All Answers 358131 Answers

If cot x = 2/3 , what is tan x ?

$$\boxed{\mathbf{ \tan{(x)}=\dfrac{1}{\cot{(x)}} }}\\\\\\
\tan{x}=\dfrac{1}{ \frac23 } = 1\cdot \frac32 = \frac32\\\\\\
\boxed{ \mathbf{ \cot{(x)}= \frac23 \qquad \tan{(x)}= \frac32 }}\\\\\\
\boxed{\mathbf{
\cot{(x)}=\frac{a}{b} \qquad \tan{(x)}= \frac{b}{a}
}}\\\\\\
\boxed{\mathbf{
\cot{(x)}=\frac{\cos{(x)}}{\sin{(x)}} \qquad \tan{(x)}= \frac{\sin{(x)}}{\cos{(x)}}
}}$$

Happy birthday! 

why are you posting this on a math forum

Good luck to you and Zegroes too! 

1/2.

$$\sqrt{6}=6^{1/2}$$

See - a square root IS an exponent :)

It is the probability that they all fail.

$${\mathtt{0.13}}{\mathtt{\,\times\,}}{\mathtt{0.05}}{\mathtt{\,\times\,}}{\mathtt{0.01}} = {\mathtt{0.000\: \!065}}$$

$${\frac{{\mathtt{65}}}{{\mathtt{1\,000\,000}}}}$$

Oh, system updates, I know nothing about the system updates, sorry MG.

That is really impressive coding MG, the maths for it is good too. :))   

-----------------------

Back to your equation though it is still not quite right.

$$\begin{array}{rrlll}\\
-6x+4&=&-2\qquad&(-4\;to\;get\;X\;alone)\\\\
-6x-4&=&-2-4\\\\
-6x&=&-6\qqquad&Divide\;6\;on\;both\;sides\\\\
\frac{-6x}{6}&=&\frac{6}{6}\qquad&simplify\\\\
\frac{-6x}{\not{6}}&=&\frac{6}{\not{6}}\\\\
-x&=&\frac{1}{1}
\end{array}$$

In your second row you 'forgot' to put the  +4 in on the LHS   (left hand side)

The 3rd row is good but you could have divided by -6 to get x on its own straght away.

Row 4   Your cancelling is not correct.

$$\\\frac{-\not{6}x}{\not{6}}=\frac{\not{6}}{\not{6}}\\\\
-x=\frac{1}{1}\\\\
-x=1\qquad $Multiply both sides by -1$\\
-x*-1=1*-1\\
x=-1$$

Mine is not lined up beautifully like yours but I have done some corrections :)

You must end up with x by itself on the LHS when you are finished  :)

The / sign suffices as the fraction sign. Writing 1/2 is the same as writing $${\frac{{\mathtt{1}}}{{\mathtt{2}}}}$$.

There isn't one. Let's say you want to write one-thirds. You would write it 1/3, or one divided by three.

Think about what is a reasonable size for 30 inches as a circumference of an object.

A car tire is too big. It would probably be way more than 30 inches.

A volleyball seems like a reasonable size. Let's check the last one to be sure.

A marble is definitely less than 30 inches. It is way smaller than that.

The only reasonable choice would be B. Volleyball

*Hint.*

If you took a piece of paper 30 in. long and wrapped it around the object, what would be the object that would fit inside the paper?

This should help you :)

$${\mathtt{2}}{\mathtt{\,\times\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{8}}}}\right) = {\frac{{\mathtt{1}}}{{\mathtt{4}}}} = {\mathtt{0.25}}$$

Fixed it.

Look at the place you want to round to. I'll call it the place. I'm using the tenths place as an example.

    4.1431      7.7923

Look one place directly to the right of it. I'll call that the digit.

    4.1431     7.7923

If the digit is 0, 1, 2, 3, or 4, the place stays the same. If the digit is 5, 6, 7, 8, or 9, the place increases by 1. All numbers behind the place become zeroes.

    4.1431 ----> 4.100 or 4.1      7.7923 ----> 7.800 or 7.8

You may have to do this. 

    1.9503 ----> 2.0000 or 2

Most problems specify the place you need to round to.

Note that if you aren't rounding to a decimal, the zeroes cannot be omitted (ignored).

$$Thanks!$$

With exponents. 

1. Parentheses.

2. Exponents/Square Roots.

etc.