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It will have a horizontal asymptote  at   y = 3

The range will be  ( - infinity, 3) U ( 3 , infinity )

y=2e^x find y if x=3 e is 2.71828 how do I solve this.

y=2e^x find y, if x=3

Just sub 3 for x as follows:

y=2e^3

y=2* 20.0855..........

y=40.1710................

This is a parabola that turns upward

Its minimum y value occurs at its multiple root of x = -2 

So.....the range is [ 0, infinity )

Yes they are identical :)

\(y^x=x^y\\ log(y^x)=log(x^y)\\ xlogy=ylogx\)

I just realized that the wording in your description is erroneous.

It says

x log base y = y log base x

You need another number or variable after the word 'base'.

Or you can assume they are the same base, which is what I did and leave the word base out altogether.

P (x) = a (x - 2) (x - 4)     and we need to find "a"

And when x =  3  we have that 

 a( 3 - 2) (3 - 4)  = 6    simplify   and solve for "a"

a (-1) = 6  →   a =  -6

So we have

-6 [ x^2 - 6x + 8]  =

P(x) =  -6x^2 + 36x - 48

I don't think that's correct

Call the  smallest angle measure, N

And the largest  N + 44

And the sum of the interior angle measures of a pentagon  = 540°

So

N + (N + d) + (N + 2d) + (N + 3d) + (N + 4d)  =  540 

Where d is the common difference....simplifying, we have that

5N + 10d  = 540     (1)    and

N + 44  = N + 4d   →  44 = 4d →  11 = d    sub this into (1)

5N + 10 (11) = 540

5N + 110  = 540

5N  = 430       divide through by 5

N  = 86°

(b + e) + 4 + ( b + d) + 6 +  (a + c) + 5 + (a + d) + 8 + ( c + e) + 7  = 28

(b + e)  + ( b + d)  +  (a + c)  + (a + d)  + ( c + e) + 30  =  28

(b + e)  + ( b + d)  +  (a + c)  + (a + d)  + ( c + e) = -2

2 [ a + b + c + d + e]  =  -2          divide both sides by 2

a + b + c + d + e  =   -1

                             1/5

                             4/5

                             1/5

                             4/5

 4/5

P ( one 6)  =  P( of not 6 on first roll and 6 on the second)  + P(6 on the first roll)  = 

                                    4/5 *  1/5                      +                              1/5

                                         4/25                        +                             5/25

                                                                      9/25

P.S.  -  Pretty sure about the first part.....unsure about the second....!!!!!

The Art of Problem Solving has begun selling a cookbook called "What Would Euler Eat?" If the price of the cookbook is n dollars n <= 72, then it will sell 720 - 10n copies. What price (in dollars) will maximize the total revenue we receive for the books?

A = n(720 - 10n)

A = -10n^2 + 720n

A = -10(n^2 - 72n)

-A/10 = n^2 - 72n

Complete the square.

-A/10 + 36^2  = n^2 - 72n + 36^2 = (n - 36)^2

Subtract 36^2 from both sides.

-A/10 = (n - 36)^2 - 36^2

and multiplying by -10 gives:

A = -10(n - 36)^2 + 12960

The total revenue is maximized when the price is $36.

I see you are taking Introduction to Algebra B... 

12x^2 + 12xy + 3y^2

3 [ 4x^2 + 4xy + y^2]

3 [ (2x + y) (2x + y) ]

3 ( 2x + y) ^2

Small note - I used y^x = x^y, but either way, it comes out the same. I tested both of them, and they are (to the limits of Desmos's calculations) identical.

Well....if the student is to keep only the amount deposited the last day, Option B is by far better

But.....I think the problem's intent is to show the total amount in both accounts after the last deposit and to compare those...

Guest is correct as to the amount on deposit at the end for option A = $ 32500

But for Option B....the total amount on deposit at the end for Option B  is :

$ [2^15  - 1 ] =  $32767

Thus.....Option B is better by $267

I think your teacher made a mistake in Option 2. I believe he/she computed 2^15 =$32,768 as the amount of money in the account on the 15th day. But that is NOT so, if you begin with a deposit of $1 on day 1 and on the 15th day, you will have:2^(15-1) =$16,384. You can easily list them and see for yourself: 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1,024, 2,048, 4096, 8,192, 16,384(15th day deposit). I think what they did was simply raise 2^15 =$32,768. This would have been true if the first day's deposit was $2 instead of $1.

Optio A is a straightforward as: $750 x 30 + $10,000 =$32,500(on 30th day)

Because a mistake appears to have been made, then Option A is obviously the better deal.

It appears that we are just multplying

767  x  .7   =  536.9

Thanks, Melody.....!!!!!

Type  "x"

Hold down the shift key and hit the "6" key

Release shift key and type "2"

Should look like this →   x^2

3x-2y=2  (1)

5x-5y=10   divide through by 5 → x - y  =  2 ..add y to both sides → x = 2 + y  (2) 

Sub (2) into (1)

3(2 + y) - 2y  = 2

6 + 3y - 2y  = 2

6 + y  = 2    subtract 6 from both sides

y = -4

And using  (2),   x = 2 + y →  2 + (-4) →  -2

x^4+5x^3-15x^2-5x+14  = 0       

Write -15x^2  as   -14x^2  - x^2

x^4+5x^3-14x^2 - x^2 -5x+14 = 0

x^2(x^2+ 5x - 14) - 1(x^2 + 5x - 14) = 0

(x^2 - 1) ( x^2 + 5x - 14) = 0

(x + 1) (x - 1) (x + 7) ( x - 2)  = 0

Set each factor to 0  and solve for x  and we get that x = -1, 1, - 7 or 2

3.2 = 8e^-0.51t  

Take logs of both sides

ln(3.2) = ln(8e^-0.51t)

Use the properties of logs to express this as

ln(3.2) = ln(8) - 0.51t

Rearrange:

0.51t = ln(8) - ln(3.2)

0.51t = ln(8/3.2)

t = ln(8/3.2)/0.51 ≈ 1.8 s

This factorises as:

(x - 1)(x + 1)(x - 2)(x + 7) = 0

I'm sure you can take it from there!

\(x^2+mx+n\\ =(x+3)^2\\ =x^2+6x+9 \)

So   m=6

"If someone could, can you tell me how i should approach a similar problem like this. I know how to plot most of them but the infinites are throwing me off."

Here is one possibility:

.

The formula at the top confused me too.

I don't know for sure what it is supposed to mean, but if it said

\(m_{sec}=\frac{f(x)-f(2)}{x-2}\)                  then I think it would make sense.

See i was doing this exact thing, but why does that formula have the (4), where does it come from? Other than thanks for the break down, i was trying to figure this out, and if it wasnt for that for it would have been fine. cause no matter what i put in it it, i was getting answer of 1.