Question

(^4√32)-(√49)-(^4√162)+(√64)

Answer 1

$\sqrt[4]{32}-\sqrt{49}-\sqrt[4]{162}+\sqrt{64}=\sqrt[4]{2^{4}*2}-7-\sqrt[4]{3^{4}*2}+8=2\sqrt[4]{2} +1-3\sqrt[4]{2}=1-\sqrt[4]{2}$

Answer 2

To simplify the expression, let's first evaluate each part individually:

1. (^4√32): This means the fourth root of 32. (^4√32) = 2, because 2^4 = 16, and 2^5 = 32, so the fourth root of 32 is 2.

2. (√49): This means the square root of 49. (√49) = 7, because 7^2 = 49.

3. (^4√162): This means the fourth root of 162. (^4√162) is approximately 3.14, rounded to two decimal places.

4. (√64): This means the square root of 64. (√64) = 8, because 8^2 = 64.

Now, let's put everything together:

(^4√32) - (√49) - (^4√162) + (√64) = 2 - 7 - 3.14 + 8

Now, perform the arithmetic:

= -5 - 3.14 + 8

= -0.14

Therefore, the result of the expression is approximately -0.14.