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3 years ago

Pleaseee help I keep getting undefined

Mathematics

2 answers:

stiks02 [169]3 years ago

8 0

Answer:

4 (Answer B)

Step-by-step explanation:

Note that we can rewrite the given expression as

{ ∛(-8)}² In words: the square of the cube root of -8

This simplifies to (-2)² which in turn is 4 (Answer B)

Zanzabum3 years ago

5 0

Answer:

Second option "B. 4" is the correct choice.

Step-by-step explanation:

$(-8)^{\frac{2}{3}}\\\\\mathrm{Apply\:exponent\:rule}:\quad \left(-a\right)^{\frac{m}{n}}=\left(-1\right)^{\frac{m}{n}}\left(a\right)^{\frac{m}{n}}\\\\=\left(-1\right)^{\frac{2}{3}}\cdot \:8^{\frac{2}{3}}\\\\=1\cdot \:8^{\frac{2}{3}}\\\\=8^{\frac{2}{3}}\\\\=\left(2^3\right)^{\frac{2}{3}}=2^{3\cdot \frac{2}{3}}=2^2\\\\=4$

Best Regards!

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A. The solution is (−2,−3)

Cerrena [4.2K]

9514 1404 393

Answer:

D. (-3, -2)

Step-by-step explanation:

The equations have different coefficients for x and y, so will have one solution. The solutions offered are easily tested in either equation.

Using (x, y) = (-2, -3):

x = y -1 ⇒ -2 = -3 -1 . . . . False

Using (x, y) = (-3, -2):

x = y -1 ⇒ -3 = -2 -1 . . . .True

2x = 3y ⇒ 2(-3) = 3(-2) . . . . True

The solution is (-3, -2).

If you'd like to solve the set of equations, substitution for x works nicely.

2(y -1) = 3y

2y -2 = 3y . . eliminate parentheses

-2 = y . . . . . . subtract 2y

x = -2 -1 = -3

The solution is (x, y) = (-3, -2).

5 0

3 years ago

Find the work done by F= (x^2+y)i + (y^2+x)j +(ze^z)k over the following path from (4,0,0) to (4,0,4)

babunello [35]

$\vec F(x,y,z)=(x^2+y)\,\vec\imath+(y^2+x)\,\vec\jmath+ze^z\,\vec k$

We want to find $f(x,y,z)$ such that $\nabla f=\vec F$ . This means

$\dfrac{\partial f}{\partial x}=x^2+y$

$\dfrac{\partial f}{\partial y}=y^2+x$

$\dfrac{\partial f}{\partial z}=ze^z$

Integrating both sides of the latter equation with respect to $z$ tells us

$f(x,y,z)=e^z(z-1)+g(x,y)$

and differentiating with respect to $x$ gives

$x^2+y=\dfrac{\partial g}{\partial x}$

Integrating both sides with respect to $x$ gives

$g(x,y)=\dfrac{x^3}3+xy+h(y)$

Then

$f(x,y,z)=e^z(z-1)+\dfrac{x^3}3+xy+h(y)$

and differentiating both sides with respect to $y$ gives

$y^2+x=x+\dfrac{\mathrm dh}{\mathrm dy}\implies\dfrac{\mathrm dh}{\mathrm dy}=y^2\implies h(y)=\dfrac{y^3}3+C$

So the scalar potential function is

$\boxed{f(x,y,z)=e^z(z-1)+\dfrac{x^3}3+xy+\dfrac{y^3}3+C}$

By the fundamental theorem of calculus, the work done by $\vec F$ along any path depends only on the endpoints of that path. In particular, the work done over the line segment (call it $L$ ) in part (a) is

$\displaystyle\int_L\vec F\cdot\mathrm d\vec r=f(4,0,4)-f(4,0,0)=\boxed{1+3e^4}$

and $\vec F$ does the same amount of work over both of the other paths.

In part (b), I don't know what is meant by "df/dt for F"...

In part (c), you're asked to find the work over the 2 parts (call them $L_1$ and $L_2$ ) of the given path. Using the fundamental theorem makes this trivial: