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liberstina [14]
3 years ago
8

What is the best way to classify each equation? 22-3x+7x=4(x+5) ,6(2x-3=3(3-5), 6x-14=2(3x-7), 6x+3(x-4)=8(x-3)

Mathematics
1 answer:
Romashka-Z-Leto [24]3 years ago
8 0
1. 22 - 3x + 7x = 4(x + 5)
   22 - 3x + 7x  = 4x + 20
  <u>      +3x  +3x   +3x          </u>
         22 + 10x = 7x + 20
        <u>          -7x   -7x        </u>
           22 + 3x = 20
          <u>-22         -22</u>
                   <u>3x</u> = -<u>2</u>
                    3      3
                     x = -2/3

2. 6(2x - 3) = 3(3 - 5)
    12x - 18 = 3(-2)
    12x - 18 = -6
    <u>       +18   +18</u>
           <u>12x</u> = <u>12</u>
            12     12
               x = 1

3. 6x - 14 = 2(3x - 7)
    6x - 14 =  -3x
    6x - 14 = 2(-7)
    6x - 14 = -14
    <u>     +14   +14</u>
           <u>6x</u> = <u>0</u>
            6     6
             x = 0

4. 6x + 3(x - 4) = 8(x - 3)
    6x + 3x - 12 = 8x - 24
            9x - 12 = 8x - 24
           <u>-8x          -8x        </u>
              x - 12 = -24
              <u>   +12    +12</u>
                     x = -12
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Answer:

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Step-by-step explanation:

So we have the expression:

(x-y)^2+10wz

And we want to evaluate it when x=25, y=10, w=11, and z=18.

So, substitute these values for the variables:

=(25-10)^2+10(11)(18)

First, subtract within the parentheses:

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Square 15:

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Now, multiply the terms on the right. 10 times 11 is 110:

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Multiply:

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Finally, add:

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Answer:

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Step-by-step explanation:

(see attached for reference)

Given the number 1627187, we can see that the number in the tens place is the number 8.  

How we round this depends on the number immediately to the right of this number. (i.e the digit in the ones place)

Case 1: If the digit in the ones place is less less than 5, then the number in the tens place remains the same and replace all the digits to its right with zeros

Case 2: If the digit in the ones places is 5 or greater, then we increase the digit in the tens place and replace all the digits to its right with zeros.

In our case, the digit in the ones places is 7, this greater than 5, hence according to Case 2 above, we increase the digit in the tens place by one (from 8 to 9) and replace all the digits to its right by zeros giving us:

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Answer:

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Step-by-step explanation:

LET:

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