NinjaDevo

There's actually an option to switch the calculator to "fractions."

The steps to what? To graphing? To solving for x?

$$\begin{array}{l}
\mbox{Depends on what n is.}\\
\mbox{If $n = 0$, then $n^\infty = 0$}\\
\mbox{If $n = 1$, then $n^\infty = 1$}\\
\mbox{If n equals anything else though, I believe the answer would be infinity.}
\end{array}$$

Start by doing the first part.

(4x10^-5)^{^2}(2.63x10¹⁷)÷6.7x10³ 

Raising something to the second power means to multiply it by itself

(4x10^-5)(4x10^-5)(2.63x10¹⁷)÷6.7x10³        

Now we can do 4 x 4 and 10^-5 x 10^-5, as long as we multiply those two results together at the end.

For example, say we have (10x5)(10x5) Here we can multiply 10x10, then 5x5, then multiply them together.

10x10=100

5x5=25

100x25=2500

We'll get the same result if we multiply in the parentheses first.

(10x5)(10x5)

(50)(50) 

2500

So, this shows that we can multiply 4 x 4 and 10^-5 x 10^-5 next.

(4x10^-5)(4x10^-5)(2.63x10¹⁷)÷6.7x10³  

(4x4 x 10^-5x10^-5)(2.63x10¹⁷)÷6.7x10³    

(16 x 10^-5x10^-5)(2.63x10¹⁷)÷6.7x10³  

Remember, it we're multiplying numbers with exponets and they have the same base, we can add the exponets.

(16 x 10^-5x10^-5)(2.63x10¹⁷)÷6.7x10³

(16 x 10^(-5)+(-5))(2.63x10¹⁷)÷6.7x10³

(16 x 10^-10)(2.63x10¹⁷)÷6.7x10³

Now we can do the same thing with the next set of parentheses.

(16 x 10^-10)(2.63x10¹⁷)÷6.7x10³

(16x2.63  x 10^-10x10¹⁷)÷6.7x10³

(16x2.63  x 10^(-10)+(17))÷6.7x10³

(16x2.63  x 10^(-10)+(17))÷6.7x10³

(42.08x10⁷)÷6.7x10³

Now we still have basically the same rule, but we subtract the exponets because we are dividing

(42.08x10⁷)÷6.7x10³

(42.08÷6.7 x 10⁷÷10³)

6.280597 x 10^(7)-(3)

6.280597 x 10⁴

So that's our final answer, or, we could multipy this out.

6.280597 x 10⁴

6.280597 x (10)(10)(10)(10)

6.280597 x (10000)

62805.97

Do you mean solve this for k?

If this is what you mean, you just have to get "k" alone, or in other words, k = something.

p=k/t              ---Multiply both sides by t

pt = k

k = pt

For some reason I don't think ths is what you meant.

Reply back if it's not, supply a better question, and we will surely help you out. :)

Yeah, it's from the DragonBall TV series.

I used to watch them alot..not so much anymore though. :)

Ok.

Thumbs up if you get the reference.

I like

$$\begin{array}{l}
$$\mathbf{P}arachute\\\mathbf{E}xpert\\\mathbf{M}y\\\mathbf{D}ear\\\mathbf{A}unt\\\mathbf{S}ally
\end{array}$$

:)

Here, we just need to get x alone, or in other words, x = something.

$$\begin{array}{l}
\frac{4}{3}x-\frac{1}{2}=0\qquad\mbox{Add $\frac{1}{2}$ to both sides}\\
\frac{4}{3}x=\frac{1}{2}\qquad\mbox{Divide both sides by $\frac{4}{3}$}\\
x=\frac{1}{2}\div\frac{4}{3}\qquad\mbox{Remember, when diving two fractions, multiply by the reciprocal (flip-flop the second fraction)}\\
x=\frac{1}{2}\times\frac{3}{4}\qquad\mbox{Multiply the left side out}\\
x=\frac{3}{8}

\end{array}$$

To check this, put in $$\frac{3}{8}$$ for x!

$$\begin{array}{l}
\frac{4}{3}x-\frac{1}{2}=0\\
\frac{4}{3}\times\frac{3}{8}-\frac{1}{2}=0\\
\frac{12}{24}-\frac{1}{2}=0\qquad\mbox{Reduce the first fraction}\\
\frac{1}{2}-\frac{1}{2}=0\\
0=0

\end{array}$$

It works, because 0 does equal 0!