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Nothing else to do with your time? Obviously!!!!.

How about 3^2. 

5/10 = 1/2

6/9  = 2/3

3/8

2/6   = 1/3

Only 3/8 cannot be reduced any further

"Of" means multiply....so we have

(1 / 16)  * 28  =

(1 / 16) *  (28 / 1)   =

28  / 16  =

7 / 4

W*F 

3\4* (-3\2)  =  -9 / 8

(-6)(-2)(-1)  =

12 *  -1  =

-12

 0  4       1  0

-6  5       0  1         add row 2 to row 1

-6  9       1  1

-6  5       0  1       multiply row 1 by -1 and add it to row 2

-6  9       1  1

 0  -4     -1  0       multiply row 2 by -1/4

-6  9       1    1

 0   1      1/4  0     multiply row 2 by -9 add it to row 1

-6  0      -5/4  1

 0  1       1/4   0    multiply row 1 by -1/6

1  0       5/24  -1/6

0  1       1/4     0

The resulting matrix on the right is the same as Melody's result  !!!!

Help please: (1-sqrt(2i))*(2 + sqrt(3i) )

Square root: sqrt(2i) = 1+i

Subtract: 1 - (1+i) = -i

Square root: sqrt(3i) = 1.2247449+1.2247449i

Add: 2 + (1.2247449+1.2247449i) = 3.2247449+1.2247449i

Multiply: (-i) * (3.2247449+1.2247449i) = 1.2247449 - 3.2247449i

Please solve ln(2*10^(-7)*sqrt(D))

Factor the following:
9 x^4 y^2-12 x^3 y^3+4 x^2 y^4

Factor x^2 y^2 out of 9 x^4 y^2-12 x^3 y^3+4 x^2 y^4:
x^2 y^2 (9 x^2-12 x y+4 y^2)

The coefficient of x^2 is 9 and the coefficient of y^2 is 4. The product of 9 and 4 is 36. The factors of 36 which sum to -12 are -6 and -6. So 9 x^2-12 x y+4 y^2  =  9 x^2-6 x y-6 x y+4 y^2  =  3 x (3 x-2 y)-2 (3 x-2 y) y:
x^2 y^2 3 x (3 x-2 y)-2 y (3 x-2 y)

Factor 3 x-2 y from 3 x (3 x-2 y)-2 y (3 x-2 y):
x^2 y^2 (3 x-2 y) (3 x-2 y)

(3 x-2 y) (3 x-2 y) = (3 x-2 y)^2:
Answer: | 
| x^2 y^2 (3x - 2y)^2

(1-sqrt(2i))*(2 + sqrt(3i) )   =

2 - 2sqrt(2i) + sqrt(3i)  - sqrt( 2 * 3 * i^2)  =

2 - 2sqrt(2i) + sqrt(3i)  - sqrt(6)i

9x^4 y^2   - 12x^3 y^3 + 4x^2 y^4    take out the greatest common factor, first

x^2 y^2  ( 9x^2  - 12xy + 4y^2)    factor the expression in the parentheses

x^2 y^2  ( 3x - 2y) (3x - 2y)  =

x^2 y^2  (3x  - 2y) ^2

And that's it  !!!

If D is positive we have the result as I've given above (with 10⁹ not 10^-9).  If D is negative, then srqt(D) = sqrt(|D|)*i, where i is sqrt(-1). The terms inside the absolute value then become, say k*i, where k is a real number.  The absolute value of k*i is given by sqrt(k²), namely just k. so |ln(...)sqrt(D)| =|ln(...)|*sqrt(|D|).  D can't be negative though, because if it were we would have D*|ln(...)|*sqrt(|D|) = - |D|*|ln(...)|*sqrt(|D|) = a negative number; but the LHS is a positive number, so D must be positive.

Here's a graph calculated by Mathcad (which is quite happy with imaginary numbers) for a range of values of D:

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Hey, adylyn_rae.......!!!!

Which equation represents a line that passes through (5, 1) and has a slope of ?????

3.00222*10^(+9)= (D)*abs([ln(2*10^(-7))*sqrt(D)])

D=?

3.00222*10^(+9)= (D)*abs[ln(2*10^(-7))*sqrt(D)]

D=?

Please pay attention that abs[ln(2*10^(-7))*sqrt(D)]

I could not solve by using the calculator because of the inside of squared brackets! 

Well, again assuming that D is real and positive (if it were negative we'd have a complex number for sqrt(D)), the result is as I gave above, except that the 10^-9 in the numerator becomes 10⁺⁹.

I should perhaps add that the vertical lines refer to taking the absolute value of whatever is the result of what is between them.  If you need a numerical result, you can use the calculator on the home page here. 

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Sorry about that; Could you please solve this? Now brackets are right:)

3.00222*10^(+9)= (D)*abs[ln(2*10^(-7))*sqrt(D)]

D=?

(2^5*3^-2)/(2^7*3^-4) =

\(\frac{2^5*3^{-2}}{2^7*3^{-4}}\\ =\frac{1*3^{-2-\;-4}}{2^{7-5}}\\ =\frac{3^{-2+4}}{2^{2}}\\ =\frac{3^{2}}{2^{2}}\\ =\frac{9}{4}\\ =2\frac{1}{4}\)

You don't have a matching number of brackets!  However, I've made an assumption below (and also assumed you are only dealing with real numbers so that D is positive).

\(\\ 3.00222\times10^{-9}=D\times |\ln{(2\times10^{-7})\times D^{0.5}}|\\\\ 3.00222\times10^{-9}=D\times |(\ln2+\ln{10^{-7}})\times D^{0.5}|\\\\ 3.00222\times10^{-9}=D\times |(\ln2-7\ln{10})\times D^{0.5}|\\\\ D^{1.5}=3.00222\times10^{-9}/|(\ln2-7\ln{10})|\\\\ D=(3.00222\times10^{-9}/|(\ln2-7\ln{10})|)^{2/3}\)

16 = 2 x 2 x 2 x 2

11 = 2 x 2  x 11

Take the common factors  from both lists   2 x 2  = 4

The highest common factor is 4

Depends which direction y is acting!  Assuming you mean y is acting along the slope, then it must equal the component of the weight of the particle that acts down the slope:   y = mg*sin(theta) where theta is the angle of the slope to the horizontal.