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3 years ago

The figure is made out of rectangular prisms. what is the volume?

Mathematics

2 answers:

nikklg [1K]3 years ago

6 0

to find the volume you have to multiply the length width and height of the prism

ehidna [41]3 years ago

4 0

Answer:

its volume would be the shape of a rectangle

Step-by-step explanation:

You might be interested in

Complete the table of values for the equation, 16x + 4y = 12.

Ann [662]

The x-y coordinates for the given equation are: (-2,11),(-1,7),(0,3), (1,-1) and (2,-5).

<h3>Linear Function</h3>

A linear function can be represented by a line. The standard form for this equation is: ax+b , for example, y=2x+7. Where:

a= the slope;
b=the constant term that represents the y-intercept.

The given equation is 16x + 4y = 12. For solving this question, you should replace the given values of x for finding the values of y.

Thus,

For x= -2, the value of y will be:

16*(-2)+4y=12

-32+4y=12

4y=12+32

4y=44

y=11

For x= -1, the value of y will be:

16*(-1)+4y=12

-16+4y=12
4y=12+16
4y=28
y=7

For x= 0, the value of y will be:

16*(0)+4y=12

0+4y=12
4y=12
4y=12
y=3

For x= 1, the value of y will be:

16*(1)+4y=12

16+4y=12
4y=12-16
4y=-4
y= -1

For x= 2, the value of y will be:

16*(2)+4y=12

32+4y=12
4y=12-32
4y=-20
y= -5

Read more about the linear equation here:

brainly.com/question/1884491

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7 0

2 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: