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3 years ago

Problem

Mathematics

1 answer:

gregori [183]3 years ago

6 0

Answer:

62.5%

Step-by-step explanation:

150/240

Divide top and bottom by 10

15/24

Divide by 3

5/8

.625

Multiply by 100%

62.5%

You might be interested in

Find the area bounded by the graphs of the indicated equations over the given interval.

NeX [460]

Answer:

Step-by-step explanation:

$y = 0$

is basically the horizontal axis.

First, find the integral of x^2-25.

Remember that

integral of a constant is that constant times x.

Also that

to take the integral of a power function, add 1 to the degree and divide by that same degree.

$y = {x}^{2} - 25$

We then get

$\frac{ {x}^{3} }{3} - 25x$

Evaluate at -3

$\frac{ - 3 {}^{3} }{3} - 25( - 3)$

$- 27 + 75 = 48$

Then we evaluate at 0

$\frac{0 {}^{3} }{3} - 25(0) = 0$

Next, we subtract the the answer then we get

$48 - 0 = 48$

8 0

2 years ago

Is a tray with a circumference of 32 inches is big enough to

yKpoI14uk [10]

Answer:

Yes

Step-by-step explanation:

Circumference formula is C=2 pi r, if you divide the pizza by 2 to get the radius,5, then input it into the formula to get 31.41592654, and since this number is smaller than 32 it will fit. If 10 is the radius it will not fit.

7 0

3 years ago

Use the distance formula to slove how far the points are on the a coordinate plane. Points are (7,5)(3,2)

maksim [4K]

Answer:

distance is 5 units

Step-by-step explanation:

plug in values

4 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: