Somebody helppppp !!!

Rationalize the denominator and simplify. 6 /5-3

sattari [20]

Answer:

-9/5

Step-by-step explanation:

Step 1: Take the LCM in the denominator

6/5-3

multiply the numerator and denominator by 5 to make the denominator 5.

6-3 (x5)

5 1 (x5)

=6-15

5

= -9

5

!!

6 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:

$\displaystyle\int_{L_1}\vec F\cdot\mathrm d\vec r=f(0,0,0)-f(4,0,0)=-\frac{64}3$

$\displaystyle\int_{L_2}\vec F\cdot\mathrm d\vec r=f(4,0,4)-f(0,0,0)=\frac{67}3+3e^4$

8 0

3 years ago

One serving of milk has 120 Calories. If one serving is equal to 1 cup, how many Calories are in 1 fluid ounce? (1 cup = 8 fluid

aleksandrvk [35]

Answer:

15

Step-by-step explanation:

the answer should be 15 because, if one cup is 120 calories or 8 fluid ounces is 120 calories, divide 120 by 8 so you can find how many calories one fluid ounce is.

8 0

3 years ago

Simplify (6 + 8i) (4 - 7i)

erica [24]

The answer is 80-10i

8 0

3 years ago

Read 2 more answers

Delia measured her bathtub to be 2 meters long. Which of these is an equivalent measurement?

snow_lady [41]

The answer would be D) 78.7 inches is equivalent to 2 meters.

6 0

3 years ago