All Answers 358391 Answers

Divide both by 8

You need to use brackets - the meanings are far from clear

$$\\\frac{sec 3\pi}{7}\\\\
=\frac{sec \pi}{7}\\\\
=\frac{1}{7cos\pi}\\\\
=\frac{1}{7*-1}\\\\
=-\frac{1}{7}\\\\$$

WOW heureka, you gave yourself something to chew on there!    

My answer is no where near as impressive.   

sinx=cosx          When cosx=0,    sinx ≠0     so    cosx ≠ 0  therefore I can divide by it.

$$\\\frac{sinx}{cosx}=1\\\\
tanx=1\\\\
$tan is pos in the 1st an 3rd quad$\\\\
x=\frac{\pi}{4},\;\;\frac{5\pi}{4}\;\;etc\\\\
x=n\pi+\frac{\pi}{4}\qquad n\in Z\qquad$ (n is an integer)$$

A general rule when working with complex numbers is , for addition and subtraction use the algebraic form of the number, for multiplication powers and roots use the polar or exponential form.

So, first putting the number into polar form,

$$-\sqrt{2}-\imath\sqrt{2}=-2(1/\sqrt{2}+\imath/\sqrt{2})=-2\angle45\deg$$

and now raising to the power eight using De Moivre's theorem,

$$(-2\angle45\deg)^{8}=(-2)^{8}\angle(8\times45\deg)=256\angle360\deg=256.$$

anon is correct but i is an imaginary number.

There are no real solutions because no real number squared = a negative number.

2x - 3 = 6    add 3 to both sides

2x  = 9        divide by 2 on both sides

x = 9/2

$$\\(0.569e^{-r*3})+(0.559e^{-r*4})=1\\\\
(0.569e^{-3r})+(0.559e^{-4r})=1\\\\$$

Before I go any further, are you sure those 2 coefficients aren't supposed to be the same?

there are no super neat solutions for this

2x^2+3x-100=0     this will not factor....using the on-site solver, we have the following solutions.....

$${\mathtt{2}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{100}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{809}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}\right)}{{\mathtt{4}}}}\\
{\mathtt{x}} = {\frac{\left({\sqrt{{\mathtt{809}}}}{\mathtt{\,-\,}}{\mathtt{3}}\right)}{{\mathtt{4}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = -{\mathtt{7.860\: \!731\: \!326\: \!663\: \!945\: \!9}}\\
{\mathtt{x}} = {\mathtt{6.360\: \!731\: \!326\: \!663\: \!945\: \!9}}\\
\end{array} \right\}$$

Use  [2nd]

From your description, I think that I've been in this shop.

He's the guy behind the sausage counter.

You say to him "I'll have six of those sausages and a couple of the beef burgers", and he weighs them for you.

Given that z=-√2-√2i, calculate z^8

$$\small{\text{
$z= -\sqrt2-\sqrt2\cdot i \qquad
\boxed{z= -\sqrt2\cdot (1+i)}
$}}\\\\
\small{\text{
$
z^8 = \left[-\sqrt2 \cdot(1+i) \right] ^8
$}}\\
\small{\text{
$
z^8 = (-\sqrt2)^8\cdot (1+i)^8
$
}} \\
\small{\text{
$
z^8 = (-1)^8\cdot 2^{\frac82}\cdot (1+i)^8
$
}} \\
\small{\text{
$ z^8 = 1\cdot 2^4\cdot (1+i)^8 $
}} \\
\small{\text{
$
\begin{array}{l|l|l}
z^8 = 2^4\cdot (1+i)^8 \quad & \quad (1+i)^2 =1+2\cdot i+ i^2 \quad & \quad \boxed{\ i^2=-1 \ } \\
& \quad (1+i)^2 = 1+2\cdot i -1 \\
& \quad (1+i)^2 = 2\cdot i \\
z^8 = 2^4 \cdot \left[(1+i)^2\right]^4 \\
z^8 = 2^4 \cdot \left[ 2\cdot i\right]^4 \\
z^8 = 2^4 \cdot 2^4 \cdot i^4 \\
z^8 = 2^8\cdot i^4 \quad & \quad i^4=i^2\cdot i^2\\
& \quad i^4=(-1)\cdot (-1)\\
& \quad i^4 = (-1)^2 \\
& \quad i^4 = 1 \\
\boxed{\ z^8 = 2^8 = 256 \ }
\end{array}
$
}} \\$$

Here's another interpretation of the question:

However, this interpretation makes the Y(x) = x somewhat irrelevant, so it's quite possible this is not what was wanted!

.

Force is mass*acceleration.  

You can plug in the numbers.

.

I think that person caught a case of CDD .... once he surged into the "lead," that is.....

According to Mathcad:

and Wolfram Alpha won't evaluate it.

.

Call the number of 20 cent coins "x"

Let the number of 10 cent coins be 38 - x

So, the number of coins times their value each gives us the total value....so we have

20x + 10(38 - x)  = 530   simplify

20x + 380 - 10x = 530

10x + 380 = 530    subtract 380 from each side

10x = 150    divide both sides by 10

x = 15   ....and that's the number of 20 cent coins

X=4 

y=3

4+3=7 4-3=1

"1st you dumbass"      That is just priceless.     ROF LOL        

Yes, your teacher will probably get the answers correct :)

If you know the base and the height...it is

Area  =  (1/2)base x height

If you know two sides and the included angle between them, it is

Area = (1/2) ab sin θ    where a,b are the known sides and θ is the angle between them

There is also a formula if you only know the sides and nothing else....but....it's a little complicated......if this is the case....post back and I'll run through it for you.... 

Thanks Heureka,

Like Melody, I'm on shaky ground here....so.....you might have to take this with a grain of salt....

By the Integral Test, the function in the integrand must be eventually positive and decreasing on the interval of interest - in this case, 0 to ∞.

Here's the graph of  y = cos(x)*e^(-sin(x))

Standard form is the English way of saying scientific notation.

37000000 in standard form = 3.7*10^7