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3 years ago

X^2 - 9x -10 =0 has a discriminant of: a. 121 b. 72 c. -359 d. 41 e. 136

1 answer:

7 0

In the standard form of quadratic
$ax^{2} +bx+c$
the discriminant is
$b^{2} -4ac$
In your quadratic, a = 1, b = -9 and c = -10
Now you need to plug these values into the expression for the discriminant.

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I can't quite remember how to approach this one, I have three others, but I think I could solve them if someone helped walked me

Nimfa-mama [501]

Https://us-static.z-dn.net/files/d8d/e008ced388704d59896d3bf37158f465.jpeg

3 0

3 years ago

Solve using Fourier series.

Olin [163]

With $2L=\pi$ , the Fourier series expansion of $f(x)$ is

$\displaystyle f(x)\sim\frac{a_0}2+\sum_{n\ge1}a_n\cos\dfrac{n\pi x}L+\sum_{n\ge1}b_n\sin\dfrac{n\pi x}L$
$\displaystyle f(x)\sim\frac{a_0}2+\sum_{n\ge1}a_n\cos2nx+\sum_{n\ge1}b_n\sin2nx$

where the coefficients are obtained by computing

$\displaystyle a_0=\frac1L\int_0^{2L}f(x)\,\mathrm dx$
$\displaystyle a_0=\frac2\pi\int_0^\pi f(x)\,\mathrm dx$

$\displaystyle a_n=\frac1L\int_0^{2L}f(x)\cos\dfrac{n\pi x}L\,\mathrm dx$
$\displaystyle a_n=\frac2\pi\int_0^\pi f(x)\cos2nx\,\mathrm dx$

$\displaystyle b_n=\frac1L\int_0^{2L}f(x)\sin\dfrac{n\pi x}L\,\mathrm dx$
$\displaystyle b_n=\frac2\pi\int_0^\pi f(x)\sin2nx\,\mathrm dx$

You should end up with

$a_0=0$
$a_n=0$
(both due to the fact that $f(x)$ is odd)
$b_n=\dfrac1{3n}\left(2-\cos\dfrac{2n\pi}3-\cos\dfrac{4n\pi}3\right)$

Now the problem is that this expansion does not match the given one. As a matter of fact, since $f(x)$ is odd, there is no cosine series. So I'm starting to think this question is missing some initial details.

One possibility is that you're actually supposed to use the even extension of $f(x)$ , which is to say we're actually considering the function

$\varphi(x)=\begin{cases}\frac\pi3&\text{for }|x|\le\frac\pi3\\0&\text{for }\frac\pi3$

and enforcing a period of $2L=2\pi$ . Now, you should find that

$\varphi(x)\sim\dfrac2{\sqrt3}\left(\cos x-\dfrac{\cos5x}5+\dfrac{\cos7x}7-\dfrac{\cos11x}{11}+\cdots\right)$

The value of the sum can then be verified by choosing $x=0$ , which gives

$\varphi(0)=\dfrac\pi3=\dfrac2{\sqrt3}\left(1-\dfrac15+\dfrac17-\dfrac1{11}+\cdots\right)$
$\implies\dfrac\pi{2\sqrt3}=1-\dfrac15+\dfrac17-\dfrac1{11}+\cdots$

as required.

5 0

3 years ago

What’s the correct answer for this ?

TiliK225 [7]

Answer:

5.25 gallson

Step-by-step explanation:

5 0

2 years ago

For fun question

ZanzabumX [31]

Consider the function $f(x)=x^{1/3}$ , which has derivative $f'(x)=\dfrac13x^{-2/3}$ .

The linear approximation of $f(x)$ for some value $x$ within a neighborhood of $x=c$ is given by

$f(x)\approx f'(c)(x-c)+f(c)$

Let $c=64$ . Then $(63.97)^{1/3}$ can be estimated to be

$f(63.97)\approxf'(64)(63.97-64)+f(64)$
$\sqrt[3]{63.97}\approx4-\dfrac{0.03}{48}=3.999375$

Since $f'(x)>0$ for $x>0$ , it follows that $f(x)$ must be strictly increasing over that part of its domain, which means the linear approximation lies strictly above the function $f(x)$ . This means the estimated value is an overestimation.

Indeed, the actual value is closer to the number 3.999374902...

4 0

3 years ago

A box contains 95 pink rubber bands and 90 brown rubber bands. You select a rubber band at random from the box. Find each probab

vivado [14]

Answer:A. The theoretical probability for pink is 95/185 or about 51.35%

B. The theoretical probability for brown is 90/185 or about 47.82%.

C. To find the experimental probability, we will make another fraction. The number of outcomes will be the numerator and the total will be the denominator.

Pink 36/69 = 52.17%

Brown 33/69 = 47.82%

Step-by-step explanation:

4 0

2 years ago