All Answers 359166 Answers

4(a - 3b)

Replaceing a with -4 and b with 6:

4( -4 - 3(6) )

4( -4 - 18 )

4(-22)

-88

2x + 3y  =  5

If you are solving for y, move the x-term fromm the left side by subtracting 2x from both sides:

     3y = 5 - 2x

Now, divide both sides by 3:

      y  =  (5 - 2x) / 3

Thc calculator at this sie will evaluate this for you.

But first, is the missing parenthesis important?

The answer is close to 0.07. 

I'm not sure what you're trying to do.

If you are trying to make one fraction out of it, change the + 1 into + x²/x²,

then 1/x² + 1 becomes 1/x² + x²/x²  = (1 + x²) / x²

More questions -  315 you can do these so perhaps you can tell me how please.   

Otherwise maybe Chris or Alan would be able to assist please.

b)  I think I can do this one    

If it approaches from above then the limit will be 1

Ig it approaches from below then the limit will be -1

therefore the limit is undefined.

I just worked out a.   you have to multiply top and bottom by the conjugate of the top and then it just falls out.

But what about c?    

2.5 MV  =  2.5 megavolts

d.n.e. = does not exist

okay.

I am really bad at limits.  I'll state that up front.

d)   I agree that d is undefined thaough I do not know what d.n.e stands for.   

f)   I get the same as Chris too.

  $$\displaystyle\lim_{x\rightarrow\infty}\;\;\frac{x+1}{2x^2+1}\\\\
=\displaystyle\lim_{x\rightarrow\infty}\;\;\frac{1}{4x+1}\\\\
=0\\\\\\
=\displaystyle\lim_{x\rightarrow\infty}\;\;e^0\\\\
=1$$

e) I'll admit to learning this one straight from CPhill.  Thanks Chris

$$\displaystyle\lim_{x\rightarrow\infty}\;\frac{3-cosx}{x^2+10}}\\\\
=\displaystyle\lim_{x\rightarrow\infty}\;\frac{3}{x^2+10}}\;\;-\;\;\displaystyle\lim_{x\rightarrow\infty}\;\frac{cosx}{x^2+10}}\\\\
=\;0\;-\;\;\displaystyle\lim_{x\rightarrow\infty}\;\frac{cosx}{x^2+10}}\\\\
=\;-\;\;\displaystyle\lim_{x\rightarrow\infty}\;\frac{cosx}{x^2+10}}\\\\
$Now $-1\le cosx\le 1\\\\
$So $\\\\
\displaystyle\lim_{x\rightarrow\infty}\;\frac{-1}{x^2+10}\;\;\le \displaystyle\lim_{x\rightarrow\infty}\;\frac{cosx}{x^2+10}\;\le\displaystyle\lim_{x\rightarrow\infty}\;\frac{1}{x^2+10}\\\\
0\;\;\le \displaystyle\lim_{x\rightarrow\infty}\;\frac{cosx}{x^2+10}\;\le\;0\\\\
$therefore$\\\\
\displaystyle\lim_{x\rightarrow\infty}\;\frac{cosx}{x^2+10}\;=\;0\\\\$$

$$\\$therefore$\\\\
\displaystyle\lim_{x\rightarrow\infty}\;\frac{3-cosx}{x^2+10}}=0-0=0\\\\$$

Thank you for showing me that Chris       (This answer is exactly the same as Chris's)

 dangerous

97a^3+0.564a=0  ....  factoring, we have

a(97a^2 + .564) = 0     the only real solution is a = 0

Note substituting x for a and using the on-site solver, we have

97x^2 + .564 = 0

$${\mathtt{97}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{0.564}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{{\sqrt{{\mathtt{141}}}}{\mathtt{\,\times\,}}{i}}{\left({\mathtt{5}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{970}}}}\right)}}\\
{\mathtt{x}} = {\frac{{\sqrt{{\mathtt{141}}}}{\mathtt{\,\times\,}}{i}}{\left({\mathtt{5}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{970}}}}\right)}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = -{\mathtt{0.076\: \!252\: \!429\: \!401\: \!893\: \!3}}{i}\\
{\mathtt{x}} = {\mathtt{0.076\: \!252\: \!429\: \!401\: \!893\: \!3}}{i}\\
\end{array} \right\}$$

9 = 3²

No, it doesn't work on mine either!

