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3 years ago

Which polynomial equations have -i as one of their roots?

1 answer:

8 0

I = sqrt(-1)
i^2 = -1
(-i)^2 = (-i)(-i) = i^2 = -1
(-i)^3 = (-i)(-i)(-i) = (-i)(i^2) = i

use these facts to find which equation is correct if x = -i

i bet on 2,3 or 4

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GaryK [48]

Answer:

Step-by-step explanation:

Does it say what x and y equals

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3 years ago

Read 2 more answers

Segment FG begins at point F(-2, 4) and ends at point G (-2, -3). Segment FG is translated by (x, y) → (x – 3, y + 2) and then r

Mariana [72]

Answer:

The length of the segment F'G' is 7.

Step-by-step explanation:

From Linear Algebra we define reflection across the y-axis as follows:

$(x',y')=(-x, y)$ , $\forall\, x, y\in \mathbb{R}$ (Eq. 1)

In addition, we get this translation formula from the statement of the problem:

$(x',y') =(x-3,y+2)$ , $\forall \,x,y\in \mathbb{R}$ (Eq. 2)

Where:

$(x, y)$ - Original point, dimensionless.

$(x', y')$ - Transformed point, dimensionless.

If we know that $F(x,y) = (-2, 4)$ and $G(x,y) = (-2,-3)$ , then we proceed to make all needed operations:

Translation

$F''(x,y) = (-2-3,4+2)$

$F''(x,y) = (-5,6)$

$G''(x,y) = (-2-3,-3+2)$

$G''(x,y) = (-5,-1)$

Reflection

$F'(x,y) = (5, 6)$

$G'(x,y) = (5,-1)$

Lastly, we calculate the length of the segment F'G' by Pythagorean Theorem:

$F'G' = \sqrt{(5-5)^{2}+[(-1)-6]^{2}}$

$F'G' = 7$

The length of the segment F'G' is 7.

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3 years ago

A cylinder with a height of 4 m has a volume of approximately

tigry1 [53]

Answer:

18m

Step-by-step explanation:

113.04m/3.14 = 36m/4m = 9m*2 = 18m

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3 years ago

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rjkz [21]

Answer:

Step-by-step explanation:

Part A

Solve 4,500 ÷ 90. Enter your answer in the box.

Answer : 50

Part B

Select all the expressions that are equal to 4,500 ÷ 90.

A. 4,500 ÷ 9 tens ✔️

B. 450 ÷ 90 ❌

C. 450 ÷ 9 ❌

D. 4,500 ÷ 9 ❌

E. 4,500 ÷ 90 tens ❌

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3 years ago

No Spam. I will mark Brainliest, if you get it right. A tank's length is 4 feet longer than its width. The height is 2 feet more

dmitriy555 [2]

Answer: 3

Step-by-step explanation:

Let say x the width of the tank,

x+4 its length,

x+2 its height

$Volume=105=x(x+4)(x+2)\\\\P(x)=x^3+6x^2+8x-105=0\\\\105=3*5*7\\\\P(1)\neq 0\\P(-1)\neq 0\\P(3)=0\\\\\begin{array}{|c|cccc|}&x^3&x^2&x&1\\&1&6&8&-105\\x=3&&3&27&105\\----&--&--&--&--\\&1&9&35&0\\\end{array}\\\\\\P(x)=x^3+6x^2+8x-105=(x-3)(x^2+9x+35)\\And \ x^2+9x+35\ is\ not \ factorizable\ (\Delat=9^2-4*35 < 0)\\$

<u>width=3</u>

length=3+4=7

height=3+2=5

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3 years ago