All Answers 358087 Answers

Not just 2's, both of them are also divisible by 3.

.

$$6-4(x+3)$$  (expand brackets)

$$6-4x-12$$   (simplify)

$$-4x-6$$ 

This is simply (n+10)!

.

Be great if we could!  This is a still unresolved question that no mathematician in the world has yet been able to answer.  Fame (if not fortune) awaits the person who eventually does so!

.

$${\mathtt{0.6}}{\mathtt{\,\times\,}}{\mathtt{4}}{\mathtt{\,\small\textbf+\,}}{\mathtt{0.2}}{\mathtt{\,\times\,}}\left(-{\mathtt{1}}\right){\mathtt{\,\small\textbf+\,}}{\mathtt{0.2}}{\mathtt{\,\times\,}}\left(-{\mathtt{10}}\right) = {\frac{{\mathtt{1}}}{{\mathtt{5}}}} = {\mathtt{0.2}}$$

 Expected win = $0.20 

Solve for each variable:

-4x-6y+5z=21

3x+4y-2z=-15

-7x-5y+3z=15

$$\small{\text{$
\begin{array}{rrrrcrl}
(1) & -4x&-6y&+5z&=&21 \quad &\\
(2) & 3x&+4y&-2z&=&-15 \quad & | \quad \cdot \frac{4}{3}\\
(3) &-7x&-5y&+3z&=&15 \quad & | \quad \cdot \frac{4}{-7}\\
\\
\hline
\\
(1) & -4x&-6y&+5z&=&21 \quad &\\
(2) & 3\cdot\frac{4}{3}x &+4\cdot \frac{4}{3} y&-2\cdot \frac{4}{3} z&=&-15\cdot \frac{4}{3} \quad &\\
(3) &-7\cdot \frac{4}{-7}x&-5\cdot \frac{4}{-7}y&+3\cdot \frac{4}{-7}z&=&15\cdot \frac{4}{-7} \quad &\\
\\
\hline
\\
(1) & -4x&-6y&+5z&=&21 \quad &\\
(2) & 4x &+\frac{16}{3} y&-\frac{8}{3} z&=&-20 \quad &\\
(3) & 4x &+\frac{20}{7}y&-\frac{12}{7}z&=&-\frac{60}{7}\quad &\\
\\
\hline
\\
(1) & -4x&-6y&+5z&=&21 \quad &\\
(2)+(1) & 0 &-\frac{2}{3} y&+\frac{7}{3} z&=&1 \quad &\\
(3)+(1) & 0 &-\frac{22}{7}y&+\frac{23}{7}z&=&\frac{87}{7}\quad &\\
\\
\hline
\\
(1) & -4x&-6y&+5z&=&21 \quad &\\
(2) & 0 &-\frac{2}{3} y&+\frac{7}{3} z&=&1 \quad &\\
(3) & 0 &-\frac{22}{7}y&+\frac{23}{7}z&=&\frac{87}{7}\quad & | \quad \cdot \frac{-7}{22}\cdot \frac{2}{3}\\
\\
\hline
\\
(1) & -4x&-6y&+5z&=&21 \quad &\\
(2) & 0 &-\frac{2}{3} y&+\frac{7}{3} z&=&1 \quad &\\
(3) & 0 &-\frac{22}{7}\cdot \frac{-7}{22}\cdot\frac{2}{3}y&+\frac{23}{7} \cdot \frac{-7}{22}\cdot \frac{2}{3}z&=&\frac{87}{7} \cdot \frac{-7}{22}\cdot\frac{2}{3}\quad & \\
\\
\hline
\\
(1) & -4x&-6y&+5z&=&21 \quad &\\
(2) & 0 &-\frac{2}{3} y&+\frac{7}{3} z&=&1 \quad &\\
(3) & 0 & +\frac{2}{3}y& -\frac{23}{22} \cdot \frac{2}{3}z&=&-\frac{87}{22}\cdot \frac{2}{3}\quad & \\
\\
\hline
\\
(1) & -4x&-6y&+5z&=&21 \quad &\\
(2) & 0 &-\frac{2}{3} y&+\frac{7}{3} z&=&1 \quad &\\
(3)+(2) & 0 & 0 & \frac{108}{66}z&=&-\frac{108}{66}\quad & \\
\end{array}
$}}$$

