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alexdok [17]
3 years ago
11

Find a cubic function that has the roots 5 and 3-2i

Mathematics
1 answer:
Zolol [24]3 years ago
4 0

Answer:

P(x)=x^3-11x^2+43x-65

Step-by-step explanation:

If the complex number 3-2i is a root of a cubic function, then the complex number 3+2i is a root too. Thus, the cubic function has three known roots 5,\ 3-2i,\ 3+2i and can be written as

P(x)=(x-5)(x-(3-2i))(x-(3+2i)),\\ \\P(x)=(x-5)(x^2-x(3-2i+3+2i)+(3-2i)(3+2i)),\\ \\P(x)=(x-5)(x^2-6x+9-4i^2),\\ \\P(x)=(x-5)(x^2-6x+9+4),\\ \\P(x)=(x-5)(x^2-6x+13),\\ \\P(x)=x^3-11x^2+43x-65.


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How can you simplify 24 to 96 ​
Nezavi [6.7K]

Step-by-step explanation:

Reduce 24/96 to lowest terms

Find the GCD (or HCF) of numerator and denominator. GCD of 24 and 96 is 24.

24 ÷ 2496 ÷ 24.

Reduced fraction: 14. Therefore, 24/96 simplified to lowest terms is 1/4.

7 0
2 years ago
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20x^2=256+128x. Solve by Factoring.
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Hey there,
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3 years ago
If d= 2, what is the value of 8/(4-d)? A.1 <br>B. 4/3<br>C. 4<br>D. 0​
zimovet [89]
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3 years ago
Which polygons are NOT quadrilaterals? Select THREE that apply.
Afina-wow [57]

Answer:

a cylinder, a sphere, and that little half sphere thing, oh and a triangular pyramid

Step-by-step explanation:

6 0
3 years ago
Solve the following differential equations or initial value problems. In part (a), leave your answer in implicit form. For parts
shepuryov [24]

Answer:

(a) (y^5)/5 + y^4 = (t^3)/3 + 7t + C

(b) y = arctan(t(lnt - 1) + C)

(c) y = -1/ln|0.09(t + 1)²/t|

Step-by-step explanation:

(a) dy/dt = (t^2 + 7)/(y^4 - 4y^3)

Separate the variables

(y^4 - 4y^3)dy = (t^2 + 7)dt

Integrate both sides

(y^5)/5 + y^4 = (t^3)/3 + 7t + C

(b) dy/dt = (cos²y)lnt

Separate the variables

dy/cos²y = lnt dt

Integrate both sides

tany = t(lnt - 1) + C

y = arctan(t(lnt - 1) + C)

(c) (t² + t) dy/dt + y² = ty², y(1) = -1

(t² + t) dy/dt = ty² - y²

(t² + t) dy/dt = y²(t - 1)

(t² + t)/(t - 1)dy/dt = y²

Separating the variables

(t - 1)dt/(t² + t) = dy/y²

tdt/(t² + t) - dt/(t² + t) = dy/y²

dt/(t + 1) - dt/(t(t + 1)) = dy/y²

dt/(t + 1) - dt/t + dt/(t + 1) = dy/y²

Integrate both sides

ln(t + 1) - lnt + ln(t + 1) + lnC = -1/y

2ln(t + 1) - lnt + lnC = -1/y

ln|C(t + 1)²/t| = -1/y

y = -1/ln|C(t + 1)²/t|

Apply y(1) = -1

-1 = ln|C(1 + 1)²/1|

-1 = ln(4C)

4C = e^(-1)

C = (1/4)e^(-1) ≈ 0.09

y = -1/ln|0.09(t + 1)²/t|

8 0
3 years ago
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