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Question 24

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Liula [17]2 years ago

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Is there a photo? Or anything

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Solve the system of equations<br><br> system

Korolek [52]

Answer:

(3, 3 )

Step-by-step explanation:

Given the 2 equations

3x - y = 6 → (1)

6x + y = 21 → (2)

Adding the 2 equations term by term will eliminate y, that is

(6x + 3x) + (y - y) = (21 + 6), that is

9x = 27 (divide both sides by 9 )

x = 3

Substitute x = 3 into either (1) or (2) and solve for y

Using (2), then

(6 × 3) + y = 21

18 + y = 21 ( subtract 18 from both sides )

y = 3

Solution is (3, 3 )

8 0

3 years ago

Read 2 more answers

2 x 6/7 <br><br> SEND HELP!!!!!!!!!!!

Allushta [10]

Answer: C. 2 6/7

Hope that helped! :)

8 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

2 years ago

12(a−4) simplified I am confused with this question

Ksenya-84 [330]

Answer:

-48

Step-by-step explanation:

8 0

2 years ago

stiv31 [10]

Answer:

I think it would be 34 because I divided 68.2 by 4 and got 17.05, so I multiplied that by 2, and it's closest to 34.

<em>So I think it's A.</em>

But y'all don't be mad if I'm wrong cuz I said I think.

5 0

2 years ago