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yarga [219]
3 years ago
5

Find a degree 3 polynomial with real coefficients having zeros 2 and 2 − 2 i and a lead coefficient of 1. Write P in expanded fo

rm. Be sure to write the full equation, including P ( x ) = .
Mathematics
1 answer:
Zanzabum3 years ago
6 0

Answer:

The polynomial with real coefficients having zeros 2 and 2 - 2i is

x³ - 6x² + 16x - 16 = 0

Step-by-step explanation:

Given that a polynomial has zeros at 2 and 2 - 2i, we want to write this polynomial.

We have

x - 2 = 0

x - (2 - 2i) = 0

=> x - 2 + 2i = 0

Since the polynomial has real coefficients, and 2 - 2i is a zero of the polynomial, the conjugate of 2 - 2i, which is 2 + 2i is also a polynomial.

x - (2 + 2i) = 0

=> x - 2 - 2i = 0

Now,

P(x) = (x - 2)(x - 2 + 2i)(x - 2 - 2i) = 0

= (x - 2)((x - 2)² - (2i)²) = 0

= (x - 2)(x² - 4x + 8) = 0

= x³ - 4x² + 8x - 2x² + 8x - 16 = 0

= x³ - 6x² + 16x - 16 = 0

This is the polynomial required.

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Answer:

0

Step-by-step explanation:

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3 years ago
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A 2-column table with 7 rows. The first column is labeled x with entries negative 3, negative 2, negative 1, 0, 1, 2, 3. The sec
DochEvi [55]

The statements that are true about the intervals of the continuous function are Options 2, 4  and 5

  • f(x) ≤ 0 over the interval [0, 2].
  • f(x) > 0 over the interval (–2, 0).
  • f(x) ≥ 0 over the interval [2, ).

<h3>What is the statement about?</h3>

Looking at the values given, the intervals which satisfies the condition are known to be:

f(x)<=0 over the interval [0,2]

f(x)>0 over the interval (-2,0)

f(x)>=0 over the interval [2,∞)

Because:

Since the table with x  and f(x) values, we have to examine analyze the table and see the each option that is in line with f(x) or not .

Examine the values of x that is from -3 to 3, the f(x) values are both positive and negative . hence f(x)>0 is false over the interval (-∞,3)

Looking at the the interval from 0 to 2, the f(x) values are 0 and negative. Hence, f(x)<=0 over the interval [0,2]

When you look over the interval (-1,1), the f(x) values are said to be both positive and negative and as such, f(x)<0 is false over the interval (-1,1)

When you look at the interval (-2,0) , the f(x)  is positive and as such, f(x)>0 over the interval (-2,0)

Looking at the interval  [2,∞), f(x) is positive and as such, f(x)>=0 over the interval [2,∞)

Therefore, Option 2, 4 and 5 are correct.

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6 0
2 years ago
Use the equation xy= 12.6. Find the value of y when x=7
Marat540 [252]

Answer: y = 1.8

Step-by-step explanation:

xy = 12.6   If x is 7  but 7  in the equation for x

7y = 12.6   Divide both sides by 7

y = 1.8  

6 0
3 years ago
The prism shown has a volume of 35 cm^3
artcher [175]

Answer:

Height = 2.5

Step-by-step explanation:

Given

Volume = 35

Length =7

Base=4 cm

Required

Determine the height

The volume is calculated as:

Volume = Height * Base\ Area

Since, it has a triangular cross-section, the expression becomes:

Volume = Height * \frac{1}{2} * Base * Length

So, we have:

35 = Height * \frac{1}{2} * 4 * 7

35 = Height * 14

Height = 35/14

Height = 2.5

7 0
3 years ago
Where is the blue point on the number line -5 0 5
noname [10]

The blue point on the number line is 5 units away from zero, which is: 5.

<h3>What is a Number Line?</h3>

A number line is an arrangement of numbers from point zero, on both sides of the starting point, which is zero.

In the number line given, the blue point is 5 units away from the starting point, zero.

Therefore, the blue point is 5 on the number line.

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3 0
2 years ago
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