can someone check my work??

- anonymous

can someone check my work??

- Stacey Warren - Expert brainly.com

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- schrodinger

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- anonymous

sqrt3(sqrt6+sqrt3)+(sqrt8-5)

- anonymous

|dw:1354245450082:dw|

- Lime

I believe it is much easier to use a calculator for Square root, but let's break it down.
\[3 + 3 \sqrt2 +(\sqrt 8 - 5)\]

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## More answers

- anonymous

well from there i got 3+3sqrt2= 6sqrt2 + sqrt8-5

- anonymous

and the sqrt of 8 is 4 and sqrt of 4 is 2 so it will be 6sqrt2+sqrt2-5

- anonymous

whoah where did that come from

- anonymous

|dw:1354246002329:dw|

- anonymous

ohh i see i took the sqaure root of 8 wrong

- anonymous

can i ask ya'll another question??

- anonymous

yep.

- anonymous

Part 1: Find the perimeter of a rectangular object which has a length of sqrt 128 feet and a width of sqrt 200 feet.
Part 2: Explain, in complete sentences, how you arrived at the answer and give the final solution in simplified radical form.
Part 3: What type of object in your home or school might this be?
i am not going to lie I don't even know where to start....

- anonymous

|dw:1354246494891:dw|
Just add up the sides, for part one. Simplify the square roots a little.

- anonymous

sqrt of 128= 8sqrt 2?

- anonymous

That's right.

- anonymous

and the sqrt of 200 = 10sqrt 2???

- anonymous

yep.

- anonymous

what do I do from there?

- anonymous

do I add them together?

- anonymous

perimeter of a rectangle.

- anonymous

how do i do that?

- anonymous

You want to know how long the edge is...
|dw:1354247036064:dw|

- anonymous

so multiply 8sqrt2*8sqrt2 and 10sqrt2*10sqrt2? and then add them together?

- anonymous

are you still there?

- anonymous

am I still doing this right? hello?

- anonymous

Yeah to find the perimeter of a rectangle, you add the lengths of all sides together.

- anonymous

after you multiply?

- anonymous

Just take the lengths of the sides and add them together.

- anonymous

so 8sqrt2*8sqrt2=64sqrt2
10sqrt2*10sqrt2= 100sqrt 2
64sqrt2+100sqrt2= 164sqrt2?

- anonymous

am i right so far?

- anonymous

was I just suppose to add them all together or multiply? like i did above?

- jim_thompson5910

|dw:1354248044017:dw|

- jim_thompson5910

|dw:1354248083431:dw|

- jim_thompson5910

|dw:1354248126981:dw|
Perimeter = Sum of all sides
P = s1 + s2 + s3 + s4
P = 8*sqrt(2) + 10*sqrt(2) + 8*sqrt(2) + 10*sqrt(2)
P = 36*sqrt(2)

- jim_thompson5910

So the perimeter is \[\Large 36\sqrt{2}\]

- anonymous

oh so I was suppose to add them not multiply

- jim_thompson5910

yes, or you can add the length and width and multiply by 2
this is because
P = L + W + L + W
P = 2L + 2W
P = 2*(L + W)

- anonymous

oh ok that what i was thinking of

- jim_thompson5910

Explain, in complete sentences, how you arrived at the answer and give the final solution in simplified radical form
you essentially simplify both sqrt(128) and sqrt(200) to get 8*sqrt(2) and 10*sqrt(2)
you can either add up the four sides
or
you can add up the length and width and multiply by 2
either way, you get the answer of \[\Large 36\sqrt{2}\]

- anonymous

ok thats what i wrote so far what is the next step?

- jim_thompson5910

What type of object in your home or school might this be?
\[\Large 36\sqrt{2} \approx 36(1.41421356)\]
\[\Large 36\sqrt{2} \approx 50.91168816\]
so the perimeter is roughly 50.91168816 feet

- jim_thompson5910

\[\Large 8\sqrt{2} \approx 8(1.41421356)\]
\[\Large 8\sqrt{2} \approx 11.31370848\]
the width is roughly 11.31370848 feet

- anonymous

the funny equal sign is a approximate sign right?

