Ask question

Ask question

All categories

3 years ago

10

What is 9 and 5/6 minus 5 and 3/4

2 answers:

dezoksy [38]3 years ago

5 0

4.0833 sorry if i’m wrong luv did you want it in a fraction?

uysha [10]3 years ago

4 0

Answer:

Step-by-step explanation:

Rewrite with common denominators. 9 10/12 - 5 9/12 = 4 1/12.

You might be interested in

Evaluate if x = 1 , y = 3 , and z = 5<br><br> |5z - 2y|

Aleks04 [339]

Answer:

1

Step-by-step explanation:

4 0

2 years ago

Simplify the following expression.

Allushta [10]

Answer:

I got C.)5

Step-by-step explanation:

5 0

2 years ago

What is the system of inequalities associated with the following graph?

slamgirl [31]

One line passes through the points (-2,3) and (0,-3) , it means the y intercept is b=-3

and slope m = $\frac{(-3-3)}{0-(-2)}$

$= \frac{-6}{2}= -3$

So the equation of line will be $y= -3x -3$

And the inequality should be $y\geq -3x-3$

Or $3x +y \geq -3$

And the other line passes through (-2,3) and (0,2)

So the y intercept is b=2

and the slope is $m= \frac{2-3}{0-(-2)} = \frac{-1}{2}$

So the equation of line will be $y= \frac{-1}{2} x +2\\and the inequality should be y \leq \frac{-1}{2} x+2$

Or

$2y +x \leq 4$

So answer is $3x +y \geq -3, x+2y\leq 4$

6 0

3 years ago

Read 2 more answers

What is the expanded form of this number? 204.017

Anna [14]

Answer:

(2×100) + (4×1) + (1÷10) + (7÷100)

5 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:

$\displaystyle\int_{L_1}\vec F\cdot\mathrm d\vec r=f(0,0,0)-f(4,0,0)=-\frac{64}3$

$\displaystyle\int_{L_2}\vec F\cdot\mathrm d\vec r=f(4,0,4)-f(0,0,0)=\frac{67}3+3e^4$

8 0

2 years ago