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1.\(The/ function /f/ satisfies/ f(\sqrt{x + 1}) = \frac{1}{x} /for/ all/x \ge -1, x\neq 0. Find f(2).\)

2.\(Let/ f(x) = 3x^2 - 4x. Find/ the/ constant/ k/ such/ that/ f(x) = f(k - x) for/ all/ real /numbers/ x.\)

3.\(Find/ all/ complex /numbers/ z/ such /that/ z^2 = 2i. Write /your /solutions/ in/ a+bi /form/, separated/ by/ commas./ So,/ "1+2i, 3-i" /is/ an/ acceptable/ answer/ format,/ but/ "2i+1; -i+3"/ is/ not./ (Don't/ include/ quotes/ in/ your/ answer.)/ Note: /This/ problem /is /not /about/ functions.\)

4.\(Let/ f/ be /a /function/ such/ that/ f(x+y) = x + f(y) /for/ any/ two/ real/ numbers/ x/ and/ y/. If f(0) = 2, then/ what/ is/ f(2012)?\)

thanks

The surface area  is given by :

2 * [ 29 * 19  + 29 * 6.5  + 19 * 6.5 ]  =

2 * [ 551  + 188.5  + 123.5 ]  =

2 * [  863 ]  =

1726 cm^2 

OK, Rick  !!!!

[ You may have to remind me of your name from time-to-time  !!! ]

Try this, Sam  

Thanks a ton!!!

OK, Pheonix.....!!!

Here again, GM....a graph might be the best way to go

y  = 0.7x^2 - 4.3x + 5.5

See the graph, here :

https://www.desmos.com/calculator/gguuvypnfx

This occurs in about the 8th year after 1985    [ in 1993 ]

Thank you so much! 😊

This time I am no use. I don't understand this one.

I was looking around and saw that this question has been solved before. The previous answer stated

"Begin by moving F a little to the left so that the square has begun it's rotation.

Call the angle EFB theta, call the side of the square b, and BF and FC x and y respectively.

Fill in the angles of the two triangles BEF and FGC and apply the sine rule in both triangles.

Use the fact that x + y = a, the length of the side of the equilateral triangle to deduce that

.

So the size of the square varies as it rotates.

Now look at your larger diagram, the one with the dotted line running from D.

Call the point where this line meets BC N, then  DN/DF = sin(angle NFD).

DF  is the diagonal of the square = b.sqrt(2) and angle NFD = theta + 45 deg, so you can now work out DN.

You should find that it's a constant, equal to  a.sqrt(3)/(1 + sqrt(3))." 

Made by Guest.

Sector Area  =  (1/2)r^2 * theta (in rads)

So we have :

(1/2) (10^2) (pi/5)  =

(1/2) (100/5) pi =

(1/2) (20) pi  =

10 pi  units^2 

Here's a few :

1. 

A(2)  - A(1)  =

[ 3  - 2(2)^2 + 4^2 ]  - [ 3 - 2(1)^2 + 4^1]

[ 3 - 8 + 16]  -  [ 3 - 2 + 4]

[ 11 ]  - [ 5]

6

3.     [ 3x - 7]  / [ x + 1]

We can put any x into the function  except  x  =  -1

So....the domain is  (-infinity , -1) U (-1, infinity )

4. [ 3x - 7 ] / [ x + 1 ]

Since we have a same power polynomial/ same power polynomial  we will have a horizontal asymptote   at    [ 3x ] / x   =    3

So....the range  is   (-infinity , 3) U (3 , infinity )

5.  

3  =  [ 2 - 3a ]  /  [ 5 - 2a ]     multiply both sides by 5 - 2x

3 [ 5 - 2a]  = 2 - 3a

15 - 6a  = 2 - 3a      add 6a to both sides, subtract 2 from each side

13 = 3a  divide both sides by 

13/3  =   a

Yeah thanks for that.

Because when you double the dimensions of the box, the volume goes up as cube or 3rd power of the dimensions. So, assume that the dimensions of the small box were: 1, 2, 3 inches. Then the volume would be: 1 x 2 x 3 = 6 in^3. Now, suppose you double the dimensions of the jumbo box to: 2, 4, 6. Then the new volume of the jumbo box would be: 2 x 4 x 6 = 48 in^3. So now, you divide the volume of the jumbo box by the volume of the regular box and you get: 48 / 6 = 8 times larger than the smaller box. Do you see it now?.

How did you get that answer?

The volume of the jumbo box will go up by: 2^3 = 8 times. Therefore, it should hold:

8 x 750 =6,000 toothpicks.

Let's solve this with a graph, GameMaster

Look here :    https://www.desmos.com/calculator/gt2fmgddhr

Note that 164 billion was reached after about 4.978 years ≈ 5 years  =  about 1995

I know. I saw that your image had changed.

its right thanks also I am that guest you were helping with the volume questions by the way 

I'm not sure if I am right but I think the answer is 1500.

No problem. Just trying to help.

you are the best thanks