Hello,

Let's assume

$t=z- \dfrac{1}{z} \\\\

t^2=z^2-2+ \dfrac{1}{z^2}= 11-2=9==\ \textgreater \ t=3\\\\

t^3=z^3-3z^2* \dfrac{1}{z} +3z* \dfrac{1}{z^2} - \dfrac{1}{z^3} \\\\
=z^3 - \dfrac{1}{z^3} -3(z-\dfrac{1}{z})\\\\
z^3 - \dfrac{1}{z^3}=t^3+3t=3^3+9=27+9=36


$

7 0

3 years ago

HELP!!!!!!!!!!! !!!!!!!!!!

Sonja [21]

Answer:

second one is 3, the last one is 4560

Step-by-step explanation:

5 0

2 years ago

Mike can make 7 holiday cards with the cardstock he already has on hand. He is also buying more paper, and he can make 12 additi

ipn [44]

Answer:y = 12x + 7

Step-by-step explanation:

8 0

1 year ago

Solving equations with one variable

kupik [55]

Answer:

1) b= -20

Step-by-step explanation:

You bring the variable you want to find to one side by either bringing over the second term to the other side of the equation and change its sign or if it's a fraction you multiply both sides to remove the denominator and then solve

5 0

3 years ago