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

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.

5 0

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

Plisss to na dzisiaj

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

6 0

3 years ago

Find the Fourier series of f on the given interval. f(x) = 1, ?7 < x < 0 1 + x, 0 ? x < 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