All Answers 358181 Answers

NG is correct that   sin(45°)  =  \(\frac{\sqrt{2}}{2}\)

However,   arcsin(45)  ≠  \(\frac{2}{\sqrt{2}}\)

If     a  =  arcsin(45)     then     sin( a )  =  45

But there is no possible angle  a  that can make   sin( a )  =  45

The biggest possible value of  sin( a )  is  1 .

So there is no real answer to this.

Mathway (Website) says nothing can be done further.

:P your welcome!!!

Ohh I remembered it now, yes I know these :) ty

What is factoring tricks? 

\(\ \phantom{=\quad}\frac{6}{\sqrt{36} + \sqrt{27}} + \frac{6}{\sqrt{27} + \sqrt{18}} + \frac{6}{\sqrt{18} + \sqrt{9}}\\~\\~\\ =\quad\Big(\frac{6}{\sqrt{36} + \sqrt{27}}\cdot\frac{\sqrt{36}-\sqrt{27}}{\sqrt{36}-\sqrt{27}}\Big) + \Big(\frac{6}{\sqrt{27} + \sqrt{18}}\cdot\frac{\sqrt{27}-\sqrt{18}}{\sqrt{27}-\sqrt{18}}\Big) + \Big(\frac{6}{\sqrt{18} + \sqrt{9}}\cdot\frac{\sqrt{18}-\sqrt9}{\sqrt{18}-\sqrt9}\Big)\\~\\~\\ =\quad\frac{6\sqrt{36}-6\sqrt{27}}{36-27}+ \frac{6\sqrt{27}-6\sqrt{18}}{27-18} + \frac{6\sqrt{18}-6\sqrt9}{18-9}\\~\\~\\ =\quad\frac{6\sqrt{36}-6\sqrt{27}}{9}+ \frac{6\sqrt{27}-6\sqrt{18}}{9} + \frac{6\sqrt{18}-6\sqrt9}{9}\\~\\~\\ =\quad\frac{6\sqrt{36}-6\sqrt9}{9}\\~\\~\\ =\quad\frac{36-18}{9}\\~\\~\\ =\quad2 \)_

This probably isn't the most sophisticated way, but we can easily solve this just by plugging in different numbers.

Let   f(x)  =  x² + 6x + 9

f(1)  =  1² + 6(1) + 9  =  1 + 6 + 9  =  16

f(2)  =  2² + 6(2) + 9  =  4 + 12 + 9  =  25

f(3)  =  3² + 6(3) + 9  =  9 + 18 + 9  =  36

f(4)  =  4² + 6(4) + 9  =  16 + 24 + 9  =  49

So we can see that there are only  2  positive integers  x  that satisfy  20 < f(x) < 40

They are:     x = 2     and     x = 3

(3^x - 2^y)(3^x + 2^y)  =  9^x - 4^y

                                             We know  3^x - 2^y  =  2   and   9^x - 4^y  =  10

( 2 )( 3^x + 2^y )  =  10

                                             Divide both sides of the equation by  2

3^x + 2^y  =  5

(4x + 2y)²  =  64

                                             Expand the left side of the equation.

(4x + 2y)(4x + 2y)  =  64

16x² + 16xy + 4y²  =  64

                                             Divide both sides of the equation by  4

4x² + 4xy + y²  =  16

                                             Subtract  4xy  from both sides of the equation.

4x² + y²  =  16 - 4xy

                                             Since  xy = 2  we can substitute  2  in for  xy

4x² + y²  =  16 - 4(2)

                                             Simplify the right side of the equation.

4x² + y²  =  8

The list of distinct, positive integers that has a median of 10 and the smallest average is:

1, 2, 3, 4, 9, 11, 12, 13, 14, 15

We know the first four numbers have to be  1, 2, 3, 4  because they are the four smallest distinct, positive integers.

Then the two middle numbers (9 and 11) must sum to 20 to make the median 10. And we can see why we should use 9 and 11 to minimize the average because, for example, if we used 5 and 15, that would force all of the last four numbers to be greater.

Then the last four numbers are the four smallest, distinct positive integers greater than 11.

average  =  (1 + 2 + 3 + 4 + 9 + 11 + 12 + 13 + 14 + 15) / 10  =  84 / 10  =  8.4

Sorry, can't read your question!.

n = 201

201 / 2010 =0.1 = 1/10

I had the computer list them for you:

(1, 1/2, 1/4, 3/4, 1/2, 1/3, 2/3, 1/2, 3/8, 5/8, 1/2, 2/5, 3/5, 1/2, 5/12, 7/12, 1/2, 3/7, 4/7, 1/2, 7/16, 9/16, 1/2, 4/9, 5/9, 1/2, 9/20, 11/20, 1/2, 5/11, 6/11, 1/2, 11/24, 13/24, 1/2, 6/13, 7/13, 1/2, 13/28, 15/28, 1/2, 7/15, 8/15, 1/2, 15/32, 17/32, 1/2, 8/17, 9/17, 1/2, 17/36, 19/36, 1/2, 9/19, 10/19, 1/2, 19/40, 21/40, 1/2, 10/21, 11/21, 1/2, 21/44, 23/44, 1/2, 11/23, 12/23, 1/2, 23/48, 25/48, 1/2, 12/25, 13/25, 1/2, 25/52, 27/52, 1/2, 13/27, 14/27, 1/2, 27/56, 29/56, 1/2, 14/29, 15/29, 1/2, 29/60, 31/60, 1/2, 15/31, 16/31, 1/2, 31/64, 33/64, 1/2, 16/33, 17/33, 1/2, 33/68, 35/68, 1/2, 17/35, 18/35, 1/2, 35/72, 37/72, 1/2, 18/37, 19/37, 1/2, 37/76, 39/76, 1/2, 19/39, 20/39, 1/2, 39/80, 41/80, 1/2, 20/41)

P.S. I may have misread your question, so be careful with this result. 

Using the advanced setup and registry editing, these default operational settings you describe are changeable.  However, it’s not necessary to isolate the image in Edge.

In MS Edge, the magnifier will magnify images while they are in context.  FireFox and Chrome require isolation to magnify images because it magnifies everything except images.

hecticar got it right! Thank you!