anonymous
  • anonymous
2√6 + 3√96 (8+√11)(8-√11) 4=√p -2 √q-9 =7 the rad sign is over the q and -9 √5x-1 =√4x+9 √p=-1 # –1 is a solution of the original equation. 1 is an extraneous solution. # 1 is a solution of the original equation. # 1 is a solution of the original equation. –1 is an extraneous solution. # There is no solution. y=√3x+3 # x ≤ – 1 # x > 1 # x ≥ –1
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
are there 7 questions.. or you showed your work !! that you attempted !!
anonymous
  • anonymous
no 7 questions i dont get these ones at all
anonymous
  • anonymous
what don't you get? have you tried at all ? ... where are you stuck, Let me know so I can help you ! :S

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anonymous
  • anonymous
i dont understand how to solve any of these or how to begin to find the answers its on my homework and ive completed all the rest besides these ones
anonymous
  • anonymous
\[\Large 2\sqrt{6}+3\sqrt{96}=2\sqrt{6}+3\sqrt{16\cdot 6}\] \[\Large 2\sqrt6+3\cdot 4\sqrt6 =2\sqrt6+12\sqrt6\] now sum them... and you'll have \[\Large 14\sqrt6\] This was a hint regarding to the first question
anonymous
  • anonymous
the second one .. just multiply, and tell me if you'll get stuck, so I can give a hint again... You'll forget these kinds of problems very soon , if you don't try them yourself !! ;)
anonymous
  • anonymous
im not understanding the second one though like i see how you did the first but the second is different what do i multiply
anonymous
  • anonymous
\[\Large (8+\sqrt{11})(8-\sqrt{11})=(8\cdot 8)-8\sqrt{11}+8\sqrt{11}-(\sqrt{11})^2\]
anonymous
  • anonymous
64-11=53 is the answer ;) ... NOTE: \[\Large -8\sqrt{11}+8\sqrt{11}=0\] so they cancel out ! :) try the next one ;)
anonymous
  • anonymous
ok so is the next one 36? i think
anonymous
  • anonymous
how do you get there .. show your work ! :P LOL
anonymous
  • anonymous
..never mind, that's correct ;) well done ! \[\Large 4=\sqrt p-2 \longrightarrow \sqrt p=4+2\] now square both sides ... \[\Large p=36\] If this is what you've done... that's correct ;) .. try the next one
anonymous
  • anonymous
ok yay! let me try the next one
anonymous
  • anonymous
is it 58?
anonymous
  • anonymous
58-9= 49 squared is 7?
anonymous
  • anonymous
\[\Large \sqrt{q-9}=7 \longrightarrow q-9=49\] q=49+9 q=58
anonymous
  • anonymous
is that what you got ?
anonymous
  • anonymous
I mean by calculating .. the result is it? 1=58 ?
anonymous
  • anonymous
q=58
anonymous
  • anonymous
yah thats what im saying
anonymous
  • anonymous
.. then that's correct... y=√3x+3 # x ≤ – 1 # x > 1 # x ≥ –1 \[\large y=\sqrt{3x+3}\] we have...\[\large \sqrt{3x+3}\geq 0\] square both sides... \[\Large 3x+3\geq 0\] \[\Large 3x\geq -3 \longrightarrow x\geq \frac{-3}{3}\] \[\Large x\geq -1\]
anonymous
  • anonymous
this was the last one...
anonymous
  • anonymous
no the √p=-1
anonymous
  • anonymous
tell me a number who when you square it, it becomes negative ??
anonymous
  • anonymous
√p=-1 # –1 is a solution of the original equation. 1 is an extraneous solution. # 1 is a solution of the original equation. # 1 is a solution of the original equation. –1 is an extraneous solution. # There is no solution.
anonymous
  • anonymous
\[\Large (-1)^2 \neq -1\] so.. there's no number which when we square it.. it becomes NEGATIVE ....it doesn't exist.
anonymous
  • anonymous
so there is no solution
anonymous
  • anonymous
exactly ! :)
anonymous
  • anonymous
thankyouu!!!
anonymous
  • anonymous
wait is there anyway you could just look at his last question????
anonymous
  • anonymous
https://www.connexus.com/content/media/459431-3172011-25122-PM-2041438739.png
anonymous
  • anonymous
@haileystowers sorry... the lat one, you'll do it yourself , I've to go :) ... Glad to help
anonymous
  • anonymous
about triangle... use Tangent since \[\Large \tan x=\frac{\sin x}{\cos x}\] we have... \[\Large \frac{\sin 42^{\circ}}{\cos 42^{\circ}}=\frac x5 \] use cross multiplication and the answer should be ... \[\Large x\approx 4.5\] sorry .. got to go... use calculator or something, Have a nice day . Bye :)
anonymous
  • anonymous
thankyou!

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