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14 - 1/2(14) = 7 

14 - (14/2) = 7

14 - 7 = 7

7 = 7

How would you do 1.0183 to the negative third power?

1.0183^-3 =0.9470.....

12-(x+3)=10

12 - x - 3 =10

9 - x = 10 subtract 9 from both sides.

-x =10 - 9

-x = 1 multiply both sides by -1

x = - 1

I DID IT!

but how do you show a step by step derivation of that equation from pythagorean's theorem? Do I do AB^2=AC^2+BC^2 and then put in (x-h)^2+(y+k)^2=r^2

divide by 100.

22x-6=184+3x subtract 3x and add 6 to both sides

19x = 190 divide both sides by 19

x =190/19

x = 10

(6/7) / ( -13/2)

(6/7) x (-2/13

-12/91

your fathers son would be your brother. so how do you have none?

 Help

\(\begin{array}{|rcll|} \hline 8^{ \sqrt{4x+2} } > \dfrac{512^{2x}} { 8^{ \sqrt{16x+8} } } \\ 8^{ \sqrt{4x+2} } > \dfrac{512^{2x}} { 8^{ 2\cdot \sqrt{4x+2} } } \quad & | \quad512 = 8^3 \\ 8^{ \sqrt{4x+2} } > \dfrac{8^{3\cdot 2x}} { 8^{ 2\cdot \sqrt{4x+2} } } \\ 8^{ \sqrt{4x+2} } > \dfrac{8^{6x}} { 8^{ 2\cdot \sqrt{4x+2} } } \\ 8^{ \sqrt{4x+2} } > 8^{6x-2\cdot \sqrt{4x+2} } \\ \sqrt{4x+2} > 6x-2\cdot \sqrt{4x+2} \\ 3\cdot \sqrt{4x+2} > 6x \quad & | \quad : 3\\ \sqrt{4x+2} > 2x \\ \hline \end{array}\)

Let's find the domain:

\( \begin{array}{|rcll|} \hline 4x+2 &\ge & 0 \quad & | \quad -2\\ 4x &\ge & -2 \quad & | \quad : 4 \\ x &\ge & -\frac12 \\ \hline \end{array} \)

The domain is \([-\frac12, \infty)\)

case differentiation

1. \( x\ge 0\)

\(\begin{array}{|rcll|} \hline \sqrt{4x+2} &>& 2x \quad & | \quad \text{square both sides}\\ 4x+2 &>& 4x^2 \quad & | \quad -4x \\ 2 &>& 4x^2-4x \quad & | \quad :4 \\ \frac12 &>& x^2-x \\ \frac12 &>& \left(x-\frac12 \right)^2-\frac14 \quad & | \quad + \frac14\\ \frac12+ \frac14 &>& \left(x-\frac12 \right)^2 \\ \frac34 &>& \left(x-\frac12 \right)^2 \quad & | \quad \sqrt{ } \\ \frac{\sqrt{3}}{2} &>& | x-\frac12 | \\ | x-\frac12 | &<& \frac{\sqrt{3}}{2} \\ \Rightarrow \quad - \frac{\sqrt{3}}{2} &<& x-\frac12 < \frac{\sqrt{3}}{2} \quad & | \quad +\frac12 \\ - \frac{\sqrt{3}}{2} +\frac12 &<& x-\frac12 +\frac12 < \frac{\sqrt{3}}{2} +\frac12 \\ \frac{1-\sqrt{3}}{2} &<& x < \frac{1+\sqrt{3}}{2} \\ \hline \end{array}\)

case differentiation
2. x < 0 , because the domain is \(x \ge -\frac12\)

\(\begin{array}{|rcll|} \hline x &<& 0 \\ \text{the domain is } x &\ge& -\frac12 \text{ or } \\ -\frac12 & \le& x \\ \Rightarrow \quad -\frac12 & \le & x < 0 \\\\ \underbrace{\sqrt{4x+2}}_{\ge 0} &>& \underbrace{2x}_{<0} \quad & | \quad \text{always true}\\ \hline \end{array} \)

together:

\(\begin{array}{|rcll|} \hline \frac{1-\sqrt{3}}{2} &<& x < \frac{1+\sqrt{3}}{2} \\ -\frac12 & \le & x < 0 \\\\ \mathbf{ -\frac12 } & \mathbf{ \le } & \mathbf{ x < \frac {1+\sqrt{3}}{2} } \\ \hline \end{array}\)

1+1=2

1² x 0.25 - 4 = -3.75  

I'm not sure if I entirely understand your question, but what it comes down to is that division is spliting a certian number of something into a certain amount of groups, and saying how much of that object is left in each group. Because you can not put objects in 0 or no groups, then you can also not divide by 0.

Why is 1/0 undefined?

Division is the inverse of multiplication, therefore,

we know that  6/2 = 3  because 2 x 3  =  6

and that       18/3  =  6  because  3 x 6  =  18.

Now, if we try  1/0:

  1/0  =  ?  if we can find a number to replace the question mark that solves this:  0 x ?  =  1.

But there is no number we can use; no number multiplied by 0 will give us an answer of 1.

We get the same problem if we try  7/0  or  -2/0  or  147/0  etc.  [No number works.]

[Dividing 0 by 0 gives us a different problem; all numbers work.

 We could argue that  0/0  =  1  because  0 x 1  =  0;

 but then, by the same argument:  0/0  =  0  because    0 x 0  =  0

                                                   and  0/0  =  27  because 0 x 27  =  0

                                                   and  0/0  =  -2  because  0 x -2  =  0.

So  0/0  is undefined because it has no single answer; every number works.]

Thank you so much. This helped alot!

if you have asymptotes plus or minus 17/18.55 do you leave them like that for your final answer since it is plus or minus a/b or should I simplify it to plus or minus 0.92?? Would it be incorrect if I left it in fraction form?

17/18.55 is not a fraction.  Fractions cannot have decimal in them.

0.92 is an approximation so I do not think it is as good as the fraction answer.

\(\pm\frac{17}{18.55} =\pm\frac{ 1700 }{ 1855 } =\pm \frac{340}{371}   \)

I think that the best answer is   \(\pm \frac{340}{371} \)

Area =[Base x Height] / 2

24 = B x 6/2

24 =B x 3

Base =24/3 =8 inches

Hypotenuse^2=8^2 + 6^2

H^2 = 64 + 36 =100 take the square root of both sides,

H = 10 inches.

The other leg has a length of 8", making the hypotenuse have a measurement of  

\(d = \sqrt{{8}^{2}+{6}^{2}}\)

\(\sqrt{100} in.\)

or 10 in.

Hi Metallica22,

You know we really like it when people ask us to explain our answers better.  :)

It means someone is really trying to learn from us :)

Maybe this pic will help.

From the picture you can see that    \(r^2=(x-h)^2+(y-k)^2\)

The question is worded badly and I am not sure if I have given the answer that is really most wanted.

I have not used the point C...... 

Oy go to EHarmony for a gf not here sir/maam

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