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telo118 [61]
4 years ago
8

What are the root of equation x2 -4x+29 in simplest a + b form

Mathematics
2 answers:
bagirrra123 [75]4 years ago
8 0

Answer:

Correct answer:  (x - 2 - 5 i) (x - 2 + 5 i)

Step-by-step explanation:

Given:

x² - 4 x + 29 = (  ) (  ) = ?  

this equation has no solution in the set of real numbers

x² - 4 x + 29 = x² - 2 · x · 2 + 2² - 2² + 29 = x² - 2 · x · 2 + 4 - 4 + 29

x² - 4 x + 29 = (x - 2)² + 25 = (x - 2)² - ( - 25) = (x - 2)² - ( - 1 · 5²)

We know that  the imaginary unit is the equal  i = √- 1 ⇒ i² = - 1

x² - 4 x + 29 = (x - 2)² - ( 5 i)²  this is the difference of the square

x² - 4 x + 29 = (x - 2 - 5 i) (x - 2 + 5 i)

these are conjugated complex numbers

God is with you!!!

Elanso [62]4 years ago
6 0
Idk hope this helps
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3 years ago
50 points! I understand A. and B. but I would really appreciate help with C.
FromTheMoon [43]

Answer:

51.72\text{ cells per hour}

Step-by-step explanation:

So, the function, P(t), represents the number of cells after t hours.

This means that the derivative, P'(t), represents the instantaneous rate of change (in cells per hour) at a certain point t.

C)

So, we are given that the quadratic curve of the trend is the function:

P(t)=6.10t^2-9.28t+16.43

To find the <em>instanteous</em> rate of growth at t=5 hours, we must first differentiate the function. So, differentiate with respect to t:

\frac{d}{dt}[P(t)]=\frac{d}{dt}[6.10t^2-9.28t+16.43]

Expand:

P'(t)=\frac{d}{dt}[6.10t^2]+\frac{d}{dt}[-9.28t]+\frac{d}{dt}[16.43]

Move the constant to the front using the constant multiple rule. The derivative of a constant is 0. So:

P'(t)=6.10\frac{d}{dt}[t^2]-9.28\frac{d}{dt}[t]

Differentiate. Use the power rule:

P'(t)=6.10(2t)-9.28(1)

Simplify:

P'(t)=12.20t-9.28

So, to find the instantaneous rate of growth at t=5, substitute 5 into our differentiated function:

P'(5)=12.20(5)-9.28

Multiply:

P'(5)=61-9.28

Subtract:

P'(5)=51.72

This tells us that at <em>exactly</em> t=5, the rate of growth is 51.72 cells per hour.

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Substitute the values for a, b, and c into b2 – 4ac to determine the discriminant. Which quadratic equations will have two real
garik1379 [7]

The complete question is

"Substitute the values for a, b, and c into b2 – 4ac to determine the discriminant. Which quadratic equations will have two real number solutions? (The related quadratic function will have two x-intercepts.) Check all that apply.

0 = 2x^2 – 7x – 9

0 = 4x^ 2 – 3x – 1

The quadratic equations that have real number solutions are; 4x^2 – 3x – 1, and 2x^2 – 7x – 9.

<h3>What is the formula for Discriminant?</h3>

The formula for finding the discriminant is

b^2 - 4ac

The solution contains the term \sqrt{b^2 - 4ac} which will be:

Real and distinct if the discriminant is positive

Real and equal if the discriminant is 0

Non-real and distinct roots if the discriminant is negative

For the quadratic equation 2x^2 - 7x - 9

b^2 - 4ac

= (-7) ^2 - 4( 2) ( -9)\\\\= 49 + 72 = 121

This equation has two real number solutions.

For the quadratic equation 4x^ 2 - 3x- 1

b^2 - 4ac

= (-3) ^2 - 4( 4) ( -1)\\\\= 9 + 16 = 25

This equation will have two real number solutions.

Learn more on discriminant here:

brainly.com/question/1537997

#SPJ1

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