All categories

3 years ago

Degree 5 zeros 8,i,6i Enter the remaining zeros

Mathematics

2 answers:

LenKa [72]3 years ago

5 0

-i and -6i

Hope it helps! :D

NARA [144]3 years ago

4 0

If a polynomial has real coefients, if a+bi is a root, then a-bi is also a root
so

8 and i and 6i
since i and 6i is zeores then -i and -6i are also zeroes

remaining zeroes are -i and -6i

You might be interested in

Two students wanted to race 100 yards across a football field. How many feet will they run? (

Molodets [167]

Answer:

300 feet

Step-by-step explanation:

6 0

2 years ago

(i) 3 (a + 5b) + a (a + 4) (ii) y (10 - y) + 3 (y - 2) (iii) 2 (8a - 5b) + 3 (5a - 12) (iv) 3 (y - 3) + ( 8 - 6y + x) (v) a (a -

AveGali [126]

Answer:

$3 (a + 5b) + a (a + 4) = a^2 + 7a + 15b$

$y (10 - y) + 3 (y - 2) = - y^2+13y - 6$

$2 (8a - 5b) + 3 (5a - 12) = 31a -10b - 36$

$3 (y - 3) + ( 8 - 6y + x) = -3y + x-1$

$a (a - 2b) + b (b + 2a - c) =a^2 + b^2 - bc$

$5 (x - y + z) + (4x + 3y) =9x - 2y +5z$

Step-by-step explanation:

Solving (i):

$3 (a + 5b) + a (a + 4)$

Open brackets

$3 (a + 5b) + a (a + 4) = 3a + 15b + a^2 + 4a$

Collect like terms

$3 (a + 5b) + a (a + 4) = a^2 + 4a+3a + 15b$

$3 (a + 5b) + a (a + 4) = a^2 + 7a + 15b$

Solving (ii)

$y (10 - y) + 3 (y - 2)$

Open bracket

$y (10 - y) + 3 (y - 2) = 10y - y^2 + 3y - 6$

Collect like terms

$y (10 - y) + 3 (y - 2) = - y^2+10y + 3y - 6$

$y (10 - y) + 3 (y - 2) = - y^2+13y - 6$

Solving (iii)

$2 (8a - 5b) + 3 (5a - 12)$

Open bracket

$2 (8a - 5b) + 3 (5a - 12) = 16a -10b + 15a - 36$

Collect like terms

$2 (8a - 5b) + 3 (5a - 12) = 16a+ 15a -10b - 36$

$2 (8a - 5b) + 3 (5a - 12) = 31a -10b - 36$

Solving (iv)

$3 (y - 3) + ( 8 - 6y + x)$

Open bracket

$3 (y - 3) + ( 8 - 6y + x) = 3y - 9 + 8 - 6y + x$

Collect like terms

$3 (y - 3) + ( 8 - 6y + x) = 3y - 6y + x- 9 + 8$

$3 (y - 3) + ( 8 - 6y + x) = -3y + x-1$

Solving (v):

$a (a - 2b) + b (b + 2a - c)$

Open bracket

$a (a - 2b) + b (b + 2a - c) =a^2 - 2ab + b^2 + 2ab - bc$

Collect like terms

$a (a - 2b) + b (b + 2a - c) =a^2 + 2ab- 2ab + b^2 - bc$

$a (a - 2b) + b (b + 2a - c) =a^2 + b^2 - bc$

Solving (vi)

$5 (x - y + z) + (4x + 3y)$

Open brackets

$5 (x - y + z) + (4x + 3y) =5x - 5y +5z+4x +3y$

Collect like terms

$5 (x - y + z) + (4x + 3y) =5x +4x +3y- 5y +5z$

$5 (x - y + z) + (4x + 3y) =9x - 2y +5z$

6 0

3 years ago

How many possible combinations of 4 numbers 1-9?

beks73 [17]

1. There are 10, 000 combinations, you have 10 choices for the four digit and there are 10 x 10 x 10 x 10 which is 4-digit combination from 1-9. This is called combinations; the calculation comes out to 10^4.

5 0

3 years ago

PLEASE HELP ME I DONT UNDERSTAND

max2010maxim [7]

Answer:

Step-by-step explanation:

the square root of 2 is not rational

5 0

2 years ago

The polynom

rodikova [14]

Answer:

Do you want the answer for all of them

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers