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For instance, if the one solving the math word problem has a limited understanding of the language (English, Spanish, etc.) they are more likely to not understand what the problem is even asking. In Example 1 (above), if one does not comprehend the definition of the word "spent," they will misunderstand the entire purpose of the word problem.
If the divisor has a fractional part, one can restate the problem by moving the decimal to the right in both numbers until the divisor has no fraction, which can make the problem easier to solve (e.g., 10/2.5 = 100/25 = 4). Division can be calculated with an abacus. [14] Logarithm tables can be used to divide two numbers, by subtracting the two ...
(For example, two-fifths is the fraction 2 / 5 and two fifths is the same fraction understood as 2 instances of 1 / 5 .) Fractions should always be hyphenated when used as adjectives. Alternatively, a fraction may be described by reading it out as the numerator over the denominator, with the denominator expressed as a cardinal ...
Long division is the standard algorithm used for pen-and-paper division of multi-digit numbers expressed in decimal notation. It shifts gradually from the left to the right end of the dividend, subtracting the largest possible multiple of the divisor (at the digit level) at each stage; the multiples then become the digits of the quotient, and the final difference is then the remainder.
In computing, the modulo operation returns the remainder or signed remainder of a division, after one number is divided by another, called the modulus of the operation.. Given two positive numbers a and n, a modulo n (often abbreviated as a mod n) is the remainder of the Euclidean division of a by n, where a is the dividend and n is the divisor.
In the division of 43 by 5, we have: 43 = 8 × 5 + 3, so 3 is the least positive remainder. We also have that: 43 = 9 × 5 − 2, and −2 is the least absolute remainder. These definitions are also valid if d is negative, for example, in the division of 43 by −5, 43 = (−8) × (−5) + 3, and 3 is the least positive remainder, while,
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The solution to this problem could be solved by drawing one bar and dividing it into two parts, with the longer part as 70 and the shorter part as 30. By visualizing these two parts, students would simply solve the above word problem by adding both parts together to build a whole bar of 100.
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