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Novay_Z [31]
3 years ago
7

What are the approximate values of the non-integral roots of the polynomial equation?

Mathematics
2 answers:
sertanlavr [38]3 years ago
6 0

Answer: D: 1.27    E: 4.73

Step-by-step explanation: D & E are right on edg

irga5000 [103]3 years ago
5 0

Answer:

-4, -2, -3-2i, -3+2i

Step-by-step explanation:

The of the polynomial equation are those values of x, for which f(x)=0.

Consider equation

f(x)=0\\ \\(x^2+6x+8)(x^2+6x+13)=0

By zero product property,

x^2+6x+8=0\ \text{or}\ x^2+6x+13=0

Solve each equation:

1. x^2+6x+8=0

D=6^2-4\cdot 8=36-32=4\\ \\x_{1,2}=\dfrac{-6\pm \sqrt{4}}{2}=\dfrac{-6\pm 2}{2}=-4,\ -2

2. x^2+6x+13=0

D=6^2-4\cdot 13=36-52=-16=16i^2\\ \\x_{1,2}=\dfrac{-6\pm \sqrt{16i^2}}{2}=\dfrac{-6\pm 4i}{2}=-3-2i,\ -3+2i

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For the function defined by f(t)=2-t, 0≤t<1, sketch 3 periods and find:
Oksi-84 [34.3K]
The half-range sine series is the expansion for f(t) with the assumption that f(t) is considered to be an odd function over its full range, -1. So for (a), you're essentially finding the full range expansion of the function

f(t)=\begin{cases}2-t&\text{for }0\le t

with period 2 so that f(t)=f(t+2n) for |t| and integers n.

Now, since f(t) is odd, there is no cosine series (you find the cosine series coefficients would vanish), leaving you with

f(t)=\displaystyle\sum_{n\ge1}b_n\sin\frac{n\pi t}L

where

b_n=\displaystyle\frac2L\int_0^Lf(t)\sin\frac{n\pi t}L\,\mathrm dt

In this case, L=1, so

b_n=\displaystyle2\int_0^1(2-t)\sin n\pi t\,\mathrm dt
b_n=\dfrac4{n\pi}-\dfrac{2\cos n\pi}{n\pi}-\dfrac{2\sin n\pi}{n^2\pi^2}
b_n=\dfrac{4-2(-1)^n}{n\pi}

The half-range sine series expansion for f(t) is then

f(t)\sim\displaystyle\sum_{n\ge1}\frac{4-2(-1)^n}{n\pi}\sin n\pi t

which can be further simplified by considering the even/odd cases of n, but there's no need for that here.

The half-range cosine series is computed similarly, this time assuming f(t) is even/symmetric across its full range. In other words, you are finding the full range series expansion for

f(t)=\begin{cases}2-t&\text{for }0\le t

Now the sine series expansion vanishes, leaving you with

f(t)\sim\dfrac{a_0}2+\displaystyle\sum_{n\ge1}a_n\cos\frac{n\pi t}L

where

a_n=\displaystyle\frac2L\int_0^Lf(t)\cos\frac{n\pi t}L\,\mathrm dt

for n\ge0. Again, L=1. You should find that

a_0=\displaystyle2\int_0^1(2-t)\,\mathrm dt=3

a_n=\displaystyle2\int_0^1(2-t)\cos n\pi t\,\mathrm dt
a_n=\dfrac2{n^2\pi^2}-\dfrac{2\cos n\pi}{n^2\pi^2}+\dfrac{2\sin n\pi}{n\pi}
a_n=\dfrac{2-2(-1)^n}{n^2\pi^2}

Here, splitting into even/odd cases actually reduces this further. Notice that when n is even, the expression above simplifies to

a_{n=2k}=\dfrac{2-2(-1)^{2k}}{(2k)^2\pi^2}=0

while for odd n, you have

a_{n=2k-1}=\dfrac{2-2(-1)^{2k-1}}{(2k-1)^2\pi^2}=\dfrac4{(2k-1)^2\pi^2}

So the half-range cosine series expansion would be

f(t)\sim\dfrac32+\displaystyle\sum_{n\ge1}a_n\cos n\pi t
f(t)\sim\dfrac32+\displaystyle\sum_{k\ge1}a_{2k-1}\cos(2k-1)\pi t
f(t)\sim\dfrac32+\displaystyle\sum_{k\ge1}\frac4{(2k-1)^2\pi^2}\cos(2k-1)\pi t

Attached are plots of the first few terms of each series overlaid onto plots of f(t). In the half-range sine series (right), I use n=10 terms, and in the half-range cosine series (left), I use k=2 or n=2(2)-1=3 terms. (It's a bit more difficult to distinguish f(t) from the latter because the cosine series converges so much faster.)

5 0
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Pepsi [2]

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Step-by-step explanation:

(3x)^2+(5)^2+2.3x.5

9x^2+25+30x

4 0
3 years ago
Solve for x: -3 + x = -22
Tanya [424]

Answer:

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Step-by-step explanation:

-3 + x = -22

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-3+3 +x = -22 +3

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3 years ago
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5 0
2 years ago
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Your answer would be 4 over x to the sixth which would look like this

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