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Alan, why can you say that theta tends to zero ?

Why can't (x, y) tend to (0, 0) along some direction other than the x-axis ?

(That's not to say that your answer is incorrect).

Here's a 3D graph to illustrate:

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The limit is zero.  See below:

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asin(0.6428) = 40.000926731461 (degree)

I'm taking this to be :

250=50e^[0.06x]  divide both sides by 50

5 = e^[0.06x ]  take the natural log of both sides

ln 5 = ln  e^[0.06x ]  and we can write this as

ln 5 = [ 0.06x ]* [ ln e  ]      and ln e = 1  .....  so we can ignore this

ln 5 =  0.06x     divide both sides by 0.06

ln5 / 0.06 =  about  26.824

-4(-4w-y)-3(y-4w)      distribute the minus 4   over the terms in the first parentheses and the -3 over the terms in the second parentheses

-4(-4w) -4(-y)   -3(y) -3(-4w)  =

16w  + 4y   - 3y   + 12w      combine like terms

16w + 12w + 4y - 3y   =

28w + 1y     or just

28w + y

3.5 * 3 = 3.5 * (1 + 7.0/100)^n Divide both sides by 3.5

3=1.07^n take the log of both sides

n=log(3) / log(1.07)

n=16.24

It is to show that the answer does not have to be answered by the one way.  There are more than one way to solve a problem.

x^2 + 4x = 12      subtract 12 from  both sides

x^2 + 4x - 12  = 0   factor

(x + 6) (x - 2)   = 0         set both factors to 0  and we get that x = -6 and x = 2

Complete the square

x^2 + 4x  =  12        take 1/2 of 4 = 2......square it   = 4   and add to both sides

x^2 + 4x + 4  = 12 + 4      simplify  the right, factor the left

(x + 2)^2    = 16              take the pos/neg square roots of both sides

x + 2    =  +/- square root (16)

x + 2  = +/- 4

We have two equations

x + 2 =  4                        and           x + 2   =  -4

Subtract 2  from both sides

x = 2                            and          x = -6                  [just as before]

Quadratic Formula  ......put the eqatuion in standard form     .....   x^2 + 4x - 12  = 0

   a = 1, b = 4 and c =  -12        and we have

  x = [ -4   +/-  square root [ 4^2 -(4)(1)(-12)]  ]   /  [2 (1)]

x = [ -4 +/-  square root [ 16 + 48] ] / 2

x =  [ -4 +\- square root [ 64]  ] / 2

x =  [ -4 + 8] / 2    =  4 / 2  = 2

And

x =  [ -4 - square root (64)] / 2   =

x = [ -4 - 8] / 2       =   -12 / 2     = -6         just as before......

gibsonj338:

You got nothing else to do? What is the point of solving a quadratic equation 3 times??

Solve for x:
250 = 53.0918 x

250 = 53.0918 x is equivalent to 53.0918 x = 250:
53.0918 x = 250

Divide both sides of 53.0918 x = 250 by 53.0918:
(53.0918 x)/53.0918 = 250/53.0918

53.0918/53.0918 = 1:
x = 250/53.0918

250/53.0918  =  4.70882:
Answer: |  x = 4.70882

Thanks so much for your help!! I am horrible at math. 

-7(4y-w-2)   distribute the -7 over the terms in the parentheses

-28y + 7w + 14

And that's all we can do......

10n = 50n     subtract 50n from both sides

10n - 50n = 0   factor out the n

n (10 - 50) = 0    simplify

n ( -40) = 0      divide both sides by -40

n = 0

-30= 6w   divide both sides by 6

-5 = w

or x=6

What is -4-+2? In a number line

-4 - +2 = -4 - (+2) = -4 -2 = -6

\({x}^{2}+4x=60\)

Two ways to do it

Way One: factoring

\({x}^{2}+4x-60=0\)

\((x+10)(x-6)=0\)

\(x=-10\) or \(x=6\)

Way Two: completing the square

\({x}^{2}+4x+4=64\)

\({(x+2)}^{2}=64\)

\(x+2=±8\)

\(x=-2±8\)

\(x=-10\)​ or 

10.5 = 3.5 1.07^n

10.5 = 21/2 and 3.5 1.07^n = 7 2^(-1-2 n) (107/25)^n:
21/2 = 7 2^(-1-2 n) (107/25)^n

21/2 = 7 2^(-1-2 n) (107/25)^n is equivalent to 7 2^(-1-2 n) (107/25)^n = 21/2:
7 2^(-1-2 n) (107/25)^n = 21/2

Divide both sides by 7:
2^(-1-2 n) (107/25)^n = 3/2

Take the natural logarithm of both sides and use the identities log(a b) = log(a)+log(b) and log(a^b) = b log(a):
log(2) (-1-2 n)+log(107/25) n = log(3/2)

Expand and collect in terms of n:
(log(107/25)-2 log(2)) n-log(2) = log(3/2)

Add log(2) to both sides:
(log(107/25)-2 log(2)) n = log(3/2)+log(2)

Divide both sides by log(107/25)-2 log(2):
Answer: |  n = (log(3/2)+log(2))/(log(107/25)-2 log(2)=16.238

Just multiply it out    379 x  . 28         Use the calculator.....

(BTW : I do not see the correct answer....did you enter answer "B" correctly? )

(x+10)(x-6) = 0

Of note.....the water will have to be under pressure or some of it will turn to steam.

If steam is allowed then the max temperature reached will be 100 degrees C and some of the water will be turned to steam at 100 degrees

How much?

The latent heat of vaoprization of water is 2230 j/gm   so if we are making steam the equation becomes:

300(.38)(800- 100) = 50 (2.05)(25) + 50(334) + 50(4.188)(100)  +  Wm(2230)   where Wm = mass of water turned to steam

Wm = 17.75 gms of steam and the entire system remains at 100 C

Assuming 25 degrees (room temperature)  Kw = 14  (self ionization of water)

[OH-]  =  Kw/[H+]  =  14/(2 x 10^-3) =  7000      I think!

Something is wrong here.....  divide both sides by n and you get   :    .10 = 50      No Go !

Maybe you mean    .1 n = 50     then n= 50/.1 = 500

-43 degrees     or    317 degrees   or   223 degrees