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jok3333 [9.3K]
3 years ago
10

Simplify the expression below you must show all work to receive full credit

Mathematics
2 answers:
Marrrta [24]3 years ago
7 0

Answer:

8w^2 I don't know why it is not on here, but I put it in an algebra calculator and everything. Good luck. :)

Step-by-step explanation:

Simply the radicals by multiplying and taking square roots.

2\sqrt[4]{8w^3} * \sqrt[4]{32w^5}

= 2\sqrt[4]{8w^3} *2w\sqrt[4]{2w}

= 4w\sqrt[4]{16w^4}

=4w(2w)

=8w^{2}

KengaRu [80]3 years ago
7 0

Answer: Uhm... None of the options seem to be correct. The answer is: 8w^2

Step-by-step explanation:

2\sqrt[4]{8w^3}*\sqrt[4]{32w^5}

This is the same as;

2\sqrt[4]{(8w^3)(32w^5)}

2\sqrt[4]{256w^8}

2\sqrt[4]{4^4w^8}

2*4\sqrt[4]{w^8}\\ 8\sqrt[4]{w^8}\\8w^2

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Step-by-step explanation:

OK, let's assume it this way:

<em>Sn=1.1!+2.2!+3.3!+...+n.n!</em><em>=</em><em>(</em><em>2</em><em>‐</em><em>1</em><em>)</em><em>.</em><em>1</em><em>!</em><em>+</em><em>(</em><em>3</em><em>-</em><em>1</em><em>)</em><em>.</em><em>2</em><em>!</em><em>+</em><em>(</em><em>4</em><em>-</em><em>1</em><em>)</em><em>3</em><em>!</em><em>+</em><em>.</em><em>.</em><em>.</em><em>+</em><em>(</em><em>(</em><em>n</em><em>+</em><em>1</em><em>)</em><em>-</em><em>1</em><em>)</em><em>.</em><em>n</em><em>!</em>

Sn=1.1!+2.2!+3.3!+...+n.n!=(2‐1).1!+(3-1).2!+(4-1)3!+...+((n+1)-1).n!<em>=</em><em>(</em><em>2</em><em>.</em><em>1</em><em>!</em><em>-</em><em>1</em><em>!</em><em>)</em><em>+</em><em>(</em><em>3</em><em>.</em><em>2</em><em>!</em><em>-</em><em>2</em><em>!</em><em>)</em><em>+</em><em>(</em><em>4</em><em>.</em><em>3</em><em>!</em><em>-</em><em>3</em><em>!</em><em>)</em><em>+</em><em>.</em><em>.</em><em>.</em><em>+</em><em>(</em><em>(</em><em>n-1</em><em>)</em><em>n</em><em>!</em><em>-</em><em>n</em><em>!</em><em>)</em><em>=</em><em>(</em><em>2</em><em>!</em><em>-</em><em>1</em><em>!</em><em>)</em><em>+</em><em>(</em><em>3</em><em>!</em><em>-</em><em>2</em><em>!</em><em>)</em><em>+</em><em>(</em><em>4</em><em>!</em><em>-</em><em>3</em><em>!</em><em>)</em><em>+</em>

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