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2 years ago

3x + 2x -x 3(x+3) 6 x - 3 + 2(x-9) All separate questions

Mathematics

2 answers:

Vesna [10]2 years ago

8 0

Answer:

1. 4x

2. 3x+9

3. 8x-21

Step-by-step explanation:

1. Combine like terms (3x)+(2x)(-x)

2. 3(x+3)

3x+9

3. 6x-3+2(x-9)

6x-3+2x-18

6x-21+2x

8x-21

d1i1m1o1n [39]2 years ago

6 0

Answer:

3x + 2x - x = 4x

3(x+3) = 3x + 9

6x - 3 + 2(x-9) = 8x - 21

Step-by-step explanation:

For the first question, we add 3x and 2x, getting 5x, then subtract x, getting 4x.

For the second question, we distribute 3 to the part in parentheses, getting us the new expression, 3x+9.

For the last question, first we need to distribute 2. We then get 6x - 3 + 2x - 18. We can then combine like terms, and we get 8x - 21.

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Read 2 more answers

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: