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I don't get the question sorry 

You're telling us the answer

Now that depends on what the question is. :)

The square root of -1 does not have a 'real'  value because 

if a negative number is squared then the answer is positive and 

if a positive number is squared the answer is positive too

SO no number squared equals a negative number!

There is another number system which include numbers that are not real.  It is called  'The complex number system.' 

It has imaginary numbers and sqrt(-1) is the base unit for the whole set of imaginary numbers.

In hre complex number system i is usually assigned the value of  sqrt(-1), although very occasionally j is used instead.

So

\(i=\sqrt{-1}\)

I hope that helps :)

-2x  <  16 - 3           Simplify the right side.

-2x  <  13                 Divide both sides of the inequality by  -2  , and, since it is negative, flip the sign.

x  >  -13/2

x  >  -6.5

x can be any  number on this blue line:

So...we can see that the smallest integer value of    x    is    -6      

9.5x  = 500       divide both sides by 9.5

x  =  500 / 9.5

Here's the first one

Since angle ACB and  angle DCE are vertical angles, they are equal

And CB = CD

And angles CDE  and CBA are  right....so....

The two triangles are  congruent by Angle-Side-Angle

And since ED  = AB, then ED  will equal the distance across the river

Let's solve this and find out what the solutions are

3 l 4 - 5x l  = 9

3|4-5x| = 9      divide both sides by  3

[ 4 - 5x l =  3

Note that we have that  either

4 -  5x  =  3              or that                       4 - 5x  =  -3

Subtract 4  from  both sides

-5x =  -1                                                  -5x    -7

Divide both sides by  -5 

x  =  1/5                                                  x   = 7/5

Our  solution   will either lie on  [ 1/5, 7/5 ]  or outside of this interval

Test an x value outside this interval in the original inequality to see if it makes the inequality true, or not.....I'll pick 0

3 l 4 - 5(0) l ≤ 9

3 l 4 l    ≤ 9

12 ≤  9

This is false......so.....the solution  comes from  [ 1/5, 7/5 ]

Both endpoints are included because of the " ≤ "  sign

\(\frac{-21}{5}\)?

Is this what you mean?

I'm assuming that  we're taking about obtaining a multiple of three on the spinner and a factor of ten on the dice

The only multiple of three possible on the spinner is 3

The factors of 10 possible on the die  are  1, 2 and 5 

So....

P( multiple of three on the spinner)  = 1/5  =  A

P(obtaining a factor of ten on the dice)  = 3/6  = 1/2   = B

So.....

P( A and B)   =   (1/5) (1/2)  =  1/10

19 3/9   -  5/9

We can use a "borrowng" trick, here

Notice that we can write  19 3/9  in a slightly different manner  =

18  +  1  +  3/9  =

18  +  9/9  + 3/9  =

18 12/9

So...we have

18 12/9  -  5/9  =

18 7/9

We can do this with something known as the Maclaurin Series

Each term is given by:

[f ⁿ (0) / n!]  * xⁿ        

Where  f ⁿ   is the nth derivative  of  (1 + x)^1/5       with n begiining at 0

1st term (n = 0) =    [ (1 + 0)^(1/5)/ 0! ] * x⁽⁰⁾ =   1

2nd term (n = 1) =  [ (1/5)(1 + 0)^(-4/5)/ 1! ] * x⁽¹⁾   =  x / 5

3rd term  (n = 2)  =  [(-4/25)(1 + 0)^(-9/5)/ 2! ] * x⁽²⁾ =   ( -2 / 25)x²

4th term (n = 3)  =  [ (36/125)(1 + 0) ^(-14/5) / 3! ] * x⁽³⁾  =  ( 6 / 125)x³

5th term ( n = 4)  = [ (-504/3125) (1 + 0)(-19/5) / 4!] * x⁽⁴⁾ =  ( -21 / 3125)x⁴ 

etc.....

Taylor series expansion is used for such series. In general, we have: (1 + x)^(n/m) = [1 +(n/m)]x - [n(m - n)/(2!m^2)]x^2 + [n(m -n)(2m - n)/(3!m^3)]x^3 - .......etc.
So, ( 1+ x)^1/5 = 1 + x/5 - 2x^2/25 + 6x^3/125 - 21x^4/625 + 399x^5/15,625 - 1,596x^6/78,125 + .......etc.

Nevermind, I feel stupid for posting something stupid. Don't listen to me...

The answer is C

We can get this quadratic equation into standard form.

y  =  x(x - 4)          Distribute the x to both terms in parenthesees.

y  =  x² - 4x

Is that all the question asks for?

true     

 (3x^2-4y)(2x^3+5y)

Distribute the first term in the first set of parentheses over the two terms in the second set

3x^2 * 2x^3  +  3x^2 *5y  =      6x^5   +  15x^2y    (1)

Distribute the second term in the first set of parenthesesover the two terms in the second set

-4y*2x^3 - 4y*5y   =    -8yx^3 - 20y^2    (2)

Combine (1) and (2)

6x^5 + 15x^2y - 8x^3y - 20y^2

Since BDA and BDC are supplemental, tan( BDA)   =  - tan (BDC) →  - tan (BDA)  = tan (BDC)

Using right angle trigonometry,  tan (BDA)  =  7/12

So  -tan(BDA)  =  -7/12  = tan (BDC)

thanks again you guys for your help, i am now able to continue working on my project !! :)

amazing thanks!!