- jim_thompson5910

\[\Large 10\sqrt{2} \approx 10(1.41421356)\]
\[\Large 10\sqrt{2} \approx 14.1421356\]
the length is roughly 14.1421356 feet

- jim_thompson5910

so you just have to name an object that's roughly 14.1421356 ft by 11.31370848 ft

- anonymous

so it could be a pool right? for the object?

- jim_thompson5910

yeah or a large table/counter of some sort

- anonymous

ohhh wow that wasnt too bad there is another one very similiar and if I post all my work will you tell me if I am right or not?

- jim_thompson5910

sure go for it

- anonymous

awesome that way I know I really understand it

- anonymous

Part 1: Find the area of a rectangular object which has a length of 4sqrt2 inches and a width of 6sqrt6 inches.
Part 2: Explain, in complete sentences, how you arrived at the simplified answer and give the final solution in simplified radical form.
Part 3: What type of object in your home or school might this be?
|dw:1354248883689:dw|

- jim_thompson5910

ok it's a bit different because they want the area this time

- jim_thompson5910

and not the perimeter

- anonymous

isn't area multiplying the sides together?

- jim_thompson5910

exactly

- jim_thompson5910

Area = Length times Width

- anonymous

ok so first i have to simplify the sqrts
6sqrt 6 = 6+3sqrt2?

- anonymous

9sqrt 2?

- jim_thompson5910

how are you getting 6sqrt 6 = 6+3sqrt2

- anonymous

uhhh i might of factored wrong
because i took the factors 2 and 3 and pulled the 3 out to add to the 6 and left the 2 inside the radical

- jim_thompson5910

you can't do that

- jim_thompson5910

6*sqrt(6) is as simple as it gets

- anonymous

i was trying to make it so both of my radicals have the same square root thing of sqrt2

- jim_thompson5910

notice that 6 has no factors that are perfect squares (ignore 1)

- jim_thompson5910

so sqrt(6) can't be simplified

- anonymous

oh..... i was wondering how that worked cause nothing times itself equals 6

- jim_thompson5910

Multiply the coefficients: 6*4 = 24
Multiply the radicands: 6*2 = 12

- Ammarah

what is ur question, i mean what r u trying to find exactly?

- Ammarah

so u multiply and then plugg it back in to the equation.

- jim_thompson5910

so 6*sqrt(6) times 4*sqrt(2) = 24*sqrt(12)
simplify 24*sqrt(12)

- Ammarah

yes dont forget!! Simplify!

- anonymous

so I can go straight to part 2 right? and find the area?

- Ammarah

yes.

- anonymous

sorry idk why that post just showed up i have a major delay

- Ammarah

?

- jim_thompson5910

yes since you can't simplify the length/width just multiply them out

- anonymous

now is 24sqrt12 the radical area that needs to be converted into the approximate length?

- jim_thompson5910

simplify that
24*sqrt(12)
24*sqrt(4*3)
24*sqrt(4)*sqrt(3)
24*2*sqrt(3)
48*sqrt(3)

- anonymous

and do you get the approximate length by putting the number in the calculator? i meant to ask you how you did that last time

- jim_thompson5910

so the area is 48*sqrt(3) square inches

- jim_thompson5910

you just type 6*sqrt(6) and you get 14.6969 roughly

- anonymous

and 4sqqrt of 2 is approximately 5.65685424949238

- jim_thompson5910

correct

- anonymous

so the object could be like a table? o ra whiite board maybe?

- jim_thompson5910

yeah that's a reasonably sized whiteboard

- anonymous

you have helped me so much and thank you for helping me with this someone was helping me but they left and I didnt know where to go from there you are definitley a life SAVER!

- jim_thompson5910

i'm glad i did and hopefully it's all making sense now

- anonymous

yeah they were a little different but there steps were very similar and I took notes while we talked so hopefully that will help me if I have a problem like this later on in the course

- anonymous

youre WELCOME

- Ammarah

no thanks

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