Question

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

Answer 1

Answer:

$\frac{41 - 3^2}{\sqrt{36} * 3 - 26} = -4$

$12 + 3\sqrt{8} * (9 - 2) = 12 + 42 \sqrt{2}$

$\frac{28 - (7^2 + 3)}{-13 + 3 * 5} = -12$

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

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

Step-by-step explanation:

Required: Solve the expressions using proper operation order

To solve this, we'll make use of BODMAS

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$\frac{41 - 3^2}{\sqrt{36} * 3 - 26}$

Evaluate all squares and square roots

$\frac{41 - 3*3}{\sqrt{36} * 3 - 26}$

$\frac{41 - 3*3}{6 * 3 - 26}$

Evaluate the numerator (Start by multiplying 3 * 3)

$\frac{41 - 9}{6 * 3 - 26}$

Subtract 9 from 41

$\frac{32}{6 * 3 - 26}$

Evaluate the denominator (Start by multiplying 6 * 3)

$\frac{32}{18 - 26}$

$\frac{32}{-8}$

Divide 32 by -8

-4

Hence;

$\frac{41 - 3^2}{\sqrt{36} * 3 - 26} = -4$

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$12 + 3\sqrt{8} * (9 - 2)$

Start by evaluating the bracket

$12 + 3\sqrt{8} * 7$

Then evaluate the multiplication

$12 + 21\sqrt{8}$

Simplify the square root

$12 + 21\sqrt{4 * 2}$

Split the square root

$12 + 21\sqrt{4} * \sqrt{2}$

Take Square root of 4

$12 + 21 * 2} * \sqrt{2}$

$12 + 42 \sqrt{2}$

Hence;

$12 + 3\sqrt{8} * (9 - 2) = 12 + 42 \sqrt{2}$

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$\frac{28 - (7^2 + 3)}{-13 + 3 * 5}$

Evaluate 7²

$\frac{28 - (49 + 3)}{-13 + 3 * 5}$

Evaluate all expression in the bracket

$\frac{28 - (52)}{-13 + 3 * 5}$

$\frac{28 - 52}{-13 + 3 * 5}$

Evaluate 3 * 5

$\frac{28 - 52}{-13 + 1 5}$

$\frac{-2 4}{2}$

$-12$

Hence;

$\frac{28 - (7^2 + 3)}{-13 + 3 * 5} = -12$

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

Evaluate all square root and cube root

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

Solve the expressions in bracket

$8 - 27 + 1$

$-18$

Hence;

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