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
Apply Pythagorus to x,12 and 2x+3
wait wat
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ok so then
\[(2x+3)^2 = 12^2+x^2\] i.e. pythagoras Thereom \[h^2 = a^2 + b^2\]
therefore, \[h = \sqrt(a^2+b^2)\] or in our case: \[2x+3 = \sqrt(12^2+x^2)\]
so then you have to put x back into the equation 2x+3 so u can find out the length of that?
can you help me with this (5x^7)(-23x^17) i need to simplify or rationalize wher appropiate
sorry, trying different tack..one minute
ok
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)..
got it at last
\[(2x+3)^2 = 12^2 +x ^2\]
multiply out to give: \[4x^2+12x+9 = x^2 +144\]
\[4x^2-x^2 +12x+9 -144 = 0\]
\[3x^2+12x-135=0\]
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: \[(2(5)+3)^2 = 12^2 + (5)^2\]
\[13^2 = 12^2 + 5^2 \]
169 = 169
the -9 answer gives 225 = 225
sorry about the delay!
its ok thank u very much! can you help me with 5\[\sqrt{8}\]
can you retype question?
yes...ratinalize this \[\sqrt[5]{8}\]
the 5 should be coming out of it though
\[\sqrt[5]{8}\] \[8^1/5\]
thanks what about solving this 2(2x+1)+4=26
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
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
this is known as the difference of two squares
(5p-q)(5p+q)=5p^2-q^2
that should read 25p^2
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.
ok now i got that answer can you help me with
go ahead!That's what I'm here for
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