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Fittoniya [83]
3 years ago
7

Will the square roots of perfect squares always be irrational numbers

Mathematics
2 answers:
pshichka [43]3 years ago
8 0

Answer:

no

Step-by-step explanation:

because it's not amig9

Reil [10]3 years ago
3 0
Nope

hopefully this helped !!
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Select the correct answer. Which expression is equivalent to 8x^2^3 sqrt 375x + 2^3 sqrt 3x^7, if x=0?
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Answer:

(8x^3)^ \frac{2}{3} \sqrt{75x^3}  + 2^3 \sqrt{ 3x^7} =28x^3\sqrt{3x}

Step-by-step explanation:

The question is poorly formatted. The original question is:

(8x^3)^ \frac{2}{3} \sqrt{75x^3} + 2^3 \sqrt{ 3x^7

We have:

(8x^3)^ \frac{2}{3} \sqrt{75x^3} + 2^3 \sqrt{ 3x^7

Open bracket

(8x^3)^ \frac{2}{3} \sqrt{75x^3} + 2^3 \sqrt{ 3x^7} =(8^ \frac{2}{3} *x^{3* \frac{2}{3}}) \sqrt{75x^3} + 2^3 \sqrt{ 3x^7}

(8x^3)^ \frac{2}{3} \sqrt{75x^3} + 2^3 \sqrt{ 3x^7} =(8^ \frac{2}{3} *x^2) \sqrt{75x^3} + 2^3 \sqrt{ 3x^7}

Express 8 as 2^3

(8x^3)^ \frac{2}{3} \sqrt{75x^3} + 2^3 \sqrt{ 3x^7} =(2^{3* \frac{2}{3}} *x^2) \sqrt{75x^3} + 2^3 \sqrt{ 3x^7}

(8x^3)^ \frac{2}{3} \sqrt{75x^3} + 2^3 \sqrt{ 3x^7} =(2^2 *x^2) \sqrt{75x^3} + 2^3 \sqrt{ 3x^7}

(8x^3)^ \frac{2}{3} \sqrt{75x^3} + 2^3 \sqrt{ 3x^7} =4x^2 \sqrt{75x^3} + 2^3 \sqrt{ 3x^7}

Express 2^3 as 8

(8x^3)^ \frac{2}{3} \sqrt{75x^3} + 2^3 \sqrt{ 3x^7}=4x^2 \sqrt{75x^3} + 8\sqrt{ 3x^7}

Expand each exponent

(8x^3)^ \frac{2}{3} \sqrt{75x^3} + 2^3 \sqrt{ 3x^7}=4x^2 \sqrt{25x^2 *3x} + 8\sqrt{ 3x * x^6}

Split

(8x^3)^ \frac{2}{3} \sqrt{75x^3} + 2^3 \sqrt{ 3x^7}=4x^2 \sqrt{25x^2} *\sqrt{3x} + 8\sqrt{3x} * \sqrt{x^6}

(8x^3)^ \frac{2}{3} \sqrt{75x^3}  + 2^3 \sqrt{ 3x^7} =4x^2 *5x *\sqrt{3x} + 8\sqrt{3x} * x^3

(8x^3)^ \frac{2}{3} \sqrt{75x^3}  + 2^3 \sqrt{ 3x^7} =20x^3\sqrt{3x} + 8x^3\sqrt{3x}

Factorize

(8x^3)^ \frac{2}{3} \sqrt{75x^3}  + 2^3 \sqrt{ 3x^7} =28x^3\sqrt{3x}

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