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rosijanka [135]
3 years ago
7

I need help on that question ​

Mathematics
2 answers:
Veseljchak [2.6K]3 years ago
7 0

Answer:

<em> </em><em>8</em><em>0</em><em>√</em><em>3</em>

Step-by-step explanation:

<em>here's </em><em>your</em><em> solution</em>

<em> </em><em> </em><em> </em><em>=</em><em>></em><em> </em><em> </em><em>First</em><em> </em><em>number</em><em> </em><em>is </em><em>4</em><em>√</em><em>5</em>

<em> </em><em> </em><em>=</em><em>></em><em> </em><em>Second</em><em> </em><em>number</em><em> is</em><em> </em><em>4</em><em>√</em><em>1</em><em>5</em>

<em> </em><em> </em><em> </em><em>=</em><em>></em><em> </em><em>we </em><em>need</em><em> to</em><em> find</em><em> </em><em>product</em>

<em> </em><em> </em><em>=</em><em>></em><em> </em><em>4</em><em>√</em><em>5</em><em>*</em><em>4</em><em>√</em><em>1</em><em>5</em>

<em> </em><em> </em><em> </em><em>=</em><em>></em><em> </em><em>4</em><em>*</em><em>4</em><em> </em><em>=</em><em> </em><em>1</em><em>6</em>

<em> </em><em> </em><em> </em><em> </em><em>=</em><em>></em><em> </em><em>√</em><em>5</em><em>*</em><em>√</em><em>1</em><em>5</em><em>. </em><em>=</em><em> </em><em>√</em><em>7</em><em>5</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em>=</em><em>></em><em> </em><em>√</em><em>7</em><em>5</em><em> </em><em>=</em><em> </em><em>5</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>now,</em><em> </em><em>multiple</em><em> </em><em>5</em><em>√</em><em>3</em><em>*</em><em>1</em><em>6</em><em> </em><em>=</em><em> </em><em>8</em><em>0</em><em>√</em><em>3</em>

<em>hope</em><em> it</em><em> helps</em>

Anika [276]3 years ago
3 0

Answer:

the required answer is

80 \sqrt{3}

Step-by-step explanation:

here's the solution : -

=》

4 \sqrt{5}  \times 4 \sqrt{15}

=》

4 \sqrt{5}  \times 4 \sqrt{5}  \times  \sqrt{3}

=》

4 \times 4 \times 5 \times  \sqrt{3}

=》

80 \sqrt{3}

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Which equation demonstrates the additive identity property?
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Answer:

Step-by-step explanation:

Solve for o in the equation (7 + 4) + (7 - 41) = 14©(7 + 4) + 0 = 7 + 41o(7 + 4)(1) = 7 + 41o(7 + 41) + ( - 7 - 41) = 0

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Step 1: Group variables: We need to group our variables (7 and 14©(7. To do that, we subtract 14©(7 from both sides (7 - 14©(7 = 14©(7 - 14©(7

Step 2: Cancel 14©(7 on the right side: 0o = 0 Step 3: Divide each side of the equation by 0

0o 0 = 0 0 o =

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