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... For any integer, the algorithm for finding the binary equivalent is given in the flow chart on the next page. Now let us look at the decimal part, that is, 0.1875. 1. Multiply 0.1875 by 2. This gives 0.375. The number before the decimal is 0 and the number after the decimal is 0.375. Since the numbe ...
... For any integer, the algorithm for finding the binary equivalent is given in the flow chart on the next page. Now let us look at the decimal part, that is, 0.1875. 1. Multiply 0.1875 by 2. This gives 0.375. The number before the decimal is 0 and the number after the decimal is 0.375. Since the numbe ...
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... For any integer, the algorithm for finding the binary equivalent is given in the flow chart on the next page. Now let us look at the decimal part, that is, 0.1875. 1. Multiply 0.1875 by 2. This gives 0.375. The number before the decimal is 0 and the number after the decimal is 0.375. Since the numbe ...
... For any integer, the algorithm for finding the binary equivalent is given in the flow chart on the next page. Now let us look at the decimal part, that is, 0.1875. 1. Multiply 0.1875 by 2. This gives 0.375. The number before the decimal is 0 and the number after the decimal is 0.375. Since the numbe ...
Example
... 1: Find a number that you can multiply by the bottom of the fraction to make it into either 10, 100 or 1,000. (Or any 1s followed by zeros.) 2: Multiply both top and bottom numbers by that amount. 3: Then write down just the top number, putting the decimal point in the right place (one place from th ...
... 1: Find a number that you can multiply by the bottom of the fraction to make it into either 10, 100 or 1,000. (Or any 1s followed by zeros.) 2: Multiply both top and bottom numbers by that amount. 3: Then write down just the top number, putting the decimal point in the right place (one place from th ...
Solving Absolute Value Inequalities
... In interval notation, this would be (−2, 6). Special Cases. If c = 0 or c < 0, the answer may be a single point, all real numbers, all real numbers except for a single point, or there may be no solutions. Some of these are special cases of the situations above. I’ll write down the possibilities for ...
... In interval notation, this would be (−2, 6). Special Cases. If c = 0 or c < 0, the answer may be a single point, all real numbers, all real numbers except for a single point, or there may be no solutions. Some of these are special cases of the situations above. I’ll write down the possibilities for ...
Mathematics Summer Session: Transition Math Chapter 1 Notes
... then add the numerator. Keep the same denominator. So, 2 ½ would become (2 x 2 + 1)/2 = 5/2. Lesson 1-8: Throw a negative sign in front of any of the positive whole numbers, and you have negative numbers. Unlike positive numbers, the closer a negative number is to zero, the bigger it is. So, -.0001 ...
... then add the numerator. Keep the same denominator. So, 2 ½ would become (2 x 2 + 1)/2 = 5/2. Lesson 1-8: Throw a negative sign in front of any of the positive whole numbers, and you have negative numbers. Unlike positive numbers, the closer a negative number is to zero, the bigger it is. So, -.0001 ...