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Let's simplify -1+i/i+1   multiply numerator and denominator by i-1...so we have

[i - 1][i-1]/[(i+1)(i-1)] = (i^2 - 2i + 1)/(i^2 -1) = (-2i) / (-2) = i =  0 + 1i

tan(1/0) = undefined = pi/2

So...arg(z) = pi/2 + n*(2pi) for n = 0, ±1,±2, ±3.......

What is between the 3^(x+2) and the 2x^(x+1)?  A plus, minus, multiply ...?

Or are you trying to find the value of 12*6^x + 3^(x+2) Given that 2x^(x+1) = 5?

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$$\frac{-1+i}{i+1}=\frac{(-1+i)(i-1)}{(i+1)(i-1)}$$

$$\frac{(-1+i)(i-1)}{(i+1)(i-1)}=\frac{-i+1+i^2-i}{i^2-i+i-1}$$

$$\frac{-i+1+i^2-i}{i^2-i+i-1}=\frac{-2i}{-2}$$   because i² = -1

$$\frac{-2i}{-2}=i$$

Now arg(i) = 1  because arg(x+iy) = √(x² + y²)

arg(i) = √(0² + (1)²) = 1

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560/100=a^17

We can write this as

28/5 = a^17       take the log of both sides

log(28/5) = log a^17    and we can write

log(28/5) = 17 log a     divide by 17 on both sides

log(28/5)/17 = log a     and exponentially, we have

10^log(28/5)/17 = a = about 1.10665

2x^2-12x+13

This polynomial is irreducible into linear factors......thus it is not "factorable"

d) lim x→3 [1/(x-3) - 9/(x² - 9)]  Notice that this simplifies to

lim x→3 [(x - 6)/(x² - 9)] 

Note that when

x → 3^- .... [x - 6)/(x² - 9)] →  ∞   and when x → 3⁺  [x - 6)/(x² - 9)] →  - ∞

Thus .......lim x→3 [x - 6)/(x² - 9)]   =  d.n.e.

e)  lim x →∞ [(3 - cosx)/ (x² + 10)] .......note that we can write this as

lim x →∞ [(3 / (x² + 10)] - lim x →∞ [( cosx)/ (x² + 10)]

The first is easy.... it just evaluates to 0

For the second, note that

-1 ≤ cosx ≤ 1      and dividing everything by x² + 10, we have

[-1/x² + 10] ≤ [cosx/x² + 10]  ≤ [1/x² + 10]

Note that, as the two outside functions appproach infinity, they both approach 0. So, the middle function is "Squeezed" and it approaches 0, as well.

Therefore......  lim x →∞ [(3 - cosx)/ (x² + 10)]  =  0

f) lim x→∞ e ^{[(x+ 1)/(2x^2 + 1)]}

This one is easy....note that.... lim x→∞ [(x+ 1)/(2x^2 + 1)] = 0

So ....lim x→∞ e ^{[(x+ 1)/(2x^2 + 1)]}  =  lim x→∞ e⁰  = 1

I'm not great at limits......but  I think these might be correct.....

Thanks 315,

That is wonderful.

Congratulations!

antilog 2.1705 = 10 ^2.1705 = about 148.08123

geno has given you the correct wavelength.  The frequency is obtained by dividing the energy by h.

$${\mathtt{frequency}} = {\frac{{\mathtt{2.93}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{\left(-{\mathtt{25}}\right)}}{\left({\mathtt{6.626\: \!069\: \!3}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{\left(-{\mathtt{34}}\right)}\right)}} \Rightarrow {\mathtt{frequency}} = {\mathtt{442\,192\,779.360\: \!155\: \!499\: \!731\: \!945\: \!2}}$$

frequency ≈ 4.422*10⁸ s^-1

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Kinetic energy is (1/2)m.v²

m is mass and v is speed.

Athlete: KE = (1/2)*50*3²    KE =  225 joules

Cheetah:  KE = (1/2)*25*30²   KE = 11250 joules

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$$\begin{array}{lcl}
\mbox{For } f(x)=x^2-1 \mbox{ and } g(x)=x-3 \mbox{ , calculate } g(f(2)).\\
\mbox{Note that } g(f(2)) \mbox{ means that we calculate } f(2) \mbox{ and we fill in this answer in function } g(x)\\
\mbox{Then}\\
f(2) = 2^2-1 = 4-1 = 3\\
\mbox{Hence } g(f(2)) = g(3)\\
g(3) = 3-3 = 0\\
\mbox{ So } g(f(2)) = 0


\end{array}$$

Note that there is another root:

(Get the other root by taking -27 as the square root of 729).

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Congratulations xvxvxv!

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