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No!  0/0  is undefined.  

This is a little bit different from the normal way of substituting, but it is still substution!

x - 2y  =  -2

3x - 2y  =  2

(x - 2y) - (3x - 2y)  =  (x - 2y) - (3x - 2y)

                                                              Since  x - 2y = -2 , we can substitute  -2  in for  x - 2y .

(x - 2y) - (3x - 2y)  =  (-2) - (3x - 2y)

                                                              Since  3x - 2y = 2 , we can substitute  2  in for  3x - 2y .

(x - 2y) - (3x - 2y)  =  (-2) - (2)

                                                   Distribute the  -1 .

x - 2y - 3x + 2y  =  -2 - 2

                                                   Combine like terms.

-2x  =  -4

                                                   Divide both sides by  -2 .

x  =  2

Now we know what  x  is. Let's plug this value into the first equation to find  y .

2 - 2y  =  -2

                      Subtract  2  from both sides.

-2y  =  -4

                      Divide both sides by  -2 .

y  =  2

No......any fraction that can be written with a denominator of 2  or  5 -   or both - raised to some  positive integer power  will always terminate

To see this....note that 1/2  = .5

But  a terminating decimal raised to any positive power will itself terminate

Likewise...... 1/5  = .2     and for the same reason, this raised to any positive integer power will also terminate

So.....their products will also terminate

1/40  =   (1/2)^3 * ( 1/5)^1  =   .5^3 * .2  =  .025

Since the domain of  f(x)  is  -1 < x < 1  ,

Then the domain of  f(x + 1)  is  -1 < x + 1 < 1 .

-1 < x + 1 < 1   simplifies to

-2 < x < 0

So the domain of  f(x + 1)  is  (-2, 0) ,

And  g(x) = f(x + 1)  so the domain of  g(x)  is  (-2, 0) .

So, you want to start by starting with 13 cookies. 

So,

OOOOOOOOOOOOO

ok, so now take away 4 away from the cookies draw above,

OOOOOOOOO

You will be left with 9 cookies. 

Hope this helped!

h (x)  = 1 / √x

y = 1 / √x     square both sides

y^2  = 1/x  

x = 1/y^2   swap x and y   and for y wriite  h^-1(x)

y =  h^-1(x)  = 1 / x^2

And     h^-1(4)  =  1 / 4^2  =  1 / 16

Then I suppose the domain is not allowed to be restricted.....so....there is no inverse in this case

Q  =  (2, 7)

Rotated 270° clock-wise  around M  ⇒  ( 0, 5 )

Reflected across   x = 1   ⇒  ( 2, 5 )  

I think that's incorrect

Not a strict inverse, since it's not one-to-one

However.....it we restict the domain to [-1, inf )   we have

y = 4x^2 + 8x + 13

y - 13 + 4  =  4[ x^2 + 2x + 1]

y - 9  =  4 ( x + 1)^2

[ y - 9] / 4   =  (x + 1)^2

√ [ (y - 9) / 4 ]  =  x + 1

√ [ (y - 9) / 4 ]  - 1  = x     swap x and y    and for y write f^-1 (x)

√ [ (x - 9) / 4 ]  - 1  = y  =  f^-1 (x)

So  f^-1 (25 )    =   √ [ (25 - 9) / 4 ]  - 1  =  √ [ 16 / 4 ]  - 1   = √4 - 1  = 2 - 1 =   1

Hi Mr O

I answered a question very simiar to this a day or two ago. I seem to recall you left a comment too..

Anyway, I'd like to see you attempt these questions, one at a time if you want, and then maybe we can talk about your answer/s.

Why don't you start with b, that looks like the easiest one because the only thing that matters is how many balls are in the boxes.

That answer is up to the individual. Just do your best!

Let be clear this question up a bit by adding my own characters.

Angela has 13 cookies, and Peter asks for 4 cookies, how many cookies does Angela have left?

Now this answer depends on what happens next.

Scenario #1: Angela is selfish and keeps all the cookies for herself. 

If Angela keeps all the cookies then Angela will still have 13 cookies and Peter will have 0 cookies :-(

Scenario #2: Anglea is kind and shares the 4 cookies with Peter.

If Anglea shares 4 cookies with Peter then Angela will have 9 cookies and Peter will have 4 cookies.

Hope that helps!

 <---- You

I think it is fair to assume that each candy must go into a lunch box.   :)

I'm pretty sure I'd still have 13 cookies.

Will do, Mr. Owl. Will do. 

How did you even lose any fingers

ah thank youu

yes thank you

If you press "enter" again it will turn it into scientific notation.

For example, If you type  20 * 300  and press enter, it gives you  6000 .

then press enter again and it becomes  6 × 10³ .

Also if you just type  35600  and press "enter" it turns it into  3.56 × 10⁴

I think what you mean is scientific notation right?

There is a formula that relates the number of sides of a regular polygon to its interior angle. It is the following:

\(\frac{180(n-2)}{n}\)

This formula will tell you the interior angle measure, if given the number of sides. However, we have the interior angle measure! So, just solve for n, the number of sides.

Therefore, this polygon has 36 sides.

The answer is two. 

Wow! That was good thinking, X² !!!  

I am ??!!??!?!?!?!?!!? Well, thank you!

I wish I could explain how I got that answer. 

I will give my best attempt, I guess.

This is just one side of the table. I have color coded the O's such that if you land on an O as your second-to-last letter, then you will have a certain number of possibilities. 

A red O means there are 3 possibilities.

A Blue O means there are 2 possibilities.

Hopefully, you can see why this is the case.

Let's start with the easiest: The "C" in the corner. Before I start, I will use my own style of Cartesian coordinates where it represents the intersection of a column and a row. For example, \((C1,R1)\) is (Column 1, Row 1), for short. In this case, this letter happens to be L.

There is only one case to consider now for the corner "C." It is the following:
 

1) \((C5,R4)\Rightarrow(C4,R3)\Rightarrow(C4,R2)\)

Look at that! I have landed on a red O, which signifies 3 possibilities. This means that the corner "C" has 3 possibilities. There are 4 instances of the corner "C," so 4*3=12

Time to consider the next case: The "C" adjacent to the corner "C"

Now, let's consider how many cases there are for the sequence \((C4,R4)\Rightarrow(C3,R3)\). Well, the number of possibilities is equal to the sum of the number of possibilities its neighbors are immediately adjacent to. 

\((C3,R3)\)  is adjacent to 2 red and 2 blue O's. Because there are 3 possibilities for a red one and 2 possibilities for a blue one, the number of possibilities is \(3+3+2+2=10\). However, this is only one sequence. Let's consider the next sequence of \((C4,R4)\Rightarrow(C4,R3)\Rightarrow(C4,R2)\). Oh look! This is a red "O," which has 3 possibilities, so let's add the number of possibilities together.

\(10+3=13\)

There are 4 of these in the diagram, so \(13*4=52\)

Now, let's consider the center "C." Well, we can use the same logic as before to know that \((C3,R4)\Rightarrow(C3,R3)\) has 10 possibilities. We know that there are 2 avenues to red O's, which equals 6 additional paths.

In total, that equates to \(10+6=16\) ways. There are two instances of there, so \(2*16=32\)

The last step is to add the numbers together. \(12+52+32=96\) ways.