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The square root of 27 plus the square root of 5 = 7.4322204

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Q: What does the square root of 27 plus the square root of 5?

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Expressed as a surd in its simplest form, sqrt(27) + sqrt(12) = 5 sqrt(3). Expressed as a decimal, rounded to two decimal places, this is equal to 8.66.

Since 5^2 is 25 and 6^2 is 36, the square root of 27 must be in between 5 and 6, This is because 27 is in between 25 and 36.

It is 49*sqrt(2).

Square root of 5 = ± 2.236068Square root of 11 = ± 3.316625

b over the square root of 5 The square root of 5b squared is 5b, and the simplified form of the square root of 125 is 5 root 5. The 5s then cancel out leaving b over the square root of 5.

Related questions

5 Square root 3. square root 27 = square root 9*3 = 3square root 3 3square root3 + 2square root3 = 5Square Root3 because both have a square root 3.

15.46482452

182.25

√-48 + 35 + √ 25 + √-27 = √[(i2)(16)(3)] + 35 + 5 + √[(i2)(9)(3)] = 4i√3 + 38 + 9i√3 = 38 + (13√3)i

If you mean vertices of (-1, 4) (2, 7) and (1, 5) then it is an isosceles triangle with a perimeter of square root of 18 plus square root of 5 plus square root of 5 which is about 8.715 rounded to three decimal places.

Square root of 25= 5. 5 + 144 + 13 = 162.

5 - x^2 = (the square root of 5 minus x)(the square root of 5 plus x)

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15

Expressed as a surd in its simplest form, sqrt(27) + sqrt(12) = 5 sqrt(3). Expressed as a decimal, rounded to two decimal places, this is equal to 8.66.

24

sqrt(4) = 2 sqrt(9) = 3 Therefore, the square root of 4 plus the square root of 9 is equal to 2 + 3 = 5.