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Mathematics 8 Online
OpenStudy (anonymous):

Find the length of the missing side of each right triangle to the nearest tenth. 1. legs of 4 cm and 6 cm

OpenStudy (anonymous):

please help or just tell em the answers or not

OpenStudy (anonymous):

a^2 + b^2 = c^2 so 4^2 + 6^2 = c^2

OpenStudy (anonymous):

16 + 36 = 52 so c^2 = 52 c = \[\sqrt{52}\] or 7.2

OpenStudy (anonymous):

thanks can u help me more?? please:)

OpenStudy (anonymous):

i'll try ;)

OpenStudy (anonymous):

leg of 9 m, hypotenuse of 15 m

OpenStudy (anonymous):

again a^2 + b^2 = c^2 this time, you know what a and c are, so we're looking for b that makes b^2 = c^2 - a^2 b^2 = 15^2 - 9^2 b^2 = 225 + 81 b^2 = 306 b = \[\sqrt{306}\] so b = 17.492 (not sure where you need to round it)

OpenStudy (anonymous):

how do u do that check thingy?

OpenStudy (anonymous):

the ^ or the \[\sqrt{?}\]

OpenStudy (anonymous):

the check

OpenStudy (anonymous):

for the square toot, you just go to equation, it's on the 4th row

OpenStudy (anonymous):

wat equation??

OpenStudy (anonymous):

where you type your reply, under that box, you have 'equation' 'drawing' and 'attach file'. If you click on equation, you'll find loads o useful things for math ;) on the 4th row you see the square root sign, click on that and then chose the one with the x under it. then you can fill in the number that has to be under it.

OpenStudy (anonymous):

srry for making u think about where u put that can you just teach me how to do this stuff clearlly and easy?:)

OpenStudy (anonymous):

explain how to calculate the sides of a triangle? i can try. it might not be always be very clear, cause i studied it all in a different languages, but math should work for everyone right? ;)

OpenStudy (anonymous):

well yea

OpenStudy (anonymous):

this only goes for right triangles btw. But you always start with: a^2 + b^2 = c^2 and then it depends on what side you need to figure out the length of. if you need to find out what the longest side is, than you canjust use the formula as it is. and fill in a and b, with the numbers that you do know. so, if you know that 2 legs are 4 and 5, you can fill in: a^2 +b^2 = c^2 4^2 + 5^2 = c^2 then calculate: 16 + 25 = c^2 41 = c^2 \[\sqrt{41}\] = c

OpenStudy (anonymous):

does that make sense?

OpenStudy (anonymous):

well yea sort of im tryin 2 copy it down

OpenStudy (anonymous):

oh, here's a video of someone that explains it really really well: http://www.khanacademy.org/math/geometry-1/core-geometry/v/the-pythagorean-theorem if you ever need to have something explained about math, that's the site to go. ;)

OpenStudy (anonymous):

might be easier to follow, than just reading what i'm typing

OpenStudy (anonymous):

no i get wat ur saying and wat about the word: hypotenuse? wat does thAT mean??

OpenStudy (anonymous):

well, i kinda had to look it up too ;) it's the longest side of the triangle, or the side that's not part of the right angle

OpenStudy (anonymous):

oh ok well i get it but can u check my answers for me?? please

OpenStudy (anonymous):

sure

OpenStudy (anonymous):

omgosh nm thank u so so so so so so so so so and it goes on.... so much i understand it now so u dont need to check my answers ur so great!!:)

OpenStudy (anonymous):

haha, you're very welcome.... it's nice to see someone actually trying to figure it out, instead of just wanting the answer. :)

OpenStudy (anonymous):

but i have one more question...... how do u round or estimate the answer?

OpenStudy (anonymous):

do you mean how i calculate that square root? i kinda use a calculator for that. ;) or do you mean, when they say that you need to round to the nearest ten? for example, if it says: 8.67, then you round it to 8.7 if it says 4.21, you round it to 4.2. if the next number is lower than 5, you round it down, if it's higher than 5, you round up.

OpenStudy (anonymous):

oh yes now i remember.... and if it is a hole number then just leave it the same?

OpenStudy (anonymous):

yes, exactly :)

OpenStudy (anonymous):

ok well thank you so much u were a great help!!:)

OpenStudy (anonymous):

no problem at all

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