Kelly likes to play a game with her friends where she stands in one spot with one end of a jump rope in her hand. She then twirls the rope around on the ground in a circle. Her friends jump over the other end of the rope as it comes near them. In this game, does the rope more closely resemble a radius or a diameter of a circle?
Question 2 options:
radius
diameter

Question

Kelly likes to play a game with her friends where she stands in one spot with one end of a jump rope in her hand. She then twirls the rope around on the ground in a circle. Her friends jump over the other end of the rope as it comes near them. In this game, does the rope more closely resemble a radius or a diameter of a circle?
Question 2 options:
	radius
	diameter

e.mccormick · Answer

OK, she is at the center and, "twirls the rope around on the ground in a circle."  So, what goes from the center to the edge? A radius or a diameter?

anonymous · Answer

RADIUS @e.mccormick

e.mccormick · Answer

Yep.  It was just looking at what was said and realizing how she would have to stand to twirl it....  Also means she has to hop.  Hehe.

anonymous · Answer

LOL your right never would have thought of that you think you could help me with something else ?

e.mccormick · Answer

Probably.

anonymous · Answer

Find the exponential Inequality

52x + 2 < 625

e.mccormick · Answer

I don't see an exponent there...

anonymous · Answer

52^x + 2 < 625

e.mccormick · Answer

Ah.

anonymous · Answer

its fine you think you can help with this though ?
256 = (1/2x - 3)2?

e.mccormick · Answer

(1/2x - 3)2
$(\frac{1}{2}x - 3)^2$
$(\frac{1}{2x} - 3)^2$

e.mccormick · Answer

Which is it?

anonymous · Answer

1
---^2
2x-3

e.mccormick · Answer

OH!  The enitere thing under the fraction....

On the $52^x + 2 < 625 $ one, have you done logs yet?  Typically, you would get it into a solvable form: $52^x < 623 $ then you would do $\log_{52}$ to both sides.

e.mccormick · Answer

$256 = \left(\cfrac{1}{2x - 3}\right)^2$

e.mccormick · Answer

Start by taking the root of both sides.  That should get you a more useful version.

anonymous · Answer

still confused

e.mccormick · Answer

The problem of the root is that it can make two anwers that need to be tested.  The squared value there could be negative or positive!

e.mccormick · Answer

Well, you have something on the left and right of the = sign.  The part on the right is squared.  So you need to start by taking the root toget rid of the square.

anonymous · Answer

it would be 16 right ?

e.mccormick · Answer

On he left, yes.

e.mccormick · Answer

$256 = \left(\cfrac{1}{2x - 3}\right)^2\implies 16 = \cfrac{1}{2x - 3}$
Now, can you solve that?  Once you do, I will show you the one thing you need to look out for because we took a root.

anonymous · Answer

still not understanding

e.mccormick · Answer

$16 = \cfrac{1}{2x - 3}$  Can you solve that for x?

magbak · Answer

It is a radius.

e.mccormick · Answer

@magbak Long past that part.  Hehe

anonymous · Answer

um im sorry but im still having a hard time

e.mccormick · Answer

Well, what is giving you a hard time.  What do you think needs to happen next?  Solving for x means getting x alone on one side of the = and everything else on the other.

anonymous · Answer

i feel like i would divide 2 into 16

e.mccormick · Answer

Well, that is kind of the right idea.  Especially if you are thinking the next issue is getting rid of the fraction.  But everything on the bottom of the fraction has to stay together.

Have you looked at multiplying by the inverse or things like that to cancel a fraction?

e.mccormick · Answer

Here are some ( ) added just to show how the bottom of the fraction is all one big thing right now.
$16 = \cfrac{1}{(2x - 3)}$

anonymous · Answer

okay

e.mccormick · Answer

So, how can you get rid of that whole fraction?  Why not look at an easier fraction and see what we can do.

$16=\cfrac{1}{a}$
If we wanted to get the a out of the bottom of this, we could do what?  As I said, there is multiplying by the inverse.  Some people also like cross multiplication.

anonymous · Answer

multiply 16 times 1

e.mccormick · Answer

Well, that would leave the a on the bottom...

anonymous · Answer

imultiply a by 16

e.mccormick · Answer

Much better choice.  In fact, you would multipy it all by a.  Let me write that up...

e.mccormick · Answer

$16=\cfrac{1}{a}\implies a\cdot 16=a\cdot \cfrac{1}{a}\implies  16a=\not{a}\cdot \cfrac{1}{\not{a}}\implies 
$

$16a=1$

e.mccormick · Answer

See how that worked?

e.mccormick · Answer

Now, you can do the same thing to this: $16 = \cfrac{1}{(2x - 3)}$

One difference.  Instead of just a, it is the entire $(2x - 3)$ you multiply through by.

e.mccormick · Answer

I'm getting called away for a work emergency.  Seeing if someone can take over helping here.

@satellite73
@jim_thompson5910

anonymous · Answer

@e.mccormick 16(2x-3)

e.mccormick · Answer

Yes, 16(2x-3)=1

Now you need to distribute the 16.  The other part I was mentioning before is this:

As you know $16^2=256$.  However, it is also true that $(-16)^2=256$!  So because we took a root, there is a second calculation that needs to be made.

$256 = \left(\cfrac{1}{2x - 3}\right)^2\implies -16 = \cfrac{1}{2x - 3}$

This will have a different, second answer.  And it is just as valid!  You can check both of them and find that they do come out to 256 if you put them in the original equation.


I pre-wrote that for when we were done with the one, but I have to go get intenet working for a few dozzen people, so I can't stay!  Urgh!  I wanted to work through all this with you.

e.mccormick · Answer

I'm getting called away for a work emergency.  Seeing if someone can take over helping here.

@ParthKohli
@mathslover

anonymous · Answer

thank you sooooo much your a great help!