = 0.9103449681.

Average this value (d) with that of step 4: (0.9103449681 + 0.9117422835) / 2 = 0.9110436258 (new guess).

Error = new guess - previous value = 0.9117422835 - 0.9110436258 = 0.0006986577.

0.0006986577 < = 0.001. As error < = accuracy, we stop the iterations and use 0.9110436258 as the square root.

So, we can say that the square root of 0.83 is 0.911 with an error smaller than 0.001 (in fact the error is 0.0006986577). this means that the first 3 decimal places are correct. Just to compare, the returned value by using the jа vascript function 'Math. sqrt (0.83) ' is 0.9110433579144299.