6. All squares are trapeziums.

7. If [tex]a = \frac{1}{2}, b = -\frac{2}{3},[/tex] and [tex]c = \frac{1}{4},[/tex] verify that [tex]a \times (b \times c) = (a \times b) \times c.[/tex]



Answer :

To verify the mathematical expression [tex]\( a \cdot (b \cdot c) = (a \cdot b) \cdot c \)[/tex] given the values [tex]\( a = \frac{1}{2} \)[/tex], [tex]\( b = -\frac{2}{3} \)[/tex], and [tex]\( c = \frac{1}{4} \)[/tex], we will follow these steps:

1. Calculate [tex]\( b \cdot c \)[/tex]:
[tex]\[ b \cdot c = \left( -\frac{2}{3} \right) \cdot \frac{1}{4} \][/tex]
Perform the multiplication:
[tex]\[ b \cdot c = -\frac{2}{3} \cdot \frac{1}{4} = -\frac{2 \times 1}{3 \times 4} = -\frac{2}{12} = -\frac{1}{6} \][/tex]

2. Calculate [tex]\( a \cdot (b \cdot c) \)[/tex]:
[tex]\[ a \cdot (b \cdot c) = \frac{1}{2} \cdot \left( -\frac{1}{6} \right) \][/tex]
Perform the multiplication:
[tex]\[ a \cdot (b \cdot c) = \frac{1}{2} \cdot \left( -\frac{1}{6} \right) = \frac{1 \times -1}{2 \times 6} = -\frac{1}{12} \][/tex]

3. Calculate [tex]\( a \cdot b \)[/tex]:
[tex]\[ a \cdot b = \frac{1}{2} \cdot \left( -\frac{2}{3} \right) \][/tex]
Perform the multiplication:
[tex]\[ a \cdot b = \frac{1}{2} \cdot \left( -\frac{2}{3} \right) = \frac{1 \times -2}{2 \times 3} = -\frac{2}{6} = -\frac{1}{3} \][/tex]

4. Calculate [tex]\((a \cdot b) \cdot c \)[/tex]:
[tex]\[ (a \cdot b) \cdot c = \left( -\frac{1}{3} \right) \cdot \frac{1}{4} \][/tex]
Perform the multiplication:
[tex]\[ (a \cdot b) \cdot c = \left( -\frac{1}{3} \right) \cdot \frac{1}{4} = \frac{-1 \times 1}{3 \times 4} = -\frac{1}{12} \][/tex]

5. Verify that [tex]\( a \cdot (b \cdot c) = (a \cdot b) \cdot c \)[/tex]:
[tex]\[ a \cdot (b \cdot c) = -\frac{1}{12} \][/tex]
[tex]\[ (a \cdot b) \cdot c = -\frac{1}{12} \][/tex]

Since both expressions evaluate to [tex]\(-\frac{1}{12}\)[/tex], the equality [tex]\( a \cdot (b \cdot c) = (a \cdot b) \cdot c \)[/tex] holds true. Therefore, we have verified that the given identity is valid for the provided values.

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