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nikitadnepr [17]
3 years ago
10

5-2+12÷4 use the order of operations

Mathematics
2 answers:
alisha [4.7K]3 years ago
6 0

Step-by-step explanation:

= 5- 2 + 12 /4

= 5 -2 + 3

= 8- 2

= 6

zhannawk [14.2K]3 years ago
6 0

Answer:

6

Explanation:

What you do is you take 12 divided by 4 and you get 3. The equation is now 5-2+3, you subtract 2 from 5 and get 3. Now you have 3 plus 3 which gets you 6.

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HELP IVE BEEN STRUGGLING FOR THE PAST 15 MIN PLEASEEEEE
joja [24]

<u>Answer:</u>

c=  -1/2

<u>Step-by-step explanation:</u>

Let's solve your equation step-by-step.

4 /3  = −6c − 5/ 3

Step 1: Simplify both sides of the equation.

4 /3  =−6c +  −5 /3

Step 2: Flip the equation.

−6c +  −5 /3  =  4/3

Step 3: Add 5/3 to both sides.

−6c +  −5 /3  +  5 /3  = 4/3  +  5 /3  −6c

=3

Step 4: Divide both sides by -6.

−6c  −6   =  3 − 6 c =  −1 /2

4 0
3 years ago
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A set of equations is shown above. Which method eliminates one of the variables? A) Multiply equation A by -1/3 and add the resu
Paraphin [41]

Answer:


Step-by-step explanation:

Multiplying Equation A by (1/3) and adding the result to Equation B will do the trick.  Let's actually solve the problem!

Equation A:  (5/3)x + 3y = 12

Equation B:    4x     -  3y = 8

                    ---------------------------

                     (5/3 + 12/3)x = 15        Note how this has eliminated the variable

                            (17/3)x = 15           y.

                                   x = (3/17)(15)

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3 years ago
Clint has to spend a minimum of $75 in order to use his coupon at the store
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Okay what’s the rest of the question ?
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According to DeMorgan 's theorem, the complement of W · X + Y · Z is W' + X' · Y' + Z'. Yet both functions are 1 for WXYZ = 1110
sergeinik [125]

Answer:

The parenthesis need to be kept intact while applying the DeMorgan's theorem on the original equation to find the compliment because otherwise it will introduce an error in the answer.

Step-by-step explanation:

According to DeMorgan's Theorem:

(W.X + Y.Z)'

(W.X)' . (Y.Z)'

(W'+X') . (Y' + Z')

Note that it is important to keep the parenthesis intact while applying the DeMorgan's theorem.

For the original function:

(W . X + Y . Z)'

= (1 . 1 + 1 . 0)

= (1 + 0) = 1

For the compliment:

(W' + X') . (Y' + Z')

=(1' + 1') . (1' + 0')

=(0 + 0) . (0 + 1)

=0 . 1 = 0

Both functions are not 1 for the same input if we solve while keeping the parenthesis intact because that allows us to solve the operation inside the parenthesis first and then move on to the operator outside it.

Without the parenthesis the compliment equation looks like this:

W' + X' . Y' + Z'

1' + 1' . 1' + 0'

0 + 0 . 0 + 1

Here, the 'AND' operation will be considered first before the 'OR', resulting in 1 as the final answer.

Therefore, it is important to keep the parenthesis intact while applying DeMorgan's Theorem on the original equation or else it would produce an erroneous result.

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3 years ago
Rafael wants to buy a Go-Cart for $2,500. He can get a loan from the bank at 5% interest for three years. By the end of the loan
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C step by step answer
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