Question

Simplify each expression using the proper order of operations. please help thank you so much :)


`(41 - 3^2)/(√(36)* 3 - 26 )`= ?

12 +
`{3} √(8)`*(9-2)= ?


`(28-(7^2+3))/(-13+3*5)`= ?


`(28)/(4)` -
`\sqrt[3]{8}`*
`(2)/(3)`= ?

7^2 - 5* 8+1= ?

(2*
`√(16)`) -(
`\sqrt[3]{27}`*
`√(81)`) + 7= ?

Answer 1

`(41 - 3^2)/(√(36) * 3 - 26) = -4`

`12 + 3√(8) * (9 - 2) = 12 + 42 √(2)`

`(28 - (7^2 + 3))/(-13 + 3 * 5) = -12`

`7^2 - 5* 8+1 = 10`

`(2 * √(16)) - (\sqrt[3]{27} * √(81)) + 1 = -18`

Explanation:

Required: Solve the expressions using proper operation order

To solve this, we'll make use of BODMAS

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`(41 - 3^2)/(√(36) * 3 - 26)`

Evaluate all squares and square roots

`(41 - 3*3)/(√(36) * 3 - 26)`

`(41 - 3*3)/(6 * 3 - 26)`

Evaluate the numerator (Start by multiplying 3 * 3)

`(41 - 9)/(6 * 3 - 26)`

Subtract 9 from 41

`(32)/(6 * 3 - 26)`

Evaluate the denominator (Start by multiplying 6 * 3)

`(32)/(18 - 26)`

`(32)/(-8)`

Divide 32 by -8

-4

Hence;

`(41 - 3^2)/(√(36) * 3 - 26) = -4`

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`12 + 3√(8) * (9 - 2)`

Start by evaluating the bracket

`12 + 3√(8) * 7`

Then evaluate the multiplication

`12 + 21√(8)`

Simplify the square root

`12 + 21√(4 * 2)`

Split the square root

`12 + 21√(4) * √(2)`

Take Square root of 4

`12 + 21 * 2} * √(2)`

`12 + 42 √(2)`

Hence;

`12 + 3√(8) * (9 - 2) = 12 + 42 √(2)`

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`(28 - (7^2 + 3))/(-13 + 3 * 5)`

Evaluate 7²

`(28 - (49 + 3))/(-13 + 3 * 5)`

Evaluate all expression in the bracket

`(28 - (52))/(-13 + 3 * 5)`

`(28 - 52)/(-13 + 3 * 5)`

Evaluate 3 * 5

`(28 - 52)/(-13 + 1 5)`

`(-2 4)/(2)`

`-12`

Hence;

`(28 - (7^2 + 3))/(-13 + 3 * 5) = -12`

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`7^2 - 5* 8+1`

Evaluate 7²

`49 - 5* 8+1`

Evaluate 5 * 8

`49 - 40 + 1`

`10`

`7^2 - 5* 8+1 = 10`

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`(2 * √(16)) - (\sqrt[3]{27} * √(81)) + 1`

Evaluate all square root and cube root

`(2 * 4) - (3 * 9) + 1`

Solve the expressions in bracket

`8 - 27 + 1`

`-18`

Hence;

`(2 * √(16)) - (\sqrt[3]{27} * √(81)) + 1 = -18`