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n200080 [17]
2 years ago
10

From the quadratic equation 3x² + X - 5 = 0, find alpha square plus beta square​

Mathematics
1 answer:
timama [110]2 years ago
4 0

Answer:

  α² +β² = 3 4/9

Step-by-step explanation:

Assuming α and β are solutions to the equation, it can be factored as ...

  (x -α)(x -β) = 0

Expanding this, we get ...

  x² -(α +β)x +αβ = 0

Dividing the original equation by 3, we find ...

  x² +(1/3)x -5/3 ≡ x² -(α+β)x +αβ   ⇒   (α+β) = -1/3, αβ = -5/3

We know that the square (α+β)² can be expanded to ...

  (α +β)² = α² +β² +2αβ

  α² +β² = (α +β)² -2αβ . . . . . . subtract 2αβ

Substituting the values for (α+β) and αβ, we find the desired expression is ...

  α² +β² = (-1/3)² -2(-5/3) = 1/9 +10/3 = 31/9

  α² +β² = 3 4/9

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Answer:

Step-by-step explanation:

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For f(x):

[ f(1) - f(0) ] / (1 - 0)

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For g(x):

[ g(1) - g(0) ] / (1 - 0)

(4 - 3) / (1 - 0)

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6 0
3 years ago
Write an equation in slope-intercept form for a line that passes through the given pair of points. (-6,5), (-9,0)
11111nata11111 [884]
Slope intercept form is y=mx+b  because m=slope and b=y-intercept hence "slope intercept form"...

m=deltay/deltax=(0-5)/(-9--6)=-5/-3=5/3 so far we have now:

y=5x/3 +b, using any point, I'll use (-9,0) we can now solve for b or the y-intercept...

0=-9(5)/3 +b

0=-15+b, so b=15 and our line is:

y=5x/3 + 15 or more neatly

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4 0
3 years ago
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.)

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3 years ago
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Elodia [21]

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3 0
2 years ago
PLEASE ANSWER ASAP!!!
Vesna [10]

Answer:

C=x (x+3)

Step-by-step explanation:

x cannot divide x+3 definitely so the denominators must be multiplied to get the least common denominator.

4 0
3 years ago
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