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3 years ago

Use the given function to find the indicated value(s) of x. (Enter your answers as a comma-separated list.)

Mathematics

1 answer:

Nady [450]3 years ago

6 0

Answer:

For f(x) = √(2·x + 2) - √(x + 18), at f(x) = -1 the possible x-values includes;

-0.757, -17.5

Step-by-step explanation:

Given that the function is f(x) = √(2·x + 2) - √(x + 18)

The value of 'x' when f(x) = -1, is given as follows;

-1 = √(2·x + 2) - √(x + 18)

-1² = (√(2·x + 2) - √(x + 18))² = 3·x + 20 - 2·√(2·x + 2)×√(x + 18)

1 = 3·x + 20 - 2·√(2·x + 2)×√(x + 18)

2·√(2·x + 2)×√(x + 18) = 3·x + 20 - 1 = 3·x + 19

2·x² + 38·x + 36 = (3·x + 19)/2

2·x² + 38·x + 36 - (3·x + 19)/2 = 0

4·x² + 73·x + 53 = 0

From which we get;

x = (-73 ± √(73² - 4 × 4 × 53))/(2 × 4)

x ≈ -0.757, and x ≈ -17.5

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A swimming pool can be filled in 18 hours if water enters through a pipe alone, or in 25 hours if water enters through a hose a

pychu [463]

I am not sure, but this is what I got:
<span>Pipe ALONE = 18 hours

Pipe ALONE in 1 hour = 1/18 of the pool

Hose ALONE in 1 hour = 1/25 of the pool

TOGETHER in 1 HOUR = 1/18 +1/25 = 43/450

TOGETHER they will fill the pool in 450/43 hours

So 5/6 of the pool will take 5/6 X 450/43 = 2250/ 258 = 8.7209302 hours OR

8 hours AND 43.26 minutes ANSWER
</span>
Hope this helps!

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: