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3 years ago

PLEASE! NEED HELP!!!

Mathematics

2 answers:

levacccp [35]3 years ago

6 0

Answer:

x^2+2x+2

Step-by-step explanation:

to find b in the equation :

b^2<4ac when having two complex roots

the required equation is x2 - Sx + P = 0

the sum of the roots :-1-i+-1-i=-2

the product=2

so the equation: x^2-(-2)x+2

x^2+2x+2

Pani-rosa [81]3 years ago

5 0

Answer:

Step-by-step explanation:

Formula for finding a quadratic equation when its roots are given is

x² - ( sum of roots)x + product of roots=0

The roots are

- 1 + i

- 1 - i

Sum of roots is = - 1 + i - 1 - i = -2

Product of roots is = ( -1 + i)(- 1 - i)

= 1 + 1 = 2

So the quadratic equation is

x² - (-2)x + 2 = 0

Hope this helps you

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Answer:

The mass of the lamina is 1

Step-by-step explanation:

Let $\rho(x,y)$ be a continuous density function of a lamina in the plane region D,then the mass of the lamina is given by:

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From the question, the given density function is $\rho (x,y)=xye^{x+y}$ .

Again, the lamina occupies a rectangular region: D={(x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1}.

The mass of the lamina can be found by evaluating the double integral:

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Since D is a rectangular region, we can apply Fubini's Theorem to get:

$I=\int\limits^1_0(\int\limits^1_0xye^{x+y}dy)dx$ .

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The inner integral is evaluated using integration by parts.

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This implies that:

$\int\limits xye^{x+y}dy=xye^{x+y}-\int\limits e^{x+y}\cdot xdy$

$\int\limits xye^{x+y}dy=xye^{x+y}-xe^{x+y}$

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We substitute the limits of integration and evaluate to get:

$I_0=xe^x$

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$I=\int\limits^1_0(xe^x)dx$ .

$I=\int\limits^1_0xe^xdx$ .

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$I=\int\limits^1_0xe^xdx=0-1(0-1)$ .

$I=\int\limits^1_0xe^xdx=0-1(-1)=1$ .

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We don't actually arrive at x = 2 itself. We simply move closer and closer.

Since we're on the positive or right hand side of 2, this means we go with the rule involving x > 2

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Plug in x = 2 to find that...

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Part B

Because $\lim_{x \to 2^{+}}f(x) \ne \lim_{x \to 2^{-}}f(x)$ this means that the limit $\lim_{x \to 2}f(x)$ does not exist.

If you are a visual learner, check out the graph below of the piecewise function. Notice the gap or disconnect at x = 2. This can be thought of as two roads that are disconnected. There's no way for a car to go from one road to the other. Because of this disconnect, the limit doesn't exist at x = 2.

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Part C

You'll follow the same type of steps shown in part A.

However, keep in mind that x = 4 is above x = 2, so we'll deal with x > 2 only.

So you'd only involve the second piece f(x) = (x/2) + 1

You should find that f(4) = 3, and that both left and right hand limits equal this value. The left and right hand limits approach the same y value. The limit does exist here. There are no gaps to worry about when x = 4.

===============================================================

Part D

As mentioned earlier, since $\lim_{x \to 4^{+}}f(x) = \lim_{x \to 4^{-}}f(x) = 3$ , this means the limit $\lim_{x \to 4}f(x)$ does exist and it's equal to 3.

As x gets closer and closer to 4, the y values are approaching 3. This applies to both directions.

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