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svlad2 [7]
3 years ago
13

Help mee pls :((

t=" {7}^{th} grade" align="absmiddle" class="latex-formula">
​

Mathematics
1 answer:
blsea [12.9K]3 years ago
5 0

Answer:

2. 3-(-2)-(+5)

⇒3-(-2)-(+5)= 3+2-(+5) = 5-(+5)  =  5-5  = <u>0</u>

3. -(-3-6)-(-4+8)

⇒ -(-9)-(-4+8) = 9-(-4+8)  =  9-4 =5

4. -8-(-3-2)

⇒ -8-5=-13

5. -(-(-5))+7

⇒ -5+7 = 2

6. -(-(-8)+3)

⇒ -(11) = -11

7. -(-6-(-3+1))

⇒6-(-3+1) = 8

8. 1-(14-(11-7))

⇒ 1-(14-4) = 1-10 = -9

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Answer:

7, 12, 13

Step-by-step explanation:

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So there are two separate problems: 9 * (11-4) and 3 * (11-4). To find the value for each expression, use order of operations. We'll solve what's in the parenthesis before multiplying that value by the other number. 11-4 is 7. 7 * 9 is 63. 11-4 is 7. 7 * 3 is 21. You can divide 63 by 21 to get 3. This means that 9 * (11 - 4) is three times as much as 3 *(11 - 4). Even just by looking at the problem, the expression in the parenthesis is the same thing, but the front numbers are different. 9 is three times as much as 3.  The answer is A.

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