Question

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

Answer 1

are there 7 questions.. or you showed your work !! that you attempted !!

Answer 2

no 7 questions i dont get these ones at all

Answer 3

what don't you get? have you tried at all ? ... where are you stuck, Let me know so I can help you ! :S

Answer 4

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

Answer 5

\[\Large 2\sqrt{6}+3\sqrt{96}=2\sqrt{6}+3\sqrt{16\cdot 6}\]

\[\Large 2\sqrt6+3\cdot 4\sqrt6 =2\sqrt6+12\sqrt6\]

now sum them... and you'll have
\[\Large 14\sqrt6\]

This was a hint regarding to the first question

Answer 6

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 !! ;)

Answer 7

im not understanding the second one though like i see how you did the first but the second is different what do i multiply

Answer 8

\[\Large (8+\sqrt{11})(8-\sqrt{11})=(8\cdot 8)-8\sqrt{11}+8\sqrt{11}-(\sqrt{11})^2\]

Answer 9

64-11=53 is the answer ;) ... NOTE: \[\Large -8\sqrt{11}+8\sqrt{11}=0\] so they cancel out ! :) try the next one ;)

Answer 10

ok so is the next one 36? i think

Answer 11

how do you get there .. show your work ! :P LOL

Answer 12

..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

Answer 13

ok yay! let me try the next one

Answer 14

is it 58?

Answer 15

58-9= 49 squared is 7?

Answer 16

\[\Large \sqrt{q-9}=7 \longrightarrow q-9=49\] q=49+9
q=58

Answer 17

is that what you got ?

Answer 18

I mean by calculating .. the result is it? 1=58 ?

Answer 19

q=58

Answer 20

yah thats what im saying

Answer 21

.. 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...

\[\Large 3x+3\geq 0\]

\[\Large 3x\geq -3 \longrightarrow x\geq \frac{-3}{3}\]

\[\Large x\geq -1\]

Answer 22

this was the last one...

Answer 23

no the √p=-1

Answer 24

tell me a number who when you square it, it becomes negative ??

Answer 25

√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.

Answer 26

\[\Large (-1)^2 \neq -1\] so.. there's no number which when we square it.. it becomes NEGATIVE ....it doesn't exist.

Answer 27

so there is no solution

Answer 28

exactly ! :)

Answer 29

thankyouu!!!

Answer 30

wait is there anyway you could just look at his last question????

Answer 31

https://www.connexus.com/content/media/459431-3172011-25122-PM-2041438739.png

Answer 32

@haileystowers sorry... the lat one, you'll do it yourself , I've to go :) ... Glad to help

Answer 33

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 :)

Answer 34

thankyou!