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Harrizon [31]
2 years ago
12

There are 380 lightbulbs lined up in a row in a long room each bulb has his own switch and it’s currently switched off each bulb

is number constitutionally from 1 to 380 you first flip every switch then you flip the switch on every second bulb you then flip the switch on every third bulb this continues until you have gone through the process 380 times
Mathematics
1 answer:
BARSIC [14]2 years ago
6 0

Flipping the bulbs is an illustration of sequence and patterns

19 bulbs would remain on.

<h3>How to determine the number of bulbs that will be on</h3>

The sequence of turning the bulbs off and on, would change the state of the bulbs until the 380th process.

While changing the state of the bulbs, the bulb at the perfect square positions would remain on.

There are 19 perfect square numbers between 1 and 380 (inclusive)

Hence, 19 bulbs would remain on.

Read more about sequence and patterns at:

brainly.com/question/15590116

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\sqrt{x^2y^3}+2\sqrt{x^3y^4}+xy\sqrt{y}=2xy^2\sqrt{x}+2xy\sqrt{y}

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Step-by-step explanation:

Given the expression

\sqrt{x^2y^3}+2\sqrt{x^3y^4}+xy\sqrt{y}

solving the expression

\sqrt{x^2y^3}+2\sqrt{x^3y^4}+xy\sqrt{y}

as

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so the expression becomes

\:\sqrt{x^2y^3}+2\sqrt{x^3y^4}+xy\sqrt{y}=xy\sqrt{y}+2xy^2\sqrt{x}+xy\sqrt{y}

Group like terms

                                        =2xy^2\sqrt{x}+xy\sqrt{y}+xy\sqrt{y}

Add similar elements

                                        =2xy^2\sqrt{x}+2xy\sqrt{y}

Therefore, we conclude that:

\sqrt{x^2y^3}+2\sqrt{x^3y^4}+xy\sqrt{y}=2xy^2\sqrt{x}+2xy\sqrt{y}

Hence, option B is true.

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