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MAVERICK [17]
2 years ago
11

Find a third-degree polynomial equation with the rational coefficients that has roots -3 and 1+i

Mathematics
1 answer:
jolli1 [7]2 years ago
3 0

Answer:

x³ + x² − 4x + 6 = 0

Step-by-step explanation:

Imaginary roots come in conjugate pairs.  So if 1+i is a root, then 1−i is also a root.

(x − (-3)) (x − (1+i) (x − (1−i)) = 0

(x + 3) (x² − (1+i) x − (1−i) x + (1+i) (1−i)) = 0

(x + 3) (x² − x − ix − x + ix + 1 − i²) = 0

(x + 3) (x² − 2x + 2) = 0

x (x² − 2x + 2) + 3 (x² − 2x + 2) = 0

x³ − 2x² + 2x + 3x² − 6x + 6 = 0

x³ + x² − 4x + 6 = 0

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Answer:

Rotation was used to transform the figure.

Step-by-step explanation:

Since only one transformation was used, rotating the pre-image ABC 180° counterclockwise around the origin formed the image DEF.

I hope this helped! :-)

6 0
3 years ago
(4+5)÷3×4= <br>can someone help me with this question please
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8 0
2 years ago
Emily is entering a bicycle race for charity. Her mother pledges $0.20 for every 0.5 mile she bikes. If Emily bikes 10 miles, ho
elena-s [515]

Answer: $4

Step-by-step explanation:

In this situation let's set up a proportional relationship .

\frac{0.20}{x }  = \frac{0.5}{10}     Now solve by cross multiply  

0.5x  = 2      Divide both sides by 0.5  

x = 4  

This means that if Emily bikes 10 miles, her mother will donate $4.

5 0
2 years ago
Read 2 more answers
(0,0) is a solution to y&gt; -3x + 4
solniwko [45]

Answer:

(0,0) is a solution.

Step-by-step explanation:

To see if this is true you can put (0,0) into the y and x variables like so: 0>-3(0)+4. When you solve this you get 0>4 so this statement is true

7 0
2 years ago
Write out the first few terms of the Picard iteration scheme for each of the following initial value problems. Where possible, f
Artist 52 [7]
Given an ODE x'=f(t,x) with initial condition x(t_0)=x_0, the general process is to write the ODE as an integral equation,

x(t)=x_0+\displaystyle\int_{t_0}^tf(u,x(u))\,\mathrm du

By setting x_0(t)=x_0 for all t, we get the following recurrence for n\ge1.

x_{n+1}(t)=x_0+\displaystyle\int_{t_0}^tf(u,x_n(u))\,\mathrm du

From this we work towards finding a pattern for x_n so that we can find a solution of the form x=\lim\limits_{n\to\infty}x_n.

\begin{cases}x'=x+2\\x(0)=2\end{cases}

Write this as the integral equation,

x_{n+1}=x(0)+\displaystyle\int_{t_0}^t(x_n(u)+2)\,\mathrm du

First step:

x_1=x(0)+\displaystyle\int_{t_0}^t(x_0(u)+2)\,\mathrm du
x_1=2\displaystyle\int_0^t\mathrm du
x_1=2t

Second step:

x_2=x(0)+\displaystyle\int_{t_0}^t(x_1(u)+2)\,\mathrm du
x_2=\displaystyle\int_0^t(2u+2)\,\mathrm du
x_2=t^2+2t

Third step:

x_3=x(0)+\displaystyle\int_{t_0}^t(x_2(u)+2)\,\mathrm du
x_3=\displaystyle\int_0^t(t^2+2t+2)\,\mathrm du
x_3=\dfrac13t^3+t^2+2t

Fourth step:

x_4=x(0)+\displaystyle\int_{t_0}^t(x_3(u)+2)\,\mathrm du
x_4=\displaystyle\int_0^t\left(\frac13u^3+u^2+2u+2\right)\,\mathrm du
x_4=\dfrac1{4\times3}t^4+\dfrac13t^3+t^2+2t

You should already start seeing a pattern. Recall that

e^t=\displaystyle\sum_{k=0}^\infty\frac{x^k}{k!}=1+t+\frac{t^2}2+\frac{t^3}{2\times3}+\frac{t^4}{2\times3\times4}+\cdots

Multiplying this by 2 gives

2e^t=2+2t+t^2+\dfrac{t^3}3+\dfrac{t^4}{4\times3}+\cdots

which matches the solution we have for x_4 except for that first term. So subtracting that, we find a solution of

x=2e^t-2

with a domain of t\in\mathbb R.

Hopefully this gives some insight on how to approach the other two problems.
6 0
3 years ago
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