Page 70 - Calculating Agriculture Cover 20191124 STUDENT - A

5-8 Measurements CH 5]

2) Beginning at the decimal point, separate the numbers into groups of two
figures each, called periods. If there are any numbers to the right of the
decimal, group them into periods also. If zeros are needed after the decimal,
add them so as to make groups of two figures. A bar may be placed over the
periods.
. .
√ 2 61. = √ 2 61 . 00 00 00

3) Determine the largest perfect square contained in the left-hand period, 2. The
largest perfect square that can be subtracted from 2 is 1. Write 1 under the 2,
and write 1 above the first period on the left. This is the first figure of the
answer.
1 .
√ 2 61 . 00 00 00
1

4) Subtract the largest perfect square (1) from the first period (2). Bring down the
next period (61) to the remainder to form the trial dividend.

1 .
√ 2 61 . 00 00 00
1
1 61

5) Multiply the first number of the answer (1) by 20 to obtain the trial divisor
(20). Write this number as indicated.

1 .
√ 2 61 . 00 00 00
1 x 1 = 1
20 x 1 = 20 1 61

6) Divide the trial divisor 20 into the remainder 161. We determine that 161
divided by 20 = 8.05. Adding 8 to 20 and the result trial divisor is 28; 28 x 8 =
224 which is larger than 161, so 8 is not used. Try 7 added to 20 and the
summation is 27 which is considered as a trial divisor; 27 x 7 = 189, which is
still larger than 161 and thus that number is discarded. Now let’s use 6;
adding 6 to 20 yields a trial divisor of 26 and 6 x 26 = 156 which is less than
161. Now we have the second digit in our calculation, 6. The product of 6 x 26
= 156 which is subtracted from the value 161 to yield a remainder of 5.
6
7
1 8 .
√ 2 61 . 00 00 00
1 x 1 = 1
20 x 1 = 20 1 61
8 x 28 = 224 too large
7 x 27 = 189 too large
20 + 6 = 26 x 6 = 1 56
20 x 16 = 320 5

Q: How is an artificial Christmas tree like the fourth root of — 68?
A: Neither has real roots.

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