All Answers 358506 Answers

Well Sasini, 

I only thought it might because your hair is already black.  

It may enhance the beauty that you and Rosala already possess.   

If my hair was black I would definitely use it.   

$$\\(1/64)^x = 1/512 \\
$turn both sides upside down and you get $
64^x=512\\\\
$method 1$\\
64^x=512\\
log64^x=log512\\
xlog64=log512\\
x=\frac{log512}{log64}\\
x=1.5\\\\
$method2$\\
64^x=512\\
64=2^6\quad and \quad 512=2^8\\
(2^6)^x=2^9\\
2^{6x}=2^9\\
6x=9\\
x=9/6=1.5$$

There is no (generalized) short cut.  Just do the division.   

700                 

$$\\\infty^{-0.001}=\frac{1}{\infty^{+0.001}}=\frac{1}{\sqrt[1000]{\infty}}
=\frac{1}{\infty}=0$$

$$\\1000^{1000}=(10^3)^{1000}=10^{3000}$$

so that is 1 followed by 3000 zeros   

you said add the exponents which is correct but you multiplied

$$\\(2.48*10^7)(3.5*10^{-12})\\
=2.48*10^7*3.5*10^{-12}\\
=2.48*3.5*\;\;10^{7-12}\\
=2.48*3.5*\;\;10^{-5}\\$$

I believe general form is

$$y=ax^2+bx+c$$

I do not know what canonique form is.  

tell me what it is and I will show you how to do the conversion.    

It will equal zero.

You have infinity raised to a negative power which means that you have one divided by infinity raised to some (positive) power and that will be zero.

You can divide that by any number you want, other than zero, and the result will still be zero.

Well, its easy once you get the hang of it. And trust me, you'll begin to understand why we all need math in life.

4(x+5)+3(x+3)

4(x) + 4(5) = 4x+20

3(x) + 3(3) = 3x+9

Combine the like terms

(4x) + (3x) = 7x

(20) + (9) = 29

= 7x+29

Take the time to actually look at it, its common sense.

I don't know what do you mean, but here is the dictionary definition of happy:

$${\mathtt{49}}{\mathtt{\,\times\,}}{\mathtt{5}} = {\mathtt{245}}$$

$${\frac{{\mathtt{245}}}{{\mathtt{8}}}} = {\mathtt{30.625}}$$

$${\frac{{\mathtt{30.625}}}{{\mathtt{2}}}} = {\frac{{\mathtt{245}}}{{\mathtt{16}}}} = {\mathtt{15.312\: \!5}}$$

There's your answer.

$${\frac{{\mathtt{9}}{\mathtt{\,\times\,}}{\mathtt{9}}{\mathtt{\,\times\,}}{\mathtt{8}}{\mathtt{\,\times\,}}{\mathtt{0}}{\mathtt{\,\times\,}}{\mathtt{6}}{\mathtt{\,\times\,}}{\mathtt{5}}{\mathtt{\,\times\,}}{{\mathtt{6}}}^{{\mathtt{7}}}{\mathtt{\,\times\,}}{\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{9}}{\mathtt{\,\times\,}}{\mathtt{4}}}{{\mathtt{3}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}} = {\mathtt{2}}$$

??

(2.48*10^7)(3.5*10^-12)

Multiply the decimals(2.48 & 3.5)

 2.48*3.5=8.68

Then ADD the exponents.

7*-12=-84

ANSWER: 8.68*10^-84

9*9=81

So therefor 9 is the sqaure root of 81.

The squareroot is a number you can multiply by itself (9) to get a number (81) 

(x + y + z) (xy + xz + yz) =

[x^2(y + z) + y^2(x + z) + z^2(x + y)] + 3xyz = 18

                             6                             +  3xyz = 18   subtract 6 from both sides

                                                                3xyz = 12    divide by 3 on both sides

                                                                  xyz = 4

1/64 = 1/2^6    1/512 = 1/2^9   ..so we have..

(1/2^6)^x  = (1/2)^9 =

(1/2)^(6x) = (1/2)^9      we have like bases, so equating exponents, we have

6x =9   divide by 6

x = 9/6 = 3/2

Actually Hindi means times 4 and add an x or y at the end of problem. But correct.

m + 1/m = 8    square both sides

m^2 + 2(m)(1/m) + 1/m^2  = 64

m^2 + 2 + 1/m^2 = 64    subtract 2 from both sides

m^2 + 1/m^2  = 62  ....therefore

m^2 + 1/m^2 + 4 = 62 + 4   = 66

42-3×=-9×

42=-9×-3x

42=-6x

42/-6=-6/-6

-7=x

So we have

 N^2 + N + N^(1/3) = 759

Let's assume that the number we're looking for is an integer.

And let's first take the sq root of 759 = about 27.55

Then 759  is a little larger than 27^2

Let's test 27

27^2 + 27  + 27^(1/2) =

729 + 27 + 3  = 

759

So.....27 is the integer we were looking for....

r^2 + s^2 = 1     square both sides

r^4 + 2(rs)^2 + s^4 = 1

r^4 + s^4 = 1 - 2(rs)^2

(7/8) = 1  - 2(rs)^2   rearrange

2(rs)^2  = 1 - 7/8

2(rs)^2 = 1/8    divide both sides by 2

(rs)^2 = 2/ 16 = 1/8   take the square root of both sides

(rs) = ±√(1/8) = ±√(1/2√2) = ±√2/4

And since r and s are positive real numbers, take the positive root

It still gonna be infinity. It does not matter if divide or multiple infinity by any number, it still stays infinity... 

it is different when you divide respectively small number with infinity will result in zero...