QuestionCalculate the value of $\frac{22+1^{3}+(3^{4}-7^{2})}{2^{3}}$ .

Studdy Solution

STEP 1

Assumptions1. The operations must be performed in the correct order, following the BIDMAS/BODMAS rule (Brackets, Indices/Orders, Division and Multiplication (from left to right), Addition and Subtraction (from left to right)).

STEP 2

First, we need to solve the operations inside the brackets. $^{4}-7^{2}$

STEP 3

Calculate the value of $3^{}$ and $7^{2}$ .
$3^{} =81$ $7^{2} =49$

STEP 4

Subtract the value of $7^{2}$ from $3^{4}$ .
$81 -49$

STEP 5

Calculate the value of $81 -49$ .
$81 -49 =32$

STEP 6

Now, replace the value of $3^{4}-^{2}$ in the original equation with32.
$\frac{22+1^{3}+32}{2^{3}}$

STEP 7

Calculate the value of $1^{3}$ .
$1^{3} =1$

STEP 8

Replace the value of $1^{3}$ in the equation with1.
$\frac{22+1+32}{2^{3}}$

STEP 9

Add the values in the numerator.
$22++32 =55$

STEP 10

Replace the sum of the values in the numerator in the equation.
$\frac{55}{2^{3}}$

STEP 11

Calculate the value of $^{3}$ .
$^{3} =8$

STEP 12

Replace the value of $2^{}$ in the equation with8.
$\frac{55}{8}$

STEP 13

Now, divide55 by8 to get the final answer.
$\frac{55}{8} =6.875$ So, the solution to the problem $\frac{22+^{3}+\left(3^{}-7^{2}\right)}{2^{3}}$ is $6.875$ .

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QuestionCalculate the value of $\frac{22+1^{3}+(3^{4}-7^{2})}{2^{3}}$ .

Studdy Solution

STEP 1

Assumptions1. The operations must be performed in the correct order, following the BIDMAS/BODMAS rule (Brackets, Indices/Orders, Division and Multiplication (from left to right), Addition and Subtraction (from left to right)).

STEP 2

First, we need to solve the operations inside the brackets. $^{4}-7^{2}$

STEP 3

Calculate the value of $3^{}$ and $7^{2}$ .
$3^{} =81$ $7^{2} =49$

STEP 4

Subtract the value of $7^{2}$ from $3^{4}$ .
$81 -49$

STEP 5

Calculate the value of $81 -49$ .
$81 -49 =32$

STEP 6

Now, replace the value of $3^{4}-^{2}$ in the original equation with32.
$\frac{22+1^{3}+32}{2^{3}}$

STEP 7

Calculate the value of $1^{3}$ .
$1^{3} =1$

STEP 8

Replace the value of $1^{3}$ in the equation with1.
$\frac{22+1+32}{2^{3}}$

STEP 9

Add the values in the numerator.
$22++32 =55$

STEP 10

Replace the sum of the values in the numerator in the equation.
$\frac{55}{2^{3}}$

STEP 11

Calculate the value of $^{3}$ .
$^{3} =8$

STEP 12

Replace the value of $2^{}$ in the equation with8.
$\frac{55}{8}$

STEP 13

Now, divide55 by8 to get the final answer.
$\frac{55}{8} =6.875$ So, the solution to the problem $\frac{22+^{3}+\left(3^{}-7^{2}\right)}{2^{3}}$ is $6.875$ .

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QuestionCalculate the value of 22+13+(34−72)23\frac{22+1^{3}+(3^{4}-7^{2})}{2^{3}}2322+13+(34−72)​.

Studdy Solution

STEP 1

Assumptions1. The operations must be performed in the correct order, following the BIDMAS/BODMAS rule (Brackets, Indices/Orders, Division and Multiplication (from left to right), Addition and Subtraction (from left to right)).

STEP 2

First, we need to solve the operations inside the brackets.4−72^{4}-7^{2}4−72

STEP 3

Calculate the value of 33^{}3 and 727^{2}72.3=813^{} =813=8172=497^{2} =4972=49

STEP 4

Subtract the value of 727^{2}72 from 343^{4}34.81−4981 -4981−49

STEP 5

Calculate the value of 81−4981 -4981−49.81−49=3281 -49 =3281−49=32

STEP 6

Now, replace the value of 34−23^{4}-^{2}34−2 in the original equation with32.22+13+3223\frac{22+1^{3}+32}{2^{3}}2322+13+32​

STEP 7

Calculate the value of 131^{3}13.13=11^{3} =113=1

STEP 8

Replace the value of 131^{3}13 in the equation with1.22+1+3223\frac{22+1+32}{2^{3}}2322+1+32​

STEP 9

Add the values in the numerator.22++32=5522++32 =5522++32=55

STEP 10

Replace the sum of the values in the numerator in the equation.5523\frac{55}{2^{3}}2355​

STEP 11

Calculate the value of 3^{3}3.3=8^{3} =83=8

STEP 12

Replace the value of 22^{}2 in the equation with8.558\frac{55}{8}855​

STEP 13

Now, divide55 by8 to get the final answer.558=6.875\frac{55}{8} =6.875855​=6.875So, the solution to the problem 22+3+(3−72)23\frac{22+^{3}+\left(3^{}-7^{2}\right)}{2^{3}}2322+3+(3−72)​ is 6.8756.8756.875.

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QuestionCalculate the value of $\frac{22+1^{3}+(3^{4}-7^{2})}{2^{3}}$ .

First, we need to solve the operations inside the brackets. $^{4}-7^{2}$

Calculate the value of $3^{}$ and $7^{2}$ .
$3^{} =81$ $7^{2} =49$

Subtract the value of $7^{2}$ from $3^{4}$ .
$81 -49$

Calculate the value of $81 -49$ .
$81 -49 =32$

Now, replace the value of $3^{4}-^{2}$ in the original equation with32.
$\frac{22+1^{3}+32}{2^{3}}$

Calculate the value of $1^{3}$ .
$1^{3} =1$

Replace the value of $1^{3}$ in the equation with1.
$\frac{22+1+32}{2^{3}}$

Add the values in the numerator.
$22++32 =55$

Replace the sum of the values in the numerator in the equation.
$\frac{55}{2^{3}}$

Calculate the value of $^{3}$ .
$^{3} =8$

Replace the value of $2^{}$ in the equation with8.
$\frac{55}{8}$

Now, divide55 by8 to get the final answer.
$\frac{55}{8} =6.875$ So, the solution to the problem $\frac{22+^{3}+\left(3^{}-7^{2}\right)}{2^{3}}$ is $6.875$ .