Ask your own question, for FREE!
Mathematics 22 Online
OpenStudy (volleyballlover55):

HELP!

OpenStudy (volleyballlover55):

@jim_thompson5910

OpenStudy (volleyballlover55):

@jim_thompson5910

jimthompson5910 (jim_thompson5910):

|dw:1469068266142:dw|

jimthompson5910 (jim_thompson5910):

|dw:1469068320896:dw|

OpenStudy (volleyballlover55):

ok can you give me the first two steps like all the time..

OpenStudy (volleyballlover55):

@jim_thompson5910

jimthompson5910 (jim_thompson5910):

a^2+b^2 = c^2 9^2 + 14^2 = x^2 solve for x

OpenStudy (volleyballlover55):

ok thanks!

OpenStudy (volleyballlover55):

a^2+b^2 = c^2 9^2 + 14^2 = x^2 81+196=x^2 277=x^2 x^2=277 sqrt (x^2)=sqrt(277) x=16.6433

jimthompson5910 (jim_thompson5910):

correct

jimthompson5910 (jim_thompson5910):

that's the approx length of PR. Make sure to round it to two decimal places

OpenStudy (volleyballlover55):

ok so thats it? this is the complete answer? The approximate length of PR: a^2+b^2 = c^2 9^2 + 14^2 = x^2 81+196=x^2 277=x^2 x^2=277 sqrt (x^2)=sqrt(277) x=16.6433

OpenStudy (volleyballlover55):

whats 16.6433 rounded to two decimal places? 16.64?

OpenStudy (volleyballlover55):

well it says to round it to the nearest hundredth..

jimthompson5910 (jim_thompson5910):

`whats 16.6433 rounded to two decimal places? 16.64?` correct

jimthompson5910 (jim_thompson5910):

`well it says to round it to the nearest hundredth..` yes that's the same as saying "round to 2 decimal places"

OpenStudy (volleyballlover55):

ok!!! so is this the final answer? The approximate length of PR: a^2+b^2 = c^2 9^2 + 14^2 = x^2 81+196=x^2 277=x^2 x^2=277 sqrt (x^2)=sqrt(277) x=16.6433 16.6433 rounded to the nearest hundredth is 16.64.

jimthompson5910 (jim_thompson5910):

that just takes care of part A

jimthompson5910 (jim_thompson5910):

make sure you include the name of the theorem you used

OpenStudy (volleyballlover55):

oh really? we arent done?

OpenStudy (volleyballlover55):

ohhhh!!! lol we arent done.. i didnt see there was a part b! whoopsies!

OpenStudy (volleyballlover55):

did we use the pythagorean theorem for part a?

jimthompson5910 (jim_thompson5910):

`did we use the pythagorean theorem for part a?` yes

OpenStudy (volleyballlover55):

ok! now lets move on to part b!

jimthompson5910 (jim_thompson5910):

PR is now 18 ft PQ stays at 14 ft the question they want us to find out: how long is QR? |dw:1469069287385:dw|

jimthompson5910 (jim_thompson5910):

I want you to try to set up the equation

OpenStudy (volleyballlover55):

ok so the first two steps are???

jimthompson5910 (jim_thompson5910):

what's the longest side here?

OpenStudy (volleyballlover55):

a^2+b^2 = c^2 18^2+14^2=c^2 is that correct so i can start solving?

jimthompson5910 (jim_thompson5910):

no, keep in mind that c is the longest side c = 18 is the longest side a = 14 is one of the legs b = y is unknown |dw:1469069407344:dw|

jimthompson5910 (jim_thompson5910):

a^2+b^2 = c^2 14^2+y^2 = 18^2

jimthompson5910 (jim_thompson5910):

hopefully it makes sense how I set up the equation?

OpenStudy (volleyballlover55):

oohhh yes i get it no!!!

OpenStudy (volleyballlover55):

@jim_thompson5910

OpenStudy (volleyballlover55):

let me solve now!

jimthompson5910 (jim_thompson5910):

alright, take your time

OpenStudy (volleyballlover55):

a^2+b^2 = c^2 14^2+y^2 = 18^2 196+y^2=324 y2=324−196 y^2=128 y=11.3137

jimthompson5910 (jim_thompson5910):

very good

OpenStudy (volleyballlover55):

thank you !

jimthompson5910 (jim_thompson5910):

don't forget to round this answer

OpenStudy (volleyballlover55):

ok so this is the complete answer for part b? The approximate length of QR is: a^2+b^2 = c^2 14^2+y^2 = 18^2 196+y^2=324 y2=324−196 y^2=128 y=11.3137 11.3137 rounded to the nearest hundredth is 11.31.

jimthompson5910 (jim_thompson5910):

yes it is

OpenStudy (volleyballlover55):

im not missing anything ? :)

jimthompson5910 (jim_thompson5910):

nope

OpenStudy (volleyballlover55):

yay!

jimthompson5910 (jim_thompson5910):

well actually, I would throw in the step sqrt(y^2) = sqrt(128) other than that, it looks good

OpenStudy (volleyballlover55):

ok!

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!