F(x) = 4x + 7, g(x) = 3x2 Find (fg)(x).

Mathematics

2 answers:

Brums [2.3K]3 years ago

5 0

F(x) = 4x + 7
g(x) = 3x²

(f · g)(x) = 3x²(4x + 7)
(f · g)(x) = 3x²(4x) + 3x²(7)
(f · g)(x) = 12x³ + 21x²

mario62 [17]3 years ago

4 0

Answer:

Value of (fg)(x) is:

12x³+21x²

Step-by-step explanation:

f(x) = 4x + 7, g(x) = 3x²

(fg)(x)=f(x)g(x)

=(4x+7)(3x²)

= 4x(3x²)+7(3x²)

= 12x³+21x²

Hence, value of (fg)(x) is:

12x³+21x²

You might be interested in

What is the next step

iragen [17]

Answer:

measure of angle ABC is 6 degree.

Step-by-step explanation:

After you found x, you'll plug it in the equation of the angle it's asking for.

Measure of angle ABD, and you know measure of angle ABC = 4x-1.

4(1)-1 = 4-1 = 3

Measure of angle CBD = 2x+1.

2(1) +1 = 3

Measure of angle ABC is ABD + CBD

3 + 3 = 6

6 0

2 years ago

Read 2 more answers

What's the flux of the vector field F(x,y,z) = (e^-y) i - (y) j + (x sinz) k across σ with outward orientation where σ is the po

emmasim [6.3K]

$\displaystyle\iint_\sigma\mathbf F\cdot\mathrm dS$
$\displaystyle\iint_\sigma\mathbf F\cdot\mathbf n\,\mathrm dS$
$\displaystyle\iint_\sigma\mathbf F\cdot\left(\frac{\mathbf r_u\times\mathbf r_v}{\|\mathbf r_u\times\mathbf r_v\|}\right)\|\mathbf r_u\times\mathbf r_v\|\,\mathrm dA$
$\displaystyle\iint_\sigma\mathbf F\cdot(\mathbf r_u\times\mathbf r_v)\,\mathrm dA$

Since you want to find flux in the outward direction, you need to make sure that the normal vector points that way. You have

$\mathbf r_u=\dfrac\partial{\partial u}[2\cos v\,\mathbf i+\sin v\,\mathbf j+u\,\mathbf k]=\mathbf k$
$\mathbf r_v=\dfrac\partial{\partial v}[2\cos v\,\mathbf i+\sin v\,\mathbf j+u\,\mathbf k]=-2\sin v\,\mathbf i+\cos v\,\mathbf j$

The cross product is

$\mathbf r_u\times\mathbf r_v=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\0&0&1\\-2\sin v&\cos v&0\end{vmatrix}=-\cos v\,\mathbf i-2\sin v\,\mathbf j$

So, the flux is given by

$\displaystyle\iint_\sigma(e^{-\sin v}\,\mathbf i-\sin v\,\mathbf j+2\cos v\sin u\,\mathbf k)\cdot(\cos v\,\mathbf i+2\sin v\,\mathbf j)\,\mathrm dA$
$\displaystyle\int_0^5\int_0^{2\pi}(-e^{-\sin v}\cos v+2\sin^2v)\,\mathrm dv\,\mathrm du$
$\displaystyle-5\int_0^{2\pi}e^{-\sin v}\cos v\,\mathrm dv+10\int_0^{2\pi}\sin^2v\,\mathrm dv$
$\displaystyle5\int_0^0e^t\,\mathrm dt+5\int_0^{2\pi}(1-\cos2v)\,\mathrm dv$

where $t=-\sin v$ in the first integral, and the half-angle identity is used in the second. The first integral vanishes, leaving you with

$\displaystyle5\int_0^{2\pi}(1-\cos2v)\,\mathrm dv=5\left(v-\dfrac12\sin2v\right)\bigg|_{v=0}^{v=2\pi}=10\pi$