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Mathematics 67 Online
OpenStudy (anonymous):

The hypotenuse of a right triangle is 3 cm longer than twice the length of the shorter leg. The longer leg measures 12 cm. Find the lengths of the shorter leg and the hypotenuse

OpenStudy (anonymous):

Apply Pythagorus to x,12 and 2x+3

OpenStudy (anonymous):

wait wat

OpenStudy (anonymous):

|dw:1315786118173:dw|

OpenStudy (anonymous):

ok so then

OpenStudy (anonymous):

\[(2x+3)^2 = 12^2+x^2\] i.e. pythagoras Thereom \[h^2 = a^2 + b^2\]

OpenStudy (anonymous):

therefore, \[h = \sqrt(a^2+b^2)\] or in our case: \[2x+3 = \sqrt(12^2+x^2)\]

OpenStudy (anonymous):

so then you have to put x back into the equation 2x+3 so u can find out the length of that?

OpenStudy (anonymous):

can you help me with this (5x^7)(-23x^17) i need to simplify or rationalize wher appropiate

OpenStudy (anonymous):

sorry, trying different tack..one minute

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

one tack is (2x+3)^2 = x^2 +12^2 and multiply these out.. the other tack is not to multiply out but make 2x+3 = sqrt(12^2+x^2)..

OpenStudy (anonymous):

got it at last

OpenStudy (anonymous):

\[(2x+3)^2 = 12^2 +x ^2\]

OpenStudy (anonymous):

multiply out to give: \[4x^2+12x+9 = x^2 +144\]

OpenStudy (anonymous):

\[4x^2-x^2 +12x+9 -144 = 0\]

OpenStudy (anonymous):

\[3x^2+12x-135=0\]

OpenStudy (anonymous):

using quadratic formula to solve this, you get two solutions: -9 and 5, which you can then check ny substituting in x and checking as follows: \[(2(5)+3)^2 = 12^2 + (5)^2\]

OpenStudy (anonymous):

\[13^2 = 12^2 + 5^2 \]

OpenStudy (anonymous):

169 = 169

OpenStudy (anonymous):

the -9 answer gives 225 = 225

OpenStudy (anonymous):

sorry about the delay!

OpenStudy (anonymous):

its ok thank u very much! can you help me with 5\[\sqrt{8}\]

OpenStudy (anonymous):

can you retype question?

OpenStudy (anonymous):

yes...ratinalize this \[\sqrt[5]{8}\]

OpenStudy (anonymous):

the 5 should be coming out of it though

OpenStudy (anonymous):

\[\sqrt[5]{8}\] \[8^1/5\]

OpenStudy (anonymous):

thanks what about solving this 2(2x+1)+4=26

OpenStudy (anonymous):

2(2x+1) just do it as you would a normal multipication.put (2x+1) on top and 2 at bottom and multiply...2 x 1 = 2 and 2 x 2x = 4x, so answer = (4x+2 )+4=26 4x+2+4=26 4x+6=26 4x=26-6 4x=20 x=5

OpenStudy (anonymous):

thanks i forgot all of this because i took it last year and i need a review of it...can you help me with this problem factor 25p^2-q^2

OpenStudy (anonymous):

this is known as the difference of two squares

OpenStudy (anonymous):

(5p-q)(5p+q)=5p^2-q^2

OpenStudy (anonymous):

that should read 25p^2

OpenStudy (anonymous):

try multiplying it out as I have said to you, put 5p-q on top and 5p+q on bottom and multiply -q by +q = -q^2, then +q by 5p = 5pq, then 5p by -q = -5pq, and finally 5p by 5p is 25p^2, now -5pq and +5pq cancel to give you 25p^2 - q^2.

OpenStudy (anonymous):

ok now i got that answer can you help me with

OpenStudy (anonymous):

go ahead!That's what I'm here for

OpenStudy (anonymous):

|dw:1315789209067:dw|

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