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

Write an equation of parabola in vertex form. NO LINKS!!!!!! Part 1

jolli1 [7]

Answer: lppkoijhuytfrdeswasdxcvbn

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

A ladder is leaning against a building.The distance from the bottom of the ladder to the building is 6ft shorter than the length

Akimi4 [234]

Answer:

15 ft

Step-by-step explanation:

Hi, the illustration for the problem is a right triangle with:

hypotenuse (C)= the length of the ladder = L

horizontal side(A) = distance from bottom of the ladder to the building = L - 6

vertical side(B) = distance from the top of the ladder to the bottom of the building = L - 3

So, we can use Pythagoras formula:

A2 +B2= C2

(L – 6 )² + (L-3)² = L²

L²-12L+36+L²-6L+9 =L²

L2 -18L+45 =0

APPLYING QUADRATIC FORMULA WE OBTAIN:

L =15 OR L=3

If L=3

Vertical side = L-3 = 0 (Length can´t be 0)

So L=15

6 0

3 years ago

Express (16³) 1/4 in simplest radical form.

Doss [256]

Answer:

Step-by-step explanation:

4 0

1 year ago