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Hmm,   3x + 2y/5 + 7   is not an equation because there is not an equal sign.

Is there meant to be an equal sign? Or maybe we just need to simplify it?

thank u for your help however it is x minus 9 not x to the negative ninth 

How do you find the range of 1/square root of x-9

Hello M!

\({\color{BrickRed}\frac{1}{\sqrt{x^{-9}}}}=\frac{1}{\sqrt{\frac{1}{x^9}}}=\sqrt{x^9}\color{blue}=x^4\sqrt{x}\)

sorry!
so

\({\color{BrickRed}\frac{1}{\sqrt{x-9}}}=\frac{\sqrt{x-9}}{x-9}\)

More is not possible.

Got it. The definition quantity of the function is searched for.

Thanks Melody!

\(f(x)=\frac{1}{\sqrt{x-9}}\\ \color{blue}9< x<\large ∞\)

  !

sinh^-1 (1)

\(sinh^{-1}(1°)_{deg}=asinh(1°)_{deg}\color{blue}=50.498986710526\\ sinh^{-1}(1)_{rad}=asinh(1)_{rad}\color{blue}=0.88137358702\)

  !

How do I answer r/4=(r+7)/6

Good morning Guest!

\(\color{red}\frac{r}{4}=\frac{r+7}{6}\\ 6r=4(r+7)\\ 6r=4r+28\\ 6r-4r=28\\ 2r=28\)           \(\times 6\ \rightarrow\times 4\\ multiply\\ -4r\\ subtract\\ divide\ by\ 2\)

\(r=14\)

\(\frac{14}{4}=\frac{14+7}{6}\\ \frac{14}{4}=\frac{21}{6}\\ 14\times 6=21\times 4\\ 84=84\)

  !

How do you do this:

x^-3/2

\(\frac{x^{-3}}{2}=\frac{1*x^{-3}}{2}=\frac{1}{2x^{+3}}=\frac{1}{2x^{3}}\\~\\ \text{or maybe you mean}\\~\\ x^{-3/2}=1*x^{-3/2}=\frac{1}{x^{+3/2}}=\frac{1}{x^{3/2}}\\~\\ \text{can also be written as}\\ \frac{1}{x^{3/2}}=\frac{1}{(\sqrt x)^3}=\frac{1}{\sqrt{x^3}}\)

Thank You!!!!

They appear to be in the same place, but they are on two different graphs.

Sin(pi/2) is found on a plane where both the x- and the y-axis are real-numbered axes.

The imaginary unit is found on a complex plane, where the x-axis is real-numbered but the y-axis is an imaginary axis.

(1/2)x+2/3=5/6     multiply through by the common denominator of 2,3 and 6  = 6    to clear the fractions

3x  + 4   =  5       subtract 4 from each side

3x  = 1              divide both sides by 3

x  = 1/3

Find the following limit:
lim_(x->0) (sqrt(1/2) - 1/2)

Hint: | Simplify the expression inside the limit.
sqrt(1/2) - 1/2 = 1/sqrt(2) - 1/2:
lim_(x->0) (1/sqrt(2) - 1/2)

Hint: | The limit of a constant is that constant.
Since 1/sqrt(2) - 1/2 is constant, lim_(x->0) (1/sqrt(2) - 1/2) = 1/sqrt(2) - 1/2: 
 = 1/sqrt(2) - 1/2

I love bananas

As an alternative answer  to this.....

