anonymous
  • anonymous
can someone check my work??
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
sqrt3(sqrt6+sqrt3)+(sqrt8-5)
anonymous
  • anonymous
|dw:1354245450082:dw|
Lime
  • 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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anonymous
  • anonymous
well from there i got 3+3sqrt2= 6sqrt2 + sqrt8-5
anonymous
  • anonymous
and the sqrt of 8 is 4 and sqrt of 4 is 2 so it will be 6sqrt2+sqrt2-5
anonymous
  • anonymous
whoah where did that come from
anonymous
  • anonymous
|dw:1354246002329:dw|
anonymous
  • anonymous
ohh i see i took the sqaure root of 8 wrong
anonymous
  • anonymous
can i ask ya'll another question??
anonymous
  • anonymous
yep.
anonymous
  • 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
  • anonymous
|dw:1354246494891:dw| Just add up the sides, for part one. Simplify the square roots a little.
anonymous
  • anonymous
sqrt of 128= 8sqrt 2?
anonymous
  • anonymous
That's right.
anonymous
  • anonymous
and the sqrt of 200 = 10sqrt 2???
anonymous
  • anonymous
yep.
anonymous
  • anonymous
what do I do from there?
anonymous
  • anonymous
do I add them together?
anonymous
  • anonymous
perimeter of a rectangle.
anonymous
  • anonymous
how do i do that?
anonymous
  • anonymous
You want to know how long the edge is... |dw:1354247036064:dw|
anonymous
  • anonymous
so multiply 8sqrt2*8sqrt2 and 10sqrt2*10sqrt2? and then add them together?
anonymous
  • anonymous
are you still there?
anonymous
  • anonymous
am I still doing this right? hello?
anonymous
  • anonymous
Yeah to find the perimeter of a rectangle, you add the lengths of all sides together.
anonymous
  • anonymous
after you multiply?
anonymous
  • anonymous
Just take the lengths of the sides and add them together.
anonymous
  • anonymous
so 8sqrt2*8sqrt2=64sqrt2 10sqrt2*10sqrt2= 100sqrt 2 64sqrt2+100sqrt2= 164sqrt2?
anonymous
  • anonymous
am i right so far?
anonymous
  • anonymous
was I just suppose to add them all together or multiply? like i did above?
jim_thompson5910
  • jim_thompson5910
|dw:1354248044017:dw|
jim_thompson5910
  • jim_thompson5910
|dw:1354248083431:dw|
jim_thompson5910
  • 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
  • jim_thompson5910
So the perimeter is \[\Large 36\sqrt{2}\]
anonymous
  • anonymous
oh so I was suppose to add them not multiply
jim_thompson5910
  • 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
  • anonymous
oh ok that what i was thinking of
jim_thompson5910
  • 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
  • anonymous
ok thats what i wrote so far what is the next step?
jim_thompson5910
  • 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
  • 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
  • anonymous
the funny equal sign is a approximate sign right?
jim_thompson5910
  • 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
  • jim_thompson5910
so you just have to name an object that's roughly 14.1421356 ft by 11.31370848 ft
anonymous
  • anonymous
so it could be a pool right? for the object?
jim_thompson5910
  • jim_thompson5910
yeah or a large table/counter of some sort
anonymous
  • 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
  • jim_thompson5910
sure go for it
anonymous
  • anonymous
awesome that way I know I really understand it
anonymous
  • 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
  • jim_thompson5910
ok it's a bit different because they want the area this time
jim_thompson5910
  • jim_thompson5910
and not the perimeter
anonymous
  • anonymous
isn't area multiplying the sides together?
jim_thompson5910
  • jim_thompson5910
exactly
jim_thompson5910
  • jim_thompson5910
Area = Length times Width
anonymous
  • anonymous
ok so first i have to simplify the sqrts 6sqrt 6 = 6+3sqrt2?
anonymous
  • anonymous
9sqrt 2?
jim_thompson5910
  • jim_thompson5910
how are you getting 6sqrt 6 = 6+3sqrt2
anonymous
  • 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
  • jim_thompson5910
you can't do that
jim_thompson5910
  • jim_thompson5910
6*sqrt(6) is as simple as it gets
anonymous
  • anonymous
i was trying to make it so both of my radicals have the same square root thing of sqrt2
jim_thompson5910
  • jim_thompson5910
notice that 6 has no factors that are perfect squares (ignore 1)
jim_thompson5910
  • jim_thompson5910
so sqrt(6) can't be simplified
anonymous
  • anonymous
oh..... i was wondering how that worked cause nothing times itself equals 6
jim_thompson5910
  • jim_thompson5910
Multiply the coefficients: 6*4 = 24 Multiply the radicands: 6*2 = 12
Ammarah
  • Ammarah
what is ur question, i mean what r u trying to find exactly?
Ammarah
  • Ammarah
so u multiply and then plugg it back in to the equation.
jim_thompson5910
  • jim_thompson5910
so 6*sqrt(6) times 4*sqrt(2) = 24*sqrt(12) simplify 24*sqrt(12)
Ammarah
  • Ammarah
yes dont forget!! Simplify!
anonymous
  • anonymous
so I can go straight to part 2 right? and find the area?
Ammarah
  • Ammarah
yes.
anonymous
  • anonymous
sorry idk why that post just showed up i have a major delay
Ammarah
  • Ammarah
?
jim_thompson5910
  • jim_thompson5910
yes since you can't simplify the length/width just multiply them out
anonymous
  • anonymous
now is 24sqrt12 the radical area that needs to be converted into the approximate length?
jim_thompson5910
  • jim_thompson5910
simplify that 24*sqrt(12) 24*sqrt(4*3) 24*sqrt(4)*sqrt(3) 24*2*sqrt(3) 48*sqrt(3)
anonymous
  • 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
  • jim_thompson5910
so the area is 48*sqrt(3) square inches
jim_thompson5910
  • jim_thompson5910
you just type 6*sqrt(6) and you get 14.6969 roughly
anonymous
  • anonymous
and 4sqqrt of 2 is approximately 5.65685424949238
jim_thompson5910
  • jim_thompson5910
correct
anonymous
  • anonymous
so the object could be like a table? o ra whiite board maybe?
jim_thompson5910
  • jim_thompson5910
yeah that's a reasonably sized whiteboard
anonymous
  • 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
  • jim_thompson5910
i'm glad i did and hopefully it's all making sense now
anonymous
  • 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
  • anonymous
youre WELCOME
Ammarah
  • Ammarah
no thanks

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