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2 years ago

What is the product of (2p 7)(3p2 4p – 3)? 6p3 29p2 – 34p 21 6p3 29p2 – 22p 21 6p3 29p2 22p – 21 6p3 29p2 34p – 21

1 answer:

4 0

The product of the two provided equations, obtained by multiplying each term of the first equation from the second one, is 6p³+29p²+22p-21.

<h3>What is the product of two equations?</h3>

To multiply the two equation, each term of the first equation is multiples from the second term.

The first equation provided in the form of binomial as,

$2p+ 7$

The second equation provided in the form of quadratic equation as,

$3p^2 +4p-3$

The product of these two equations are,

$P=(2p+7)\times (3p^2 +4p-3)\\P=6p^3+8p^2-6p+21p^2+28p-21$

Arrange the equation with the same power terms,

$P=6p^3+8p^2+21p^2-6p+28p-21\\P=6p^3+29p^2+22p-21$

Hence, the product of the two provided equations, obtained by multiplying each term of the first equation from the second one, is 6p³+29p²+22p-21.

Learn more about the multiplication of two equation here;

brainly.com/question/69383

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3 years ago

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3 years ago

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GaryK [48]

Answer:

[-5, 4) ∪ (4, ∞)

Step-by-step explanation:

Given functions:

$f(x)=\dfrac{1}{x-3}$

$g(x)=\sqrt{x+5}$

Composite function:

$\begin{aligned}(f\:o\:g)(x)&=f[g(x)]\\ & =\dfrac{1}{\sqrt{x+5}-3} \end{aligned}$

Domain: input values (x-values)

For $(f\:o\:g)(x)$ to be defined:

$x+5\geq 0 \implies x\geq -5$

$\sqrt{x+5}\neq 3 \implies x\neq 4$

Therefore, $-5\leq x < 4$ and $x > 4$

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2 years ago

Can I get help with finding the Fourier cosine series of F(x) = x - x^2

trapecia [35]

Assuming you want the cosine series expansion over an arbitrary symmetric interval $[-L,L]$ , $L\neq0$ , the cosine series is given by

$f_C(x)=\dfrac{a_0}2+\displaystyle\sum_{n\ge1}a_n\cos nx$

You have

$a_0=\displaystyle\frac1L\int_{-L}^Lf(x)\,\mathrm dx$
$a_0=\dfrac1L\left(\dfrac{x^2}2-\dfrac{x^3}3\right)\bigg|_{x=-L}^{x=L}$
$a_0=\dfrac1L\left(\left(\dfrac{L^2}2-\dfrac{L^3}3\right)-\left(\dfrac{(-L)^2}2-\dfrac{(-L)^3}3\right)\right)$
$a_0=-\dfrac{2L^2}3$

$a_n=\displaystyle\frac1L\int_{-L}^Lf(x)\cos nx\,\mathrm dx$

Two successive rounds of integration by parts (I leave the details to you) gives an antiderivative of

$\displaystyle\int(x-x^2)\cos nx\,\mathrm dx=\frac{(1-2x)\cos nx}{n^2}-\dfrac{(2+n^2x-n^2x^2)\sin nx}{n^3}$

and so

$a_n=-\dfrac{4L\cos nL}{n^2}+\dfrac{(4-2n^2L^2)\sin nL}{n^3}$

So the cosine series for $f(x)$ periodic over an interval $[-L,L]$ is

$f_C(x)=-\dfrac{L^2}3+\displaystyle\sum_{n\ge1}\left(-\dfrac{4L\cos nL}{n^2L}+\dfrac{(4-2n^2L^2)\sin nL}{n^3L}\right)\cos nx$

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3 years ago