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(sqrt3(3)*sqrt(10))^4

$${\left({\sqrt[{{\mathtt{3}}}]{{\mathtt{3}}}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{10}}}}\right)}^{{\mathtt{4}}}$$  

=  $${\left({{\mathtt{3}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)}\right)}^{{\mathtt{4}}}$$

=   $${{\mathtt{3}}}^{\left({\frac{{\mathtt{4}}}{{\mathtt{3}}}}\right)}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{\left({\frac{{\mathtt{4}}}{{\mathtt{2}}}}\right)}$$

=  $${\left({{\mathtt{3}}}^{\left({\mathtt{4}}\right)}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{{\mathtt{2}}}$$

=  $${{\mathtt{81}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}{\mathtt{\,\times\,}}{\mathtt{100}}$$

=  $${\mathtt{100}}{\mathtt{\,\times\,}}{\sqrt[{{\mathtt{{\mathtt{3}}}}}]{{\mathtt{81}}}}$$

Think of it like this

-2 + 1 2/3

=-2 + 1 +2/3

[ -2+1=-1     Start at -2 on the number line and go 1 unit in the positive direction that'll put you at -1 ]

=-1+2/3        Start at -1 on the number line and go 2/3 unit in the positive direction that'll put you at -1/3 ]

= -1/3

I have asked a question on the original thread - thanks Alan.   :)

How do you go about finding all of these Alan

and are there no other numbers that would work?

You made a little mistake anon.

2 and  4/7 is not equal to  16/7

Other than that your answer is really good.  Thanks for contributing.

If you were a member you would be able to change your answer and make it correct.

It is really easy, and totally free, to join up.  You don't have to give any personal information at all.  

Yes anon is right.

There are no real solutions to the square root of a negative number.

Any real number multiplied by itself must give a positive answer!

However there is another whole number system full of 'imaginary numbers'

The basic unit for imaginary numbers is  $$\sqrt{-1}$$    which is called  $$i$$    (or sometimes j)

so

if  $$a$$   is a positive number then    $$\sqrt{-a}=\sqrt{-1}\times\sqrt{a} = i\sqrt{a}$$

its 7.8

10 - 2 4/7

= 10 - 16/7

= (70 - 16)/7

= 54/7

= 7.8

arc sin is the same as inverse sine

so on this calc use asin

10-2 4/7

=10-2-4/7

=8-4/7

=7+1-4/7

=7+7/7-4/7

=7 and 3/7

If you have any trouble remembering the quadratic formula then you should listen to this a few times.

4|3x-5|-5      =-1

4|3x-5|-5+5=-1+5

        4|3x-5|=4

          |3x-5|=1  

3x-5=1          or            3x-5=-1 

   3x=6          or                3x=4 

     x=2          or                  x=4/3 

This answer is the same as Alan's just presented slightly differently.   

Any whole number that goes into 16 with out a remainder is a factor of 16    

Happyday's logic is good too except that it will not be a whole number.  

5/16*100  percent

there is no question here. 

Saseflower - Gradient is just the formal word for slope.   

You are making yourself indispensible to us saseflower.  I am most impressed with your contributions :)

just work them out backwards

$$\\x=7\\
5x=35\\
5x-3=32\\
So \\
5x-3=32\qquad $Is one example$$$

You set this out like an ordinary long division.

It is a bit hard for us to show you here.

If you google  "algebraic division" or "algebraic long division " or "synthetic division"

there are a lot of sites that will show you how to do it.

There are also a number of you tube clips that will show you exactly what to do.  They are very good. :)

Here is one example that I did for someone else. :)

I am not sure if I am answering your question but

0.2858 = sin theta 2

$$\\0.2858 = sin (\theta_2)\\\\
\theta_2=asin(0.2858)$$

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}^{\!\!\mathtt{-1}}{\left({\mathtt{0.285\: \!8}}\right)} = {\mathtt{16.606\: \!674\: \!357\: \!671^{\circ}}}$$

what is arcsin(-square root of 2 divided by 2) =

Well you could do it on a calculator.

or

arcsin is only defined for  [-pi/2, +pi/2]

It is negative so the angle will be between  -pi/2 and 0

Now I am just going to look at 

$$\\sin\theta=+\frac{\sqrt2}{2}=+\frac{1}{\sqrt2}\\\\
\theta =\frac{ \pi}{4}\\\\
so\\\\
asin\;\frac{-\sqrt2}{2}=\frac{ -\pi}{4}$$

2x² - 16x = 20

Divide each term by 2

x² - 8x = 10

Subtract 10 from both sides

x² - 8x - 10 = 0

This doesn't factorize nicely, so use the quadratic formula to solve it.

$$x=\frac{-(-8)\pm \sqrt{(-8)^2-4\times1\times (-10)}}{2\times1}$$

.

For the first solution we assume 3x is greater than 5 so the equation simplifies to:

4(3x - 5) - 5 = -1

12x -20 -5 = -1

12 x -25 = -1

12x = 24

x = 2

For the second solution we assume 3x is less than 5, so the equation simplifies to:

4(5 - 3x) - 5 = -1

20 - 12x - 5 = -1

15 - 12x = -1

12x = 16

x = 16/12

x = 4/3

So the sum of the two solutions is 2 + 4/3 = 10/3

.

The gradient of a straight line that is perpendicular to y + 3 = 2x is -(1/2), as the gradient of the given line is +2.

So the equation of the perpendicular line is y = -(1/2)x + c, where c is a constant.  We can find the value of c because we know the line passes through (-5, 2), so at that point we must have:

2 = -(1/2)*(-5)+c

or

2 = 5/2 + c

so c = 2 - 5/2 

or c = -1/2

so the equation of the line becomes

y = -(1/2)x - (1/2)

.

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