Question

1. 6-1 9/12*3+3/6*7/21= 2.(3/4)³=

Answer 1

1) = 6-19/4+1/6=72/12 - 147/12 + 2/12=- 73 /12 = - 6 1/12

2) = 27/ 64

Answer 2

Let's solve these two mathematical expressions step by step:

1. $6 - \frac{1}{12} \cdot 3 + \frac{3}{6} \cdot \frac{7}{21}$

First, let's simplify each part:

$6$ is already simplified.

$\frac{1}{12} \cdot 3 = \frac{3}{12} = \frac{1}{4}$

$\frac{3}{6} = \frac{1}{2}$

$\frac{7}{21} = \frac{1}{3}$

Now, we can substitute these simplified values back into the original expression:

$6 - \frac{1}{4} + \frac{1}{2} \cdot \frac{1}{3}$

Next, let's calculate the products:

$\frac{1}{2} \cdot \frac{1}{3} = \frac{1}{6}$

Now, we can rewrite the expression:

$6 - \frac{1}{4} + \frac{1}{6}$

To add fractions with different denominators, we need to find a common denominator. In this case, the common denominator is 12:

$6 \cdot \frac{12}{12} - \frac{3}{4} \cdot \frac{3}{3} + \frac{2}{12}$

Now, we can add the fractions:

$\frac{72}{12} - \frac{9}{12} + \frac{2}{12}$

Now, subtract the fractions:

$\frac{72 - 9 + 2}{12} = \frac{65}{12}$

So, the result of the first expression is $\frac{65}{12}$.

2. $(\frac{3}{4})^3$

To calculate the cube of a fraction, simply raise both the numerator and denominator to the power of 3:

$(\frac{3}{4})^3 = \frac{3^3}{4^3} = \frac{27}{64}$

So, the result of the second expression is $\frac{27}{64}$.