QuestionSolve using the Order of Operations: 7) $(\frac{6}{3})^{3}+8-2^{2}$ 8) $[(12 \div 3)+2]^{4}$ 9) $(9 \cdot 3)-2^{2}+10$ 10) $-3+6-(-2 \cdot-4)$

Studdy Solution

STEP 1

Assumptions1. We are solving these problems using the order of operations, which is Parentheses, Exponents, Multiplication and Division (from left to right), Addition and Subtraction (from left to right). This is often abbreviated as PEMAS. . The operations inside the parentheses are performed first.
3. If there are multiple operations inside the parentheses, we follow the order of operations within the parentheses.

STEP 2

Let's start with the first problem. $\left(\frac{6}{}\right)^{}+8-2^{2}$ First, we perform the operation inside the parentheses.
$\left(\frac{6}{}\right) =2$

STEP 3

Now, we calculate the exponent.
$2^{3} =8$

STEP 4

Then, we calculate the other exponent.
$2^{2} =4$

STEP 5

Now, we perform the addition and subtraction from left to right.
$8 +8 -4 =12$ So, the solution to the first problem is12.

STEP 6

Now, let's move on to the second problem.
$[(12 \div3)+2]^{4}$ First, we perform the operation inside the parentheses.
$12 \div3 =4$

STEP 7

Then, we perform the addition inside the parentheses.
$4 +2 =6$

STEP 8

Finally, we calculate the exponent.
$6^{4} =1296$ So, the solution to the second problem is1296.

STEP 9

Now, let's solve the third problem.
$(9 \cdot3)-2^{2}+$ First, we perform the multiplication.
$9 \cdot3 =27$

STEP 10

Then, we calculate the exponent.
$2^{2} =4$

STEP 11

Finally, we perform the addition and subtraction from left to right.
$27 -4 +10 =33$ So, the solution to the third problem is33.

STEP 12

Now, let's solve the last problem.
$-+6-(-2 \cdot-4)$ First, we perform the multiplication.
$-2 \cdot -4 =8$

STEP 13

Then, we perform the addition and subtraction from left to right.
$-3 +6 -8 = -5$ So, the solution to the last problem is -5.

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QuestionSolve using the Order of Operations: 7) $(\frac{6}{3})^{3}+8-2^{2}$ 8) $[(12 \div 3)+2]^{4}$ 9) $(9 \cdot 3)-2^{2}+10$ 10) $-3+6-(-2 \cdot-4)$

Studdy Solution

STEP 1

STEP 2

Let's start with the first problem. $\left(\frac{6}{}\right)^{}+8-2^{2}$ First, we perform the operation inside the parentheses.
$\left(\frac{6}{}\right) =2$

STEP 3

Now, we calculate the exponent.
$2^{3} =8$

STEP 4

Then, we calculate the other exponent.
$2^{2} =4$

STEP 5

Now, we perform the addition and subtraction from left to right.
$8 +8 -4 =12$ So, the solution to the first problem is12.

STEP 6

Now, let's move on to the second problem.
$[(12 \div3)+2]^{4}$ First, we perform the operation inside the parentheses.
$12 \div3 =4$

STEP 7

Then, we perform the addition inside the parentheses.
$4 +2 =6$

STEP 8

Finally, we calculate the exponent.
$6^{4} =1296$ So, the solution to the second problem is1296.

STEP 9

Now, let's solve the third problem.
$(9 \cdot3)-2^{2}+$ First, we perform the multiplication.
$9 \cdot3 =27$

STEP 10

Then, we calculate the exponent.
$2^{2} =4$

STEP 11

Finally, we perform the addition and subtraction from left to right.
$27 -4 +10 =33$ So, the solution to the third problem is33.

STEP 12

Now, let's solve the last problem.
$-+6-(-2 \cdot-4)$ First, we perform the multiplication.
$-2 \cdot -4 =8$

STEP 13

Then, we perform the addition and subtraction from left to right.
$-3 +6 -8 = -5$ So, the solution to the last problem is -5.

