What is (2^2+2)/(2^2) ?

The solution to (2^2+2)/(2^2) is 1 1/2

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(2^2+2)/(2^2)

\frac{2 ^{2} + 2}{2 ^{2}}

1 \frac{1}{2}

Solution steps

\frac{2 ^{2} + 2}{2 ^{2}}

Solve

2^{2} + 2 : 6

Follow the PEMDAS order of operations

Calculate exponents

2^{2} : 4

2^{2}

2^{2} = 4

= 4

= 4 + 2

Add and subtract (left to right)

4 + 2 : 6

4 + 2

4 + 2 = 6

= 6

= 6

Solve

2^{2} : 4

2^{2} = 4

= 4

= \frac{6}{4}

Simplify

\frac{6}{4} : 1 \frac{1}{2}

Cancel the common factor:

2

= \frac{3}{2}

Convert improper fractions to mixed numbers:

\frac{3}{2} = 1 \frac{1}{2}

\frac{3}{2} = 1

Remainder

1

\frac{3}{2}

Write the problem in long division format

2 \overline{∣ 3}

Divide

3

2

to get

1

Divide

3

2

to get

1

1 2 \overline{∣ 3}

Multiply the quotient digit

(1)

by the divisor

2

1 2 \overline{∣ 3} \underline{2}

Subtract

2

from

3

1 2 \overline{∣ 3} \underline{2} 1

1 2 \overline{∣ 3} \underline{2} 1

The solution for Long Division of

\frac{3}{2}

1

with remainder of

1

1 Remainder 1

Convert to mixed number: Quotient

\frac{Remainder}{Divisor}

\frac{3}{2} = 1 \frac{1}{2}

= 1 \frac{1}{2}

= 1 \frac{1}{2}

= 1 \frac{1}{2}

- 9 \div 3 - [(8 - 10) - (9 - 2)]

1 \times 2^{5}

\frac{5}{7} + 9 - \frac{5}{3} - \frac{9}{4}

- 4 \cdot (- 4) \cdot (- 4) \cdot (- 4)

- 1 - (- 1)^{2}