$$\small{\text{$
\begin{array}{rcl}
\frac{108}{66}z &=& -\frac{108}{66}\\\\
z &=& \dfrac{ -\frac{108}{66} } {\frac{108}{66}} \\\\
z &=& - \frac{108}{66 }\cdot \frac{66}{108} \\\\
\boxed{\, z = -1 \, }
\end{array}
$}}$$

$$\small{\text{$
\begin{array}{rcl}
-\frac{2}{3} y+\frac{7}{3} z&=&1 \\\\
-\frac{2}{3} y+\frac{7}{3} \cdot(-1)&=&1 \\\\
-\frac{2}{3} y-\frac{7}{3} &=&1 \\\\
-\frac{2}{3} y &=&1+\frac{7}{3} \\\\
-\frac{2}{3} y &=&\frac{10}{3} \\\\
y &=& \frac{10}{3}\cdot ( -\frac{3}{2} ) \\\\
\boxed{\, y = -5 \, }
\end{array}
$}}$$

$$\small{\text{$
\begin{array}{rcl}
-4x-6y+5z&=&21 \\\\
-4x-6\cdot(-5)+5\cdot(-1)&=&21 \\\\
-4x+30-5&=&21 \\\\
-4x+25&=&21 \\\\
-4x &=&-4 \\\\
4x &=&4 \\\\
\boxed{\, x= 1 \, }
\end{array}
$}}$$

cheese is 1-1 and i dont care 

nothing, life is only like 1-1=0. It doesnt get you anywhwere

you canif you use the formula :)

Thank you 7up 

ln(e^0.047t) ?

$$\ln{(e^{0.047\cdot t})}\\
= 0.047\cdot t \cdot \ln{(e)} \quad | \quad \boxed{\;\ln{(e)} =1\;}\\
= 0.047\cdot t$$

What is the sum of the geometric sequence 1, 3, 9, ... if there are 14 terms?

$$$$

$$\small{\text{
geometric sequence: $a_1 = 1 \quad r = 3$
}}\\
\small{\text{
$
\begin{array}{lcll}
\hline
\\
s_{14} &=& \textcolor[rgb]{150,0,0}{1} + & 3^1 + 3^2 + 3^3
+ 3^4 + 3^5 + 3^6
+ 3^7 + 3^8 + 3^9
+ 3^{10} + 3^{11} + 3^{12} + 3^{13} \\
3\cdot s_{14} &=& & 3^1 + 3^2 + 3^3
+ 3^4 + 3^5 + 3^6
+ 3^7 + 3^8 + 3^9
+ 3^{10} + 3^{11} + 3^{12} + 3^{13} + \textcolor[rgb]{150,0,0}{3^{14}}\\
\hline
\\
s_{14}-3\cdot s_{14} &=& \textcolor[rgb]{150,0,0}{1-}&\textcolor[rgb]{150,0,0}{3^{14}}\\
s_{14}\cdot(1-3) &=& 1- &3^{14}\\
-2\cdot s_{14} &=& 1- &3^{14}\\
\end{array}
$}}$$

$$\small{\text{
$
\begin{array}{rcl}
s_{14} & = & \dfrac { 1- 3^{14} } {-2}\\\\
s_{14} & = & \dfrac { 3^{14}-1 } {2} = 2\,391\,484 \\
\end{array}
$
}}$$

Yes. In circuit A, If the amp meter is removed and a volt meter is connected in series (the same as in circuit B) then Vaa' will equal Vab which is the “unloaded” source voltage.

_7UP_

I will give you both some points ;)

1 and 26

2 and 13

and that is it :)

I have given you the factors

And Titanium has give you additions.

I am not sure what your really want.   

Use     atan

Ok I understand :(

Sometimes I feel the same way.

I expect things will get a lot better when the forum is revised but I do not know when that will be. :(

Use     [2nd][atan]

When they started littering this forum with bad stuff, I've decided to erase all my posts and leave.

If things get better, I'll come back.

F = (9/5)C + 32   ...so

F = (9/5)35 + 32 = (9)(35/5) + 32 = (9)(7) + 32 =     63 + 32   =  95F

http://www.dose.com/lists/18990/16-Completely-Disturbing-Pictures-From-the-Internet-That-Will-Haunt-Your-Dreams-ab580-0

if u have an iPad, go 2 calculator on that, and go on temperature, and type in 35

Numbers go on forever, because you can always add 1. Oh yeah, look at this pic:

Let's call the elephant's weight before walking through the field = x.

So we have

x + 80 = 1.08x    subtract x from both sides

.08x  = 80            divide both sides by  .08

x = 1000 lbs