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2 years ago

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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Alinara [238K]

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.

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3 years ago

(4+5)÷3×4= <br>can someone help me with this question please

Darina [25.2K]

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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> -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

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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