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DENIUS [597]
2 years ago
9

Perform the multiplication. Simplify the answers. 2 √30*( √5+ √6+ √10+ √15)

Mathematics
1 answer:
Delicious77 [7]2 years ago
3 0

The simplified expression of 2\sqrt{30} \times (\sqrt5+ \sqrt6+ \sqrt{10} + \sqrt{15})is 10\sqrt{6} +  6\sqrt{20}+  20\sqrt{3} +  30\sqrt{2}

The expression is given as:

2\sqrt{30} \times (\sqrt5+ \sqrt6+ \sqrt{10} + \sqrt{15})

Expand the expression

2\sqrt{30} \times (\sqrt5+ \sqrt6+ \sqrt{10} + \sqrt{15}) = 2\sqrt{30} \times \sqrt5+  2\sqrt{30} \times \sqrt6+  2\sqrt{30} \times \sqrt{10} +  2\sqrt{30} \times \sqrt{15}

Factor out 2

2\sqrt{30} \times (\sqrt5+ \sqrt6+ \sqrt{10} + \sqrt{15}) = 2(\sqrt{30} \times \sqrt5+  \sqrt{30} \times \sqrt6+  \sqrt{30} \times \sqrt{10} +  \sqrt{30} \times \sqrt{15})

Combine the radicals

2\sqrt{30} \times (\sqrt5+ \sqrt6+ \sqrt{10} + \sqrt{15}) = 2(\sqrt{150} +  \sqrt{180}+  \sqrt{300} +  \sqrt{450})

Expand the expression

2\sqrt{30} \times (\sqrt5+ \sqrt6+ \sqrt{10} + \sqrt{15}) = 2(\sqrt{25 \times 6} +  \sqrt{9 \times 20}+  \sqrt{100 \times 3} +  \sqrt{225\times 2})

Evaluate the roots

2\sqrt{30} \times (\sqrt5+ \sqrt6+ \sqrt{10} + \sqrt{15}) = 2(5\sqrt{6} +  3\sqrt{20}+  10\sqrt{3} +  15 \sqrt{2})

Expand

2\sqrt{30} \times (\sqrt5+ \sqrt6+ \sqrt{10} + \sqrt{15}) =10\sqrt{6} +  6\sqrt{20}+  20\sqrt{3} +  30\sqrt{2}

Hence, the simplified expression of 2\sqrt{30} \times (\sqrt5+ \sqrt6+ \sqrt{10} + \sqrt{15})is 10\sqrt{6} +  6\sqrt{20}+  20\sqrt{3} +  30\sqrt{2}

Read more about simplified expressions at:

brainly.com/question/8008182

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What is the leading coefficient of a third degree function that has an output of 221 when x=2, and has zeros of −15, 3i, and −3i
Vinil7 [7]

Answer:

  1

Step-by-step explanation:

The function with the given zeros will factor as ...

  f(x) = a(x +15)(x^2 +9) . . . . with leading  coefficient 'a'

You have ...

  f(2) = 221 = a(2+15)(2^2+9) = a(17)(13) = 221a

Then a = 221/221 = 1

The leading coefficient is 1.

_____

<em>Additional comment</em>

As you know, a function with zero x=p has a factor of (x -p). The given zeros mean the function has factors (x -(-15)), (x -3i). and (x -(-3i)). The product of the last two factors is the difference of squares: (x^2 -(3i)^2) = (x^2 -(-9)) = (x^2 +9). This is how we arrived at the factorization shown above.

3 0
2 years ago
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