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3 years ago

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Construct a polynomial function with the following properties: third degree, only real coefficients, −3 and 3+i are two of the z

eros, y-intercept is −90.

1 answer:

irina [24]3 years ago

5 0

Answer:

$\boxed{-3(x+3)(x^2-6x+10)}$

Step-by-step explanation:

Hello,

As the polynomial has only real coefficients, it means that 3-i is a zero too, because we apply the Conjugate Zeros Theorem.

It means that we can write the expression as below, k being a real number that we will have to identify.

$k(x+3)(x-3-i)(x-3+i)=k(x+3)((x-3)^2-i^2)\\\\=k(x+3)(x^2-6x+9+1)\\\\=k(x+3)(x^2-6x+10)$

And for x = 0, y = -90 so we can write

-90=k*3*10, meaning that k=-3

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

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A flat circular plate has the shape of the region x squared plus y squared less than or equals 1x2+y2≤1. the plate, including t

vredina [299]

You're looking for the extreme values of $x^2+3y^2+13x$ subject to the constraint $x^2+y^2\le1$ .

The target function has partial derivatives (set equal to 0)

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$\dfrac{\partial(x^2+3y^2+13x)}{\partial y}=6y=0\implies y=0$

so there is only one critical point at $\left(-\dfrac{13}2,0\right)$ . But this point does not fall in the region $x^2+y^2\le1$ . There are no extreme values in the region of interest, so we check the boundary.

Parameterize the boundary of $x^2+y^2\le1$ by

$x=\cos u$

$y=\sin u$

with $0\le u$ . Then $t(x,y)$ can be considered a function of $u$ alone:

$t(x,y)=t(\cos u,\sin u)=T(u)$

$T(u)=\cos^2u+3\sin^2u+13\cos u$

$T(u)=3+13\cos u-2\cos^2u$

$T(u)$ has critical points where $T'(u)=0$ :

$T'(u)=-13\sin u+4\sin u\cos u=\sin u(4\cos u-13)=0$

$(1)\quad\sin u=0\implies u=0,u=\pi$

$(2)\quad4\cos u-13=0\implies\cos u=\dfrac{13}4$

but $|\cos u|\le1$ for all $u$ , so this case yields nothing important.

At these critical points, we have temperatures of

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so the plate is hottest at (1, 0) with a temperature of 14 (degrees?) and coldest at (-1, 0) with a temp of -12.

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Answer:

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Read 2 more answers

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ivolga24 [154]

Answer:

Interval of 50 on both axis

Step-by-step explanation:

Given

$(x_1,y_1) = (-50,0)$

$(x_2,y_2) = (150,100)$

$(x_3,y_3) = (200,-100)$

$(x_4,y_4) = (350,50)$

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There are several ways to do this, but I will use the observation method, since the dataset is small.

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$x = \{-50,150,200,350,-250}$

Each element of the data set is a multiple of 50.

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Each element of the data set is a multiple of 50.

Hence, an interval of 50 can be used on the y-axis

<em>So, an interval of 50 can be used on both axes</em>

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Answer:

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Step-by-step explanation:

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