Note that each  element  will be summed  5  times

So we have  the sum  (1 + 2 + 3 + 4 + 5 + 6)   summed 5 times

And the sum of  the frist  6 elements  =  (6) (7)  / 2  =  21 

So......the total sum will be  

(5) * [ (6) * ( 7)  / 2 ]   =    (N -1) (N) ( N + 1)  / 2     which is the correct "formula"

As an addendum to my answer.....I believe  that frasinscotland's  answer will be true if "Point A"  is (5,5)

Point P will be located at  ( 50/5, 0)  = (10,0)

So AP  will have a length of   sqrt   [ (10 - 5)^2  + ( 5 - 0)^2 ]  =  sqrt ( 25 + 25)  = sqrt (50)

And this will be  equal  to  OA

(x - 2) / 3  +  (x + 5) / 2   =  1/3          multiply through by the common denominator of 2, 3  = 6

2 (x -2)  + 3( x + 5)  =  2          simplify

2x  - 4  + 3x + 15  = 2             combine like terms

5x + 11  = 2                          subtract 11 from  both sides

5x  =  - 9                            divide both sides by 5

x = -9 / 5

what is 5/8 x 1 and 1/2 divided by 15/16

If you mean: [5/8 x 1 1/2] / [15/6] =

[5/8 x 3/2]  x  [6/15] =

15/16  x  6/15  =

90 / 240 = divide both sides by 30

3 / 8 = 0.375

Let's look at this a little closer......I don't know if "Point A"  is some specific point   or any random point on the circle.....but....if it's a random point.......look at this graph

Let  " Point  A "  be on the on the circle   =  ( a , sqrt (r^2 - a^2))

The slope of  the tangent line to this circle will be   =    -a / sqrt (r^2 - a^2)

So....the equation of a  line tangent to  the circle at  this point  is

y =  [ -a / sqrt (r^2 - a^2)] ( x - a) +  sqrt (r^2  - a^2)

To find "Point P"  let  y  = 0   we have that

0 =   [ -a / sqrt (r^2 - a^2)] ( x - a)  +  sqrt (r^2  - a^2)

 [ a / sqrt (r^2 - a^2)] ( x - a)   =   sqrt (r^2  - a^2)

a (x - a)   =   r^2 - a^2

ax - a^2  =  r^2  - a^2

ax =  r^2

x  = r^2 / a   ...   so  "Point P"   (  r^2/ a , 0 )

So.......  the   the base of triangle OAP will  be  r^2 / a      and the height will be sqrt ( r^2 - a^2)

So....the area of this triangle will be    (1/2) (r^2 / a) sqrt (r^2 - a^2)  =    (r^2) sqrt (r^2 - a^2) / [ 2a ]

So....suppose  r  = sqrt (50)  and  a  = 4

Then the area of triangle OAP  will  be    50 sqrt (34) / 8  ≈  36.44 sq units

But suppose that   r =sqrt (50) and  a  = 2

Then the area of triangle OAP  will be  50 sqrt ( 46) / 4  ≈ 84.78 sq units

5/2 = 2.5

(49^6)^3 / (7^7)^5        =
[(7^2)^6]^3 / (7^7)^5  =
[7^12]^3 /  (7^7)^5       =
[7^36] /  [7^35]              =
7^(36 - 35) =7^1             = 7

-2x2-5x=-50

Hello Guest!

          \(-2x^2-5x=-50\)         [ both sides + 50

\(-2x^2-5x+50=0\)               [ : (-2)  

    \(x^2+\frac{5}{2}x-25=0\)

              p          q

\(x=-\frac{p}{2}\pm\sqrt{\frac{p^2}{4}-q}\)                [ p,q transfer

\(x=-\frac{5}{4}\pm\sqrt{\frac{25}{16}+25}\)             [ calculate root

\(x=-\frac{5}{4}\pm\sqrt{26.5625}=-1.25\pm5.15388203202\)

\(x_1=3.90388203202\\ x_2=-6.40388203202\)

  !

Banana is is is isa banana

You are getting confused. They are not positioned at the same place.

i   X  sin(pi/2)  = i   X    1      

                           = i.

But  sin(pi/2) is  just  1 .       So  1    and  i  are not positioned at the same place,far from it.  I think you need to look some more at complex numbers     (they aren't really all that complex ,someone just decided to call them by that name)  and get some practise with them.

flip the second fraction and multiply

79.6666666667

Sorry Juriemagic,

I hadn't actually looked at the guests answer.  I still haven't. I should have let you know that before. 

Thanks for answering Chris :)