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QuestionSolve using the Order of Operations: 7) (63)3+8−22(\frac{6}{3})^{3}+8-2^{2}(36​)3+8−22 8) [(12÷3)+2]4[(12 \div 3)+2]^{4}[(12÷3)+2]4 9) (9⋅3)−22+10(9 \cdot 3)-2^{2}+10(9⋅3)−22+10 10) −3+6−(−2⋅−4)-3+6-(-2 \cdot-4)−3+6−(−2⋅−4)

Studdy Solution

STEP 1

STEP 2

Let's start with the first problem.(6)+8−22\left(\frac{6}{}\right)^{}+8-2^{2}(6​)+8−22First, we perform the operation inside the parentheses.(6)=2\left(\frac{6}{}\right) =2(6​)=2

STEP 3

Now, we calculate the exponent.23=82^{3} =823=8

STEP 4

Then, we calculate the other exponent.22=42^{2} =422=4

STEP 5

Now, we perform the addition and subtraction from left to right.8+8−4=128 +8 -4 =128+8−4=12So, the solution to the first problem is12.

STEP 6

Now, let's move on to the second problem.[(12÷3)+2]4[(12 \div3)+2]^{4}[(12÷3)+2]4First, we perform the operation inside the parentheses.12÷3=412 \div3 =412÷3=4

STEP 7

Then, we perform the addition inside the parentheses.4+2=64 +2 =64+2=6

STEP 8

Finally, we calculate the exponent.64=12966^{4} =129664=1296So, the solution to the second problem is1296.

STEP 9

Now, let's solve the third problem.(9⋅3)−22+(9 \cdot3)-2^{2}+(9⋅3)−22+First, we perform the multiplication.9⋅3=279 \cdot3 =279⋅3=27

STEP 10

Then, we calculate the exponent.22=42^{2} =422=4

STEP 11

Finally, we perform the addition and subtraction from left to right.27−4+10=3327 -4 +10 =3327−4+10=33So, the solution to the third problem is33.

STEP 12

Now, let's solve the last problem.−+6−(−2⋅−4)-+6-(-2 \cdot-4)−+6−(−2⋅−4)First, we perform the multiplication.−2⋅−4=8-2 \cdot -4 =8−2⋅−4=8

STEP 13

Then, we perform the addition and subtraction from left to right.−3+6−8=−5-3 +6 -8 = -5−3+6−8=−5So, the solution to the last problem is -5.

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QuestionSolve using the Order of Operations: 7) $(\frac{6}{3})^{3}+8-2^{2}$ 8) $[(12 \div 3)+2]^{4}$ 9) $(9 \cdot 3)-2^{2}+10$ 10) $-3+6-(-2 \cdot-4)$

Let's start with the first problem. $\left(\frac{6}{}\right)^{}+8-2^{2}$ First, we perform the operation inside the parentheses.
$\left(\frac{6}{}\right) =2$

Now, we calculate the exponent.
$2^{3} =8$

Then, we calculate the other exponent.
$2^{2} =4$

Now, we perform the addition and subtraction from left to right.
$8 +8 -4 =12$ So, the solution to the first problem is12.

Now, let's move on to the second problem.
$[(12 \div3)+2]^{4}$ First, we perform the operation inside the parentheses.
$12 \div3 =4$

Then, we perform the addition inside the parentheses.
$4 +2 =6$

Finally, we calculate the exponent.
$6^{4} =1296$ So, the solution to the second problem is1296.

Now, let's solve the third problem.
$(9 \cdot3)-2^{2}+$ First, we perform the multiplication.
$9 \cdot3 =27$

Then, we calculate the exponent.
$2^{2} =4$

Finally, we perform the addition and subtraction from left to right.
$27 -4 +10 =33$ So, the solution to the third problem is33.

Now, let's solve the last problem.
$-+6-(-2 \cdot-4)$ First, we perform the multiplication.
$-2 \cdot -4 =8$

Then, we perform the addition and subtraction from left to right.
$-3 +6 -8 = -5$ So, the solution to the last problem is -5.