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3 years ago

Which polynomial equations have -i as one of their roots? A) x3 + 5x2 - 20 = 0

Mathematics

2 answers:

oksian1 [2.3K]3 years ago

8 0

Answer:

Equation B, C and D have -i as one of their roots

Step-by-step explanation:

We have to check the polynomial equations that have -i as one of the roots.

Thus, we have to check whether

$f(-i) = 0$

$f(x) x^3 + 5x^2 - 20 = 0\\f(-i) = (-i)^3 + 5(-i)^2 - 20 \neq 0$

$f(x) x^3 -x^2 + x - 1 = 0\\f(-i) = (-i)^3 - (-i)^2 -i -1 = i + 1-i-1= 0$

$f(x) x^3 +3x^2 + x + 3 = 0\\f(-i) = (-i)^3 +3 (-i)^2 -i +3 = i -3-i+3= 0$

$f(x) x^3 -2x^2 + x -2= 0\\f(-i) = (-i)^3 -2 (-i)^2 -i -2 = i +2-i-2= 0$

$F(x) = x^3 - 6x^2 - 16x + 96 = 0 \\f(-i) = (-i)^3 -6(-i)^2+16i+96 \neq 0$

kirill115 [55]3 years ago

6 0

The 2nd, 3rd, and 4th ones

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otez555 [7]

Answer:

Step-by-step explanation:

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3 years ago

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A tank with a capacity of 1000 L is full of a mixture of water and chlorine with a concentration of 0.02 g of chlorine per liter

faltersainse [42]

At the start, the tank contains

(0.02 g/L) * (1000 L) = 20 g

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Pure water is pumped into the tank, so no chlorine is flowing into it, but is flowing out at a rate of

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In case it's unclear why this is the case:

The amount of liquid in the tank at the start is 1000 L. If water is pumped in at a rate of 10 L/s, then after t s there will be (1000 + 10t ) L of liquid in the tank. But we're also removing 25 L from the tank per second, so there is a net "gain" of 10 - 25 = -15 L of liquid each second. So the volume of liquid in the tank at time t is (1000 - 15t ) L. Then the concentration of chlorine per unit volume is c (t ) divided by this volume.

So the amount of chlorine in the tank changes according to

$\dfrac{\mathrm dc(t)}{\mathrm dt}=-\dfrac{5c(t)}{200-3t}$

which is a linear equation. Move the non-derivative term to the left, then multiply both sides by the integrating factor 1/(200 - 5t )^(5/3), then integrate both sides to solve for c (t ):

$\dfrac{\mathrm dc(t)}{\mathrm dt}+\dfrac{5c(t)}{200-3t}=0$

$\dfrac1{(200-3t)^{5/3}}\dfrac{\mathrm dc(t)}{\mathrm dt}+\dfrac{5c(t)}{(200-3t)^{8/3}}=0$

$\dfrac{\mathrm d}{\mathrm dt}\left[\dfrac{c(t)}{(200-3t)^{5/3}}\right]=0$

$\dfrac{c(t)}{(200-3t)^{5/3}}=C$

$c(t)=C(200-3t)^{5/3}$

There are 20 g of chlorine at the start, so c (0) = 20. Use this to solve for C :

$20=C(200)^{5/3}\implies C=\dfrac1{200\cdot5^{1/3}}$

$\implies\boxed{c(t)=\dfrac1{200}\sqrt[3]{\dfrac{(200-3t)^5}5}}$

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3 years ago

Li’s brother used the work shown to determine that he must grow at most 6 inches to be able to ride the roller coaster. Is this

miv72 [106K]

No, the work is correct, but the interpretation is not. “6 is less than or equal to h” means that h is greater than or equal to 6. He must grow at least 6 inches, not at most 6 inches.

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pickupchik [31]

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