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2 years ago

Which of the following is a like radical to RootIndex 3 StartRoot 7x EndRoot?

Mathematics

2 answers:

lana [24]2 years ago

4 0

Answer:

A. 4 (RootIndex 3 StartRoot 7 x EndRoot) or $4(\sqrt[3]{7x})$

Step-by-step explanation:

Given:

A radical whose value is, $r_1=\sqrt[3]{7x}$

Now, we need to find the like radical for $r_1$ .

Let the like radical be $r_2$ .

As per the definition of like radicals, like radicals are those that can be expressed as multiples of each other.

So, if two radicals $r_1\ and\ r_2$ are like radicals, then

$r_1 = n \times r_2
$

Where, 'n' is a real number.

Here, $r_1=\sqrt[3]{7x}$

Now, let us check all the options .

Option A:

4 (RootIndex 3 StartRoot 7 x EndRoot) or $r_2=4\sqrt[3]{7x}$

Now, we observe that $r_2$ is a multiple of $r_1$ because

$r_2=4\times \sqrt[3]{7x}\\\\ r_2=4\times r_1..............(r_1=\sqrt[3]{7x})$

Therefore, option A is correct.

Option B:

StartRoot 7 x EndRoot or $r_2=\sqrt{7x}$

As the above radical is square root and not a cubic root, this option is incorrect.

Option C:

x (RootIndex 3 StartRoot 7 EndRoot) or $r_2=x\sqrt[3]{7}$

As the term inside the cubic root is not same as that of $r_1$ , this option is also incorrect.

Option D:

7 StartRoot x EndRoot or $r_2=7\sqrt{x}$

As the above radical is square root and not a cubic root, this option is incorrect.

Therefore, the like radical is option (A) only.

Flura [38]2 years ago

3 0

Answer:

option a

Step-by-step explanation:

on edge

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2) Use spherical coordinates to find the integral of f(x,y,z) = z over x^2 + y^2 + z^2 < 4 and x,y,z < 0

777dan777 [17]

1. Denote by $\mathcal D$ the region bounded by the cylinder $x^2+y^2=9$ and the planes $z=0$ and $z=6$ . Converting to cylindrical coordinates involves setting

$\begin{cases}x=r\cos u\\y=r\sin u\\z=v\end{cases}$

and $\mathcal D$ is obtained by varying the parameters over $0\le r\le3$ , $0\le u\le2\pi$ , and $0\le v\le6$ . The volume element is

$\mathrm dV=\mathrm dx\,\mathrm dy\,\mathrm dz=r\,\mathrm dr\,\mathrm du\,\mathrm dv$

so the integral becomes

$\displaystyle\iiint_{\mathcal D}xz\,\mathrm dV=\int_{v=0}^{v=6}\int_{u=0}^{u=2\pi}\int_{r=0}^{r=3}r^2v\cos u\,\mathrm dr\,\mathrm du\,\mathrm dv=0$

2. Now let $\mathcal D$ denote the region bounded by the sphere $x^2+y^2+z^2=4$ and the coordinate planes in the octant in which each coordinate of the point $(x,y,z)$ is negative.

Converting to spherical coordinates, we set

$\begin{cases}x=\rho\cos u\sin v\\y=\rho\sin u\sin v\\z=\rho\cos v\end{cases}$

where we obtain $\mathcal D$ by varying over $0\le\rho\le2$ , $\pi\le u\le\dfrac{3\pi}2$ , and $\dfrac\pi2\le v\le\pi$ . The volume element is now

$\mathrm dV=\rho^2\sin v\,\mathrm d\rho\,\mathrm du\,\mathrm dv$

so the integral becomes

$\displaystyle\iiint_{\mathcal D}z\,\mathrm dV=\int_{v=\pi/2}^{v=\pi}\int_{u=\pi}^{u=3\pi/2}\int_{\rho=0}^{\rho=2}\rho^3\cos v\sin v\,\mathrm d\rho\,\mathrm du\,\mathrm dv=-\pi$