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Kamila [148]
4 years ago
7

Struggling with a homework question here :( spent 3 hours on it the past two days. Please please help!!!! Photo is attatched.

Mathematics
1 answer:
Alexandra [31]4 years ago
3 0

i'm sorry I have no idea and this might not help but you could use this app called photomath (if you haven't already used it)(I just responded just in case no one did with the actual answer and you still needed help)

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Find the x-intercepts of the parabola with vertex (1, -9) and y intercept at (0, -6).
MaRussiya [10]
Hello,

The equation of the parabola is
y=k(x-1)²-9
and -6=k(0-1)²-9==>k=3

x-intercepts is:
y=0==> 3(x-1)²-9=0==>(x-1)²=3
==> x=1+√3 or x=1-√3
7 0
3 years ago
Find all complex solutions of the equation z⁴-1-i=0.​
bulgar [2K]

Answer:

Step-by-step explanation:

hello :

z⁴=1+i      given : z² = t    t is the complex number  so :   z⁴= (z²)² =t²

solve for  t  this equation : t² = 1+i

if : t = x+iy     t² =(x+iy)² = x²+2xiy+(iy)² =x²-y²+2xyi ......     (  i² = -1)

t² = 1+i  means : x²-y²+2xyi = 1+i

you have this system :   x²-y² = 1.....(*)

                                      2xy  = 1.....(**)

slve for   x  and y    you ca add this equation :   ....(***)

( use : t² = 1+i

/t²/ = /1+i/   so :/t/² = /1+i/ .... (√(x²+y²))²=√(1²+1²) =√2  means x²+y² = √2)

the system is :   x²-y² = 1.....(*)

                         2xy  = 1.....(**)

                         x²+y² = √2 ....(***)

add (*) and (***) : 2x²= 1+√2

x² = (√2+1)/2

substrac  (*) and (***)  you have : y² = (√2-1)/2

use (**) the proudect  xy is positif (same sign)  so :

x=√( (√2+1)/2)  and  y = √( (√2-1)/2)   so :  t1 =√( (√2+1)/2)+i√( (√2-1)/2)  

x= -√( (√2+1)/2)  and  y = -√( (√2-1)/2) so :t2= -t1

same method for equation : z² =t1  ..(2 solution )  and z² = t2..(2 solution )

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3 years ago
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V is 60 degree because the v is in the line of 60

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1 plus 5 is six 1+5=6
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