The diagram shows a rectangular garden path.

What is 3 1/2 as a quotient of two integers?

allsm [11]

Answer:

7 divided by 2

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Please look at image

Jet001 [13]

The value of x is 12

5 0

2 years ago

!SUBSTITUTION METHOD!<br><br>6y - 5z = -8 <br>3z = -6 <br>4x - 3y - 2z= 21

hichkok12 [17]

Answer:

Step-by-step explanation:

6y - 5z = -8 .......... Equation 1

3z = -6 ................... Equation 2

4x - 3y - 2z= 21...... Equation 3

<u>First solve for z in Equation 2</u>

That's

3z = - 6

Divide both sides by 3

Next substitute the value of z into Equation 1 in order to find y

We have

6y - 5(-2) = - 8

6y + 10 = - 8

6y = - 8 - 10

6y = - 18

Divide both sides by 6

Finally substitute the values of y and z into Equation 3 to find the value of x

That's

4x - 3(-3) - 2(-2) = 21

4x + 9 + 4 = 21

4x + 13 = 21

4x = 21 - 13

4x = 8

Divide both sides by 4

So the solutions are

Hope this helps you

3 0

3 years ago

Describe the effects on the graph of y = f(x) when y = –f(x + 4)

Anni [7]

Answer:

Step-by-step explanation:

Graph is shifted 4 units to the right and reflected over the x-axis.

7 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