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hjlf
3 years ago
12

Concider the graph of the polynomial function p(x)=(x-1)(x+2)(x+5)

Mathematics
1 answer:
yulyashka [42]3 years ago
4 0
I need to see the horse??
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A rocket travels in a trajectory given by the equation s = -4.9t 2 + v0t, where s is meters above ground, t is time in seconds a
Reptile [31]

Step-by-step explanation:

s = -4.9t² + 49t

The vertex of the parabola is at t = -b/(2a).

t = -49 / (2 × -4.9)

t = 5

The rocket reaches its maximum height after 5 seconds.

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3 years ago
8. If the measure of one side of a square is increased by 2 centimeters and the measure of
GalinKa [24]

Answer:

  6 cm

Step-by-step explanation:

If the original side length is x, then the modified square has an area of ...

  A = LW

  32 = (x +2)(x -2) = x^2  -4

  36 = x^2 . . . . . . . . add 4

  6 = x . . . . . . . . . . take the square root

The original figure has a side length of 6 cm.

__

<em>Check</em>

The modified figure is 8 cm by 4 cm = 32 cm^2.

8 0
2 years ago
Will give 32 points! Read the excerpt from act 1 of The Monsters Are Due on Maple Street.
Kruka [31]

Answer:

Lots of people working

Step-by-step explanation:

4 0
3 years ago
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CaHeK987 [17]
I’m sorry i don’t know the answer but i need to ask another question but i gave up answer a question first sorry
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3 years ago
Find the Fourier series of f on the given interval. f(x) = 1, ?7 &lt; x &lt; 0 1 + x, 0 ? x &lt; 7
Zolol [24]
f(x)=\begin{cases}1&\text{for }-7

The Fourier series expansion of f(x) is given by

\dfrac{a_0}2+\displaystyle\sum_{n\ge1}a_n\cos\frac{n\pi x}7+\sum_{n\ge1}b_n\sin\frac{n\pi x}7

where we have

a_0=\displaystyle\frac17\int_{-7}^7f(x)\,\mathrm dx
a_0=\displaystyle\frac17\left(\int_{-7}^0\mathrm dx+\int_0^7(1+x)\,\mathrm dx\right)
a_0=\dfrac{7+\frac{63}2}7=\dfrac{11}2

The coefficients of the cosine series are

a_n=\displaystyle\frac17\int_{-7}^7f(x)\cos\dfrac{n\pi x}7\,\mathrm dx
a_n=\displaystyle\frac17\left(\int_{-7}^0\cos\frac{n\pi x}7\,\mathrm dx+\int_0^7(1+x)\cos\frac{n\pi x}7\,\mathrm dx\right)
a_n=\dfrac{9\sin n\pi}{n\pi}+\dfrac{7\cos n\pi-7}{n^2\pi^2}
a_n=\dfrac{7(-1)^n-7}{n^2\pi^2}

When n is even, the numerator vanishes, so we consider odd n, i.e. n=2k-1 for k\in\mathbb N, leaving us with

a_n=a_{2k-1}=\dfrac{7(-1)-7}{(2k-1)^2\pi^2}=-\dfrac{14}{(2k-1)^2\pi^2}

Meanwhile, the coefficients of the sine series are given by

b_n=\displaystyle\frac17\int_{-7}^7f(x)\sin\dfrac{n\pi x}7\,\mathrm dx
b_n=\displaystyle\frac17\left(\int_{-7}^0\sin\dfrac{n\pi x}7\,\mathrm dx+\int_0^7(1+x)\sin\dfrac{n\pi x}7\,\mathrm dx\right)
b_n=-\dfrac{7\cos n\pi}{n\pi}+\dfrac{7\sin n\pi}{n^2\pi^2}
b_n=\dfrac{7(-1)^{n+1}}{n\pi}

So the Fourier series expansion for f(x) is

f(x)\sim\dfrac{11}4-\dfrac{14}{\pi^2}\displaystyle\sum_{n\ge1}\frac1{(2n-1)^2}\cos\frac{(2n-1)\pi x}7+\frac7\pi\sum_{n\ge1}\frac{(-1)^{n+1}}n\sin\frac{n\pi x}7
3 0
3 years ago
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