For what real values of does the quadratic 12x^2+k*x+27=0 have nonreal roots

Step-by-step explanation:

Discriminant : k² - 4(12)(27) < 0

So k² - 1296 < 0, -36 < k < 36.

If a quadratic has nonreal roots, then the discriminant .  b^2 - 4ac

Plug in the values for b, a, and c into this equation to get .  k^2-4 *12*27 < 0

Rearrange to get .  k^2 < *12*27

Now factorize 4*12*27 and solve to get - 36 < k<26

Step-by-step explanation:

hope this help

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use the power rule  

( a b ) n = a n b n

to distribute the exponent.

apply the product rule to  

3 x 2 y 3 . ( 3 x 2 ) 2 ( y 3 ) 2

apply the product rule to  

3 x 2 . 3 2 ( x 2 ) 2 ( y 3 ) 2

raise  

3

to the power of  

2 . 9 ( x 2 ) 2 ( y 3 ) 2 multiply the exponents in   ( x 2 ) 2

.

apply the power rule and multiply exponents,  

( a m ) n = a m n . 9 x 2 ⋅ 2 ( y 3 ) 2

multiply  

2   by   2 . 9 x 4 ( y 3 ) 2

multiply the exponents in  

( y 3 ) 2 .

apply the power rule and multiply exponents,  

( a m ) n = a m n . 9 x 4 y 3

⋅

2

multiply  

3   by   2 . 9 x 4 y 6

step-by-step explanation:

so far i think you have the right thing going for you just remember soh cah toh

step-by-step explanation:

sine over hypotenuse

cosine   equals adjacent

tangent over hypotenuse