All Answers 358504 Answers

 4 + 4n = -12

The idea is to get the lots of n on one side and the numbers that are by themselves on the other side.

So start by subtracting 4 from BOTH sides.

That leaves you with  4n=-16

or

4 * n = -16

the opposite of multiply is divide so

Now divide BOTH sides by 4 and you will have n by itself = some number :)

Answered here......http://web2.0calc.com/questions/how-do-you-solve-4-4n-12

subtract 4 from both sides then divide by -12-4

^_^

Hi, adylyn-rae.......!!!!!!

Here's one.......

12 + 6 + 3 + 1.5 + .75 + ..............  this is an infinite sum  !!!!!

The sum of the terms =

12 / (1 - 1/2)  = 12/ (1/2)   = 24

I googled it and this is what I got:

"Open education is a collective term to describe institutional practices and programmatic initiatives that broaden access to the learning and training traditionally offered through formal education systems."

So it does not mean anything very specific it is just a term used when education is not just offered in a very formal educational institution.  That is how I interprete those words anyway. :)

4 + 4n = -12     subtract 4 from both sides

4n = -16     divide both sides by 4

n = -4

Note that  sin(x - 90)    =  sinxcos90 - sin90cos x =  -cosx

So we have :

f(x)   = -1/cosx   = -secx

f ' (x)  = -sec(x)tan(x)

Look at the graph of the derivative here : https://www.desmos.com/calculator/4mdr48rapz

Note that the derivative [gradient ] is always negative on 100 < x < 180

frequency

\(440\ \text{Hz} \cdot 2^{\frac{12}{12}} = 880\ \text{Hz}\)

It tells you how accurate the measurement is.

Can you expand on that, I don't know what you mean ://

How can I describe a function in which its value of 1 doubles the functions value, but greater than 1 does nothing

\(2\cdot f(x) *\left[ \frac{1~-~1 \pmod x}{x} \right] \qquad x \ge 1\)

Perimeter = 2 ( W + L)  ......so....in our case....we have  :

56 = 2 (W + 6W)   simplify

56 = 14W      divide by 14 on both sides

4 ft = W

And L  is 6 times this = 24 ft

Here's the procedure for terminating decimals :  http://www.mathsisfun.com/converting-decimals-fractions.html

For repeating decimals, here's what I do :

Example :    .127         with the "27" repeating

(1)    Forget the decimal.   Write 127 and from it subtract the non-repeating part, 1.  So we have 127 - 1 = 126

(2)   Write a number consisting of the same number of 9's as the digits that repeat and the same number of 0's as the non-repeating digits. So...we have two repeating digits........so write two 9s. And we have one non-repeating digit....so write one 0.........then....our number is 990

3)   Put  the result of (1) over the result of (2)   →  126 / 990

Verify for yourself that this is equal to .1272727..........

Triangle RST has vertices R (-3,4), S (-1,0), T (3,2). Find the length of the altitude from S to RT.

\(\boxed{~ \begin{array}{rcl} \vec{RT} &=& \vec{R} - \vec{T} = \binom{-3}{4} - \binom{3}{2} = \binom{-6}{2}\\\\ \overline{RT}=|\vec{RT}| &=&\sqrt{ \vec{RT}\cdot \vec{RT} }\\ \overline{RT} &=& \sqrt{ \binom{-6}{2} \binom{-6}{2} }\\ \overline{RT} &=& \sqrt{36+4 }\\ \overline{RT} &=& \sqrt{40}\\ \overline{RT} &=& \sqrt{4\cdot 10}\\ \mathbf{\overline{RT}}& \mathbf{=} & \mathbf{2\sqrt{10} }\\ \end{array} ~}\)

\(\boxed{~ \begin{array}{rcl} \vec{ST} &=& \vec{S} - \vec{T} = \binom{-1}{0} - \binom{3}{2} = \binom{-4}{-2}\\\\ \overline{RT} \cdot h &=& |\vec{RT} \times \vec{ST}|\\ h &=& \frac{|\vec{RT} \times \vec{ST}|}{\overline{RT}}\\ h &=& \frac{ \left| \binom{-6}{2} \times\binom{-4}{-2} \right| }{2\sqrt{10}}\\ h &=& \frac{(-6)\cdot (-2) - (2) \cdot (-4) }{2\sqrt{10}}\\ h &=& \frac{ 12 + 8 }{2\sqrt{10}}\\ h &=& \frac{ 20 }{2\sqrt{10}}\\ h &=& \frac{ 10 }{ \sqrt{10} } \cdot \frac{ \sqrt{10} }{ \sqrt{10} }\\ h &=& \frac{ 10 }{ 10 } \cdot \sqrt{10} \\ \mathbf{h} &\mathbf{=}& \mathbf{\sqrt{10}} \\ \end{array} ~} \)

Thanks Chris, that is really great :))             ROF  LOL   :)))

I am most impressed.  I did not know that you were a poet as well as a mathematician AND a fossil.

