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6. Simplify:
[tex]\[ 173 \% \times (6 + 8 \times 3 - 2) + 11957125 = [37 + \frac{8}{14}] \][/tex]

7. Using the rule of BODMAS, determine the answer to:
[tex]\[ 18 + \left( 10 - 4 + 3 \times 28 \div (4 + 10) \right) \times 2 - 11 \][/tex]


Sagot :

Let's solve these problems step-by-step using the rules of BODMAS/BIDMAS (Brackets, Orders, Division and Multiplication, Addition and Subtraction).

### Problem 6:
Simplify [tex]\(173 \% \times (6 + 8 \times 3 - 2) + 11957125\)[/tex]

Here is the step-by-step simplification:

1. Evaluate the expression inside the parentheses first:
[tex]\[ 6 + 8 \times 3 - 2 \][/tex]
2. According to BODMAS, we first perform the multiplication:
[tex]\[ 8 \times 3 = 24 \][/tex]
3. Now substitute this back into the expression:
[tex]\[ 6 + 24 - 2 \][/tex]
4. Perform the addition and subtraction from left to right:
[tex]\[ 6 + 24 = 30 \][/tex]
[tex]\[ 30 - 2 = 28 \][/tex]
5. Now evaluate [tex]\(173\% \times 28\)[/tex]. Here, [tex]\(173\%\)[/tex] means [tex]\( \frac{173}{100} \)[/tex]:
[tex]\[ \frac{173}{100} \times 28 = 1.73 \times 28 = 48.44 \][/tex]
6. Finally, add this result to 11957125:
[tex]\[ 48.44 + 11957125 = 11957173.44 \][/tex]

So, the simplified result is [tex]\(11957173.44\)[/tex].

### Problem 7:
Determine the answer to [tex]\(18 \div (10 - 4 + 328(4 + 10)) \div 2 - 11\)[/tex]

Here's the step-by-step simplification:

1. Evaluate the expression inside the brackets first:
[tex]\[ 4 + 10 \][/tex]
[tex]\[ 4 + 10 = 14 \][/tex]
2. Now substitute this back into the expression:
[tex]\[ 328 \times 14 \][/tex]
3. Perform the multiplication:
[tex]\[ 328 \times 14 = 4592 \][/tex]
4. Substitute this value back into the original expression:
[tex]\[ 10 - 4 + 4592 \][/tex]
5. Perform the addition and subtraction from left to right:
[tex]\[ 10 - 4 = 6 \][/tex]
[tex]\[ 6 + 4592 = 4598 \][/tex]
6. Now, perform the division:
[tex]\[ 4598 \div 2 = 2296 \][/tex]
7. Finally, subtract 11 from this result:
[tex]\[ 2296 - 11 = 2285 \][/tex]

So, the result is [tex]\(2285\)[/tex].

The detailed solutions for both problems are:
1. [tex]\(11957173.44\)[/tex]
2. [tex]\(2285\)[/tex]