Question

Still need some more help!..
Sorry, but I have just a few more I believe!

Answer 1

@whpalmer4 @Compassionate

Answer 2

And anyone else that can help!!!

Answer 3

please post the questions, then tag, not the other way around, thanks :-)

Answer 4

sorry. Still new to this...
~
A square with a side of 20 is decreased to a side of 16. By how much is the diagonal decreased?

Answer 5

Okay, our first square is 20 x 20. What is the diagonal?

Draw a picture:

Answer 6

Use the Pythagorean theorem: \[a^2+b^2=c^2\]\[a=20,b=20\]\[20^2+20^2=c^2\]\[2*20^2 = c^2\]Take the square root of both sides:\[\sqrt{2*20^2}=\sqrt{c^2}\]We know \(c\) is positive, so we can simplify that to:\[20\sqrt{2} = c\]

Answer 7

okay

Answer 8

New square has the sides of length 16 instead of 20. What will \(c=\)

Answer 9

Hint: in a 45/45/90 degree triangle such as we have here, the hypotenuse is always \(s\sqrt{2}\) where \(s\) is the side length

Answer 10

16sqrt2

Answer 11

Right. So how much did the diagonal decrease when the side length went from 20 to 16?

Answer 12

4 sqrt 2

Answer 13

There you go!

Answer 14

THANK YOU! I have a few more..
Do you feel like helping? If not I understand

Answer 15

if I didn't feel like helping, why would I hang out here? :-)

Answer 16

Haha. true.
THANK YOU!

~Simplify SQRT75x-SQRT3x+SQRT12x

Answer 17

\[\sqrt{75x} - \sqrt{3x} + \sqrt{12x}\]Is that it?

Answer 18

yes. I do not know how to do that, but yes

Answer 19

okay, your job is to factor 75, 3, and 12 down to their prime factors. do you know how to do that?

Answer 20

5 and the SQRT 3
SQRT 3
2 and the SQRt of 3

Answer 21

Okay, you did more than I asked :-)

\[\sqrt{75x} -\sqrt{3x} + \sqrt{12x} =\sqrt{3*5*5*x} - \sqrt{3*x} + \sqrt{2*2*3*x} \]\[= 5\sqrt{3x}-\sqrt{3x}+2\sqrt{3x}=\]

Answer 22

7 and the SQRT of 3x?

Answer 23

you fell into the trap :-)

\[5a - a + 2a= \]

Answer 24

ohhh o 6 and the SQRT of 3

Answer 25

getting closer...but still not completely correct :-)

Answer 26

a different way to look at it:\[5\sqrt{3x}-\sqrt{3x}+2\sqrt{3x} = \sqrt{3x}(5-1+2) =\]

Answer 27

oh I left out the x in the SQRT of 3.
I was like what am I doing wrong?1 Lol

Answer 28

first rule of intelligent tinkering: save all the parts :-)

Answer 29

THANK YOU! I have another one just like the last.

~ SQRT5x - SQRT20x = SQRT80x

Answer 30

What's the ~ about?

Answer 31

Oh it's just what I use to tell the problem

Answer 32

\[\sqrt{5x} - \sqrt{20x} = \sqrt{80x}\]Are you supposed to solve for \(x\), or what?

Answer 33

wait it should be minus SQRT 80


Sorry..

Answer 34

so another simplify problem?

Answer 35

\[\sqrt{5x} - \sqrt{20x} - \sqrt{80x}\]
What are the factors of 5, 20, 80

Answer 36

5 can go into all
so SQRT 5
2 SQRT 5
4 SQRT 5

Answer 37

you're up to your old tricks :-)

Answer 38

Lol. sorry when you say that, that's what I think about.

Answer 39

what happened to the \(x\)?

Answer 40

x is in all of the SQRTs Sorry

Answer 41

attention to detail...lack thereof will get you wrong answers very quickly!

Answer 42

Okay. Sorry. So
SQRT 5x
2 SQRT 5x
4 SQRT 5x

Answer 43

So the final result is...

Answer 44

Would the answer be -5 SQRT 5x

Answer 45

You should use the equation editor button at the bottom left, it isn't hard...

Answer 46

Yes, \[-5\sqrt{5x}\]

Answer 47

\[-5\sqrt{5x}\]

Answer 48

Or \[-5\sqrt{5}\sqrt{x}\]

Answer 49

Thank you SOOOO much! You are amazing! That is all of my questions! THANKKK YOUU!!!

Answer 50

You're welcome! I hope it sticks with you :-)

Answer 51

Remember, don't lose the variables when simplifying :-)

Might be best to factor the \(\sqrt{x}\) out up front:
\[\sqrt{5x}-\sqrt{20x}-\sqrt{80x} = \sqrt{x}(\sqrt{5}-\sqrt{20} -\sqrt{80}) \]\[= \sqrt{x}(\sqrt{5}-2\sqrt{5}-4\sqrt{5}) = -5\sqrt{5}\sqrt{x}\]