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harkovskaia [24]
4 years ago
15

What is the product of 1 1/2 and -1 1/4?

Mathematics
1 answer:
Savatey [412]4 years ago
6 0

positive number times negative number is negative number "(+)(-) = (-)".

1\dfrac{1}{2}\cdot\left(-1\dfrac{1}{4}\right)=-\left(1\dfrac{1}{2}\cdot1\dfrac{1}{4}\right)

we change the mixed numbers to the improper fractions

=-\left(\dfrac{1\cdot2+1}{2}\cdot\dfrac{1\cdot4+1}{4}\right)=-\left(\dfrac{3}{2}\cdot\dfrac{5}{4}\right)=-\dfrac{3\cdot5}{2\cdot4}=-\dfrac{15}{8}=-1\dfrac{7}{8}

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Alexander says that 3x + 4y is equivalent to (3)(4) + xy because of any order, any grouping. Is he correct why or why not
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Answer:

Alexander is incorrect because the expressions are not equivalent.

Step-by-step explanation:

If the expression is evaluated for any value of x, y; the result will not be same.

For instance, let assume x = 1 and y = 2

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(3)(4) + xy = (3)(4) + (1 * 2) = 12 + 2 = 14

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Answer:

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Step-by-step explanation:

A relation on a set A is called an equivalence relation if it is reflexive, symmetric, and transitive

We are going to analyze each one.

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Is an equivalence relation because it has all the properties.

b){ (0,0), (0,2), (2,0), (2,2), (2,3), (3,2), (3,3) }

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c){ (0,0), (1,1), (1,2), (2,1), (2,2), (3,3) }

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d){ (0,0), (1,1), (1,3), (2,2), (2,3), (3,1), (3,2) (3,3) }

Is not an equivalence relation. Not transitive: (1,3) and (3,2) are in the relation, but not (1,2)

e){ (0,0), (0,1) (0,2), (1,0), (1,1), (1,2), (2,0), (2,2), (3,3) }

Is not an equivalence relation. Not symmetric: (1,2) is present, but not (2,1)Not transitive: (2,0) and (0,1) are in the relation, but not (2,1)

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