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3 years ago

Which is the answer ?

Mathematics

1 answer:

erik [133]3 years ago

4 0

Answer:

(a) y = ±(√x)/2

Step-by-step explanation:

You want to find y such that ...

x = f(y)

x = 4y^2 . . . . . . . put in the expression for f(y)

x/4 = y^2 . . . . . . .divide by 4

±√(x/4) = y . . . . . take the square root

y = ±(√y)/2 . . . . . simplify

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Can a triangle with side lengths 7, 25, and 19 be a right triangle?

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Does 7^2+19^2=25^2?

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3 years ago

Solve -2x+3=-3x+2 and is it no solution <br>One solution or infinitely many solutions.

Alex73 [517]

Answer is =-1

implifying

-2x + 3 = -3x + 2

Reorder the terms:

3 + -2x = -3x + 2

Reorder the terms:

3 + -2x = 2 + -3x

Solving

3 + -2x = 2 + -3x

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Add '3x' to each side of the equation.

3 + -2x + 3x = 2 + -3x + 3x

Combine like terms: -2x + 3x = 1x

3 + 1x = 2 + -3x + 3x

Combine like terms: -3x + 3x = 0

3 + 1x = 2 + 0

3 + 1x = 2

Add '-3' to each side of the equation.

3 + -3 + 1x = 2 + -3

Combine like terms: 3 + -3 = 0

0 + 1x = 2 + -3

1x = 2 + -3

Combine like terms: 2 + -3 = -1

1x = -1

Divide each side by '1'.

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Simplifying

x = -1

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3 years ago

Read 2 more answers

What is the solution set of the following equation?

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3 years ago

Let f(x,y,z) = ztan-1(y2) i + z3ln(x2 + 1) j + z k. find the flux of f across the part of the paraboloid x2 + y2 + z = 3 that li

Sophie [7]

Consider the closed region $V$ bounded simultaneously by the paraboloid and plane, jointly denoted $S$ . By the divergence theorem,

$\displaystyle\iint_S\mathbf f(x,y,z)\cdot\mathrm dS=\iiint_V\nabla\cdot\mathbf f(x,y,z)\,\mathrm dV$

And since we have

$\nabla\cdot\mathbf f(x,y,z)=1$

the volume integral will be much easier to compute. Converting to cylindrical coordinates, we have

$\displaystyle\iiint_V\nabla\cdot\mathbf f(x,y,z)\,\mathrm dV=\iiint_V\mathrm dV$
$=\displaystyle\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=1}\int_{z=2}^{z=3-r^2}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta$
$=\displaystyle2\pi\int_{r=0}^{r=1}r(3-r^2-2)\,\mathrm dr$
$=\dfrac\pi2$

Then the integral over the paraboloid would be the difference of the integral over the total surface and the integral over the disk. Denoting the disk by $D$ , we have

$\displaystyle\iint_{S-D}\mathbf f\cdot\mathrm dS=\frac\pi2-\iint_D\mathbf f\cdot\mathrm dS$

Parameterize $D$ by

$\mathbf s(u,v)=u\cos v\,\mathbf i+u\sin v\,\mathbf j+2\,\mathbf k$
$\implies\mathbf s_u\times\mathbf s_v=u\,\mathbf k$

which would give a unit normal vector of $\mathbf k$ . However, the divergence theorem requires that the closed surface $S$ be oriented with outward-pointing normal vectors, which means we should instead use $\mathbf s_v\times\mathbf s_u=-u\,\mathbf k$ .

Now,

$\displaystyle\iint_D\mathbf f\cdot\mathrm dS=\int_{u=0}^{u=1}\int_{v=0}^{v=2\pi}\mathbf f(x(u,v),y(u,v),z(u,v))\cdot(-u\,\mathbf k)\,\mathrm dv\,\mathrm du$
$=\displaystyle-4\pi\int_{u=0}^{u=1}u\,\mathrm du$
$=-2\pi$

So, the flux over the paraboloid alone is

$\displaystyle\iint_{S-D}\mathbf f\cdot\mathrm dS=\frac\pi2-(-2\pi)=\dfrac{5\pi}2$

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3 years ago