High School

11. Simplify: [tex]3 + \frac{6}{2} - 4[/tex]

12. Simplify: [tex]45 - \left[38 - \left\{ \frac{60}{3} - \left(6 - \frac{9}{3}\right) \div 3 \right\} \right][/tex]

13. Simplify: [tex]27 - \left[38 - \left\{46 - \left(15 - \sqrt{3 - 2}\right)\right\} \right][/tex]

14. Simplify: [tex]25 - \frac{1}{2} \left\{ 5 + 4 - (3 + 2 - \overline{1 + 3}) \right\}[/tex]

15. Simplify: [tex]-25 + \frac{14}{(5 - 3)}[/tex]

Answer :

Below is a detailed solution for each expression.

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1. Evaluate
[tex]$$3 + 6 \div 2 - 4.$$[/tex]

Step 1. Perform the division:
[tex]$$6 \div 2 = 3.$$[/tex]

Step 2. Replace the division in the expression:
[tex]$$3 + 3 - 4.$$[/tex]

Step 3. Perform the addition:
[tex]$$3 + 3 = 6.$$[/tex]

Step 4. Finally, subtract:
[tex]$$6 - 4 = 2.$$[/tex]

Thus, the final result is [tex]$$2.$

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2. Evaluate
$$[/tex]45 - \Bigl[38 - \Bigl\{60 \div 3 - \Bigl(6 - 9 \div 3\Bigr) \div 3\Bigr\}\Bigr].[tex]$$

Step 1. Compute the innermost division:
$$[/tex]9 \div 3 = 3.[tex]$$

Step 2. Replace in the parentheses:
$$[/tex]6 - 3 = 3.[tex]$$

Step 3. Divide the result by 3:
$$[/tex]3 \div 3 = 1.[tex]$$

Step 4. Compute the other division:
$$[/tex]60 \div 3 = 20.[tex]$$

Step 5. Inside the curly braces, subtract:
$$[/tex]20 - 1 = 19.[tex]$$

Step 6. Then, inside the square brackets, subtract:
$$[/tex]38 - 19 = 19.[tex]$$

Step 7. Finally, subtract from 45:
$$[/tex]45 - 19 = 26.[tex]$$

Thus, the final result is $$[/tex]26.[tex]$

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3. Evaluate
$[/tex][tex]$27 - \Bigl[38 - \Bigl\{46 - \Bigl(15 - \sqrt{3-2}\Bigr)\Bigr\}\Bigr].$[/tex][tex]$

Step 1. Evaluate the square root:
$[/tex][tex]$\sqrt{3-2} = \sqrt{1} = 1.$[/tex][tex]$

Step 2. Inside the innermost parentheses:
$[/tex][tex]$15 - 1 = 14.$[/tex][tex]$

Step 3. Subtract this result from 46:
$[/tex][tex]$46 - 14 = 32.$[/tex][tex]$

Step 4. Inside the brackets, compute:
$[/tex][tex]$38 - 32 = 6.$[/tex][tex]$

Step 5. Finally, subtract from 27:
$[/tex][tex]$27 - 6 = 21.$[/tex][tex]$

Thus, the final result is $[/tex][tex]$21.$[/tex]

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4. Evaluate
[tex]$$25 - \frac{1}{2}\Bigl\{5+4 - \Bigl(3+2-\overline{1+3}\Bigr)\Bigr\}.$$[/tex]

Note: The notation [tex]$\overline{1+3}$[/tex] is treated as the expression [tex]$(1+3)$[/tex].

Step 1. Calculate the overline:
[tex]$$1+3 = 4.$$[/tex]

Step 2. Inside the inner parentheses compute:
[tex]$$3+2 = 5,$$[/tex]
and then subtract the result of the overline:
[tex]$$5 - 4 = 1.$$[/tex]

Step 3. Now, add the numbers inside the braces:
[tex]$$5+4 = 9.$$[/tex]

Step 4. Subtract the result from Step 2:
[tex]$$9 - 1 = 8.$$[/tex]

Step 5. Multiply by [tex]$\frac{1}{2}$[/tex]:
[tex]$$\frac{1}{2} \times 8 = 4.$$[/tex]

Step 6. Finally, subtract from 25:
[tex]$$25 - 4 = 21.$$[/tex]

Thus, the final result is [tex]$$21.$

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5. Evaluate
$$[/tex]-25 + 14 \div (5-3).[tex]$$

Step 1. Compute the expression in parentheses:
$$[/tex]5-3 = 2.[tex]$$

Step 2. Divide 14 by the result:
$$[/tex]14 \div 2 = 7.[tex]$$

Step 3. Finally, add to $$[/tex]-25[tex]$$:
$$[/tex]-25 + 7 = -18.[tex]$$

Thus, the final result is $$[/tex]-18.[tex]$

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Final Answers:

11. $[/tex][tex]$2$[/tex][tex]$
12. $[/tex][tex]$26$[/tex][tex]$
13. $[/tex][tex]$21$[/tex][tex]$
14. $[/tex][tex]$21$[/tex][tex]$
5. $[/tex][tex]$-18$[/tex]$