The product of two rational numbers is always rational because (ac/bd) is the ratio of two integers, making it a rational number.
We need to prove that the product of two rational numbers is always rational. A rational number is a number that can be stated as the quotient or fraction of two integers : a numerator and a non-zero denominator.
Let us consider two rational numbers, a/b and c/d. The variables "a", "b", "c", and "d" all represent integers. The denominators "b" and "d" are non-zero. Let the product of these two rational numbers be represented by "P".
P = (a/b)×(c/d)
P = (a×c)/(b×d)
The numerator is again an integer. The denominator is also a non-zero integer. Hence, the product is a rational number.
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Answer:
E , G, B
Step-by-step explanation:
I did it
Answer: The test contains 10 three-point questions and 14 five-point questions.
Step-by-step explanation:
Let x represent the number of 3-point questions.
Let y represent the number of 5-point questions.
The maths test consists of 24 questions. This means that
x + y = 24
Each question is worth either 3 points or 5 and the test is worth 100 points. This means that
3x + 5y = 100 - - - - - - - - - - -1
Substituting x = 24 - y into equation 1, it becomes
3(24 - y) + 5y = 100
72 - 3y + 5y = 100
- 3y + 5y = 100 - 72
2y = 28
y = 28/2 = 14
Substituting y = 14 into x = 24 - y , it becomes
x = 24 - 14 = 10
Answer:
B. {16, 19, 20}
Step-by-step explanation:
The <em>triangle inequality</em> requires for any sides a, b, c you must have ...
a + b > c
b + c > a
c + a > b
The net result of those requirements are ...
- the sum of the two shortest sides must be greater than the longest side
- the length of the third side lies between the difference and sum of the other two sides
__
If we look at the offered side length choices, we see ...
A: 8+11 = 19 . . . not > 19; not a triangle
B: 16+19 = 35 > 20; could be a triangle
C: 3+4 = 7 . . . not > 8; not a triangle
D: 5+5 = 10 . . . not > 11; not a triangle
The side lengths {16, 19, 20} could represent the sides of a triangle.
_____
<em>Additional comment</em>
The version of triangle inequality shown above ensures that a triangle will have non-zero area.
The alternative version of the triangle inequality uses ≥ instead of >. Triangles where a+b=c will look like a line segment--they will have zero area. Many authors disallow this case. (If it were allowed, then {8, 11, 19} would also be a "triangle.")