You are a man with many great talents  :)))          

Parentheses are kinda' like commas.......their use - or omission - can imply two radically different things. For instance, notice the difference in these two sentences.......

1)  Let's eat grandmother.

2)  Let's eat, grandmother.

See the difference????

x/8-(10-3/2x)=x+5

x/8-10+3/2x=x+5

x/8+3/2x=x+5+10 (Taking 10 from LHS to RHS by changing sign)

x+12x = x+15 ( Taking LCM of 8 and 2 on LHS)

   8

13x/8 = x+15

13x = 8(x+15) (Taking 8 to RHS by changing sign)

13x = 8x + 120

13x-8x = 120 (Taking 8x to LHS by changing sign)

5x = 120

x = 120/5 (Taking 5 to RHS by changing sign)

x = 24 

Play nice, Melody.....PLAY NICE.....!!!!!!

LOL!!!!!!!!

( 4 1/3 )  /  ( 5 1/6 ) = ?

\(\begin{array}{rcl} \frac{ ( 4 \frac13 ) }{ ( 5 \frac16 ) }\\ &=& \frac{ \frac{13}{3} }{ \frac{31}{6} }\\ &=& \frac{13}{3} \cdot \frac{6}{31} \\ &=& \frac{13}{3} \cdot \frac{2\cdot 3}{31} \\ &=& 13 \cdot \frac{2}{31} \\ &=& \frac{26}{31} \\ &=& 0.83870967742\\ \mathbf{\frac{ ( 4 \frac13 ) }{ ( 5 \frac16 ) } }& \mathbf{=} & \mathbf{0.83870967742}\\ \end{array}\)

Parentheses impact on the order of operation so of couse they can affect the answer.

Example:

(6+2)*5=8*5=40

6+2*5 = 6+10=16

x-y=1

y-2z=1

3x-4z=7

x=?

y=?

z=?

\(\begin{array}{lrcll} (1) & x-y&=&1 \\ & x &=& 1+y \\ \hline (2) & y-2z&=&1 \\ & y &=& 1+2z \\ \hline (3) & 3x-4z&=&7\\ & 3x &=& 7 + 4z &\qquad x = 1+y\\ & 3(1+y) &=& 7 + 4z &\qquad y = 1+2z \\ & 3[1+(1+2z)] &=& 7 + 4z \\ & 3(1+1+2z) &=& 7 + 4z \\ & 3(2+2z) &=& 7 + 4z \\ & 6+6z &=& 7 + 4z \qquad &| \qquad -4z \\ & 6+6z-4z &=& 7 \\ & 6 + 2z &=& 7 \qquad &| \qquad -6 \\ & 2z &=& 7 -6\\ & 2z &=& 1 \qquad &| \qquad :2 \\ & \mathbf{z} &\mathbf{=}& \mathbf{0.5}\\ \hline (2) & y &=& 1+2z \\ & y &=& 1+2(0.5)\\ & y &=& 1+1\\ & \mathbf{y} &\mathbf{=}& \mathbf{2}\\ \hline (1) & x &=& 1+y \\ & x &=& 1+2\\ & \mathbf{x} &\mathbf{=}& \mathbf{3} \end{array}\)

-1 1/2 + 5 7/8 = ?

\(\begin{array}{rcl} -1 \frac12 + 5 \frac78 \\ &=& - 1 - \frac12 + 5+\frac78 \qquad | \qquad \frac78 = \frac{4+3}{8} = \frac48+\frac38 = \frac12+\frac38\\ &=& - 1 - \frac12 + 5+\frac12+\frac38\\ &=& - 1 + 5+\frac38\\ &=& 4+\frac38\\ &=& 4\frac38\\ \mathbf{-1 \frac12 + 5 \frac78 } & \mathbf{=} & \mathbf{4\frac38}\\ \end{array}\)