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Mathematics 14 Online
OpenStudy (anonymous):

find the exact perimiter of the rectangle. then approximate your answer. square root 18cm and square root 72cm

OpenStudy (anonymous):

\[\sqrt{18} and \sqrt{72}\]

OpenStudy (radar):

\[2\sqrt{18}+2\sqrt{72}=6\sqrt{2}+6\sqrt{2}=12\sqrt{2}\]

OpenStudy (radar):

Any questions?

OpenStudy (anonymous):

ok I get

OpenStudy (radar):

Good

OpenStudy (anonymous):

how do I get the approx?

OpenStudy (anonymous):

doesnt seem to work right I get 0

OpenStudy (radar):

I would approximate the sq rt of 2 as 2.14 as it is an irrational value

OpenStudy (radar):

excuse me 1.414 lol

OpenStudy (anonymous):

that doesnt seem right either it says 14 and it should be in the 100s?

OpenStudy (radar):

The perimeter is approx 17

OpenStudy (radar):

How do you figure in the 100s?

OpenStudy (anonymous):

ok

OpenStudy (radar):

The sides are 4.3 cm and 8.5 cm add up the 4 sides for perimeter 8.6 +17=25.6 which is quite a bit diffrerent than 17.......thinking

OpenStudy (radar):

I see the problem\[2\sqrt{72}=12\sqrt{2}\] not 6 sq r2 2

OpenStudy (radar):

\[6\sqrt{2}+12\sqrt{2}=18\sqrt{2}\] thats more better.

OpenStudy (radar):

That is the exact , the approximate is 25.45 cm

OpenStudy (radar):

did you see where I goofed\[2\sqrt{72}=2\sqrt{36}\sqrt{2}=12\sqrt{2}\]

OpenStudy (radar):

Is it clear now SaraJ?"

OpenStudy (anonymous):

Yes I was wondering why it wasnt comming out right, LoL

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