3-3-2-27 -7-3 is equal to

The correct option is C 34√(3)/3 3√(3)+2√(27) +7/√(3) = #160;3√( 3 ) +2√( 3× 3× 3 ) + 7 √( 3 ) #160; =3√( 3 ) +6√( 3 ) + 7 √( 3 ) #16 ...

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Byju's Answer

Standard IX

Mathematics

Expansion of (a ± b ± c)²

3√3+2√27 +7/√...

Question

$3 \sqrt{3} + 2 \sqrt{27} + \frac{7}{\sqrt{3}}$ is equal to

\frac{34}{3}

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\frac{1}{3}

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\frac{34 \sqrt{3}}{3}

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\frac{\sqrt{3}}{3}

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Solution

The correct option is C $\frac{34 \sqrt{3}}{3}$
$3 \sqrt{3} + 2 \sqrt{27} + \frac{7}{\sqrt{3}}$
$= 3 \sqrt{3} + 2 \sqrt{3 \times 3 \times 3} + \frac{7}{\sqrt{3}}$
$= 3 \sqrt{3} + 6 \sqrt{3} + \frac{7}{\sqrt{3}}$
$= 9 \sqrt{3} + \frac{7}{\sqrt{3}}$
$= \frac{27 + 7}{\sqrt{3}}$
$= \frac{34}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}$
$= \frac{34 \sqrt{3}}{3}$

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Similar questions

Simplify and express the result in power notation with positive exponent.

(i) (ii)

(iii) (iv)

(v)

Q. If

1 \cdot 3 + 3 \cdot 3^{2} + 5 \cdot 3^{3} + 7 \cdot 3^{4} + \dots

up to

n

terms is equal to

3 + (n - 1) 3^{t}

, then

t

Q. Solve :

(\frac{1}{7} \times \frac{15}{34}) \times

_________=

(\frac{1}{7} \times \frac{2}{3}) \times \frac{15}{34}

Q. Solve:

45 + 3 {34 - 18 - 14} \div 3 {17 + 3 \times 4 - (2 \times 7)]

Given a = 6 + √3
⇒ (1/a) = 1 / (6 + √3)
Multiply and divide by (6 − √3)
⇒ (1/a) = (6− √3) / [(6 + √3)(6 − √3)]
⇒ (1/a) = (6 − √3) / [36 − 3]
⇒ (1/a) = (6− √3)/33
∴ a - (1/a) = (6 + √3) -(6 − √3)/33
= [33(6 + √3) - (6 - √3)]/33
= [32*6 + 34*√3]/33
= [192 + 34√3]/33

Answer : [192 + 34√3]/33

Why does 1/a come above

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3√3+2√27+7√3 is equal to

The correct option is C 34√333√3+2√27+7√3=3√3+2√3×3×3+7√3=3√3+6√3+7√3=9√3+7√3=27+7√3=34√3×√3√3=34√33

$3 \sqrt{3} + 2 \sqrt{27} + \frac{7}{\sqrt{3}}$ is equal to