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Alenkinab [10]
3 years ago
9

What is the polynomial function of least degree whose only zeros are -3, ,-2, and 4 ?

Mathematics
1 answer:
victus00 [196]3 years ago
7 0

Answer: (x³ - 2x² - 11x + 12)

Step-by-step explanation:

Recall that zeroes can be transformed into factors by subtracting them from x. This gives us the following factors:

(x - 1)(x + 3)(x - 4)

Now, if you multiply the first two factors together, you get the following:

(x² + 2x - 3)

Multiply that by the last factor, (x - 4), and you get this:

(x³ + 2x² - 3x - 4x² - 8x + 12)

This can be simplified:

(x³ - 2x² - 11x + 12)

And there's your final answer. Hope this helped! (x³ - 2x² - 11x + 12)

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Evaluate the line integral, where c is the given curve. (x + 9y) dx + x2 dy, c c consists of line segments from (0, 0) to (9, 1)
viktelen [127]
\displaystyle\int_C(x+9y)\,\mathrm dx+x^2\,\mathrm dy=\int_C\langle x+9y,x^2\rangle\cdot\underbrace{\langle\mathrm dx,\mathrm dy\rangle}_{\mathrm d\mathbf r}

The first line segment can be parameterized by \mathbf r_1(t)=\langle0,0\rangle(1-t)+\langle9,1\rangle t=\langle9t,t\rangle with 0\le t\le1. Denote this first segment by C_1. Then

\displaystyle\int_{C_1}\langle x+9y,x^2\rangle\cdot\mathbf dr_1=\int_{t=0}^{t=1}\langle9t+9t,81t^2\rangle\cdot\langle9,1\rangle\,\mathrm dt
=\displaystyle\int_0^1(162t+81t^2)\,\mathrm dt
=108

The second line segment (C_2) can be described by \mathbf r_2(t)=\langle9,1\rangle(1-t)+\langle10,0\rangle t=\langle9+t,1-t\rangle, again with 0\le t\le1. Then

\displaystyle\int_{C_2}\langle x+9y,x^2\rangle\cdot\mathrm d\mathbf r_2=\int_{t=0}^{t=1}\langle9+t+9-9t,(9+t)^2\rangle\cdot\langle1,-1\rangle\,\mathrm dt
=\displaystyle\int_0^1(18-8t-(9+t)^2)\,\mathrm dt
=-\dfrac{229}3

Finally,

\displaystyle\int_C(x+9y)\,\mathrm dx+x^2\,\mathrm dy=108-\dfrac{229}3=\dfrac{95}3
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3 years ago
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Convert them to comparable thingummys

convert to fraction

percent means parts out of 100
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Step-by-step explanation:

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