Can someone help me?<br>27 : (11-8)

After finishing her shift , a waitress wrote 2.5 under "hours worked " on her time card .How long did she work?

aev [14]

Answer:

D. 2 hours and 30 minutes.

Step by step explanation:

60 minutes is an hour.

30 minutes is half an hour.

When she wrote 2.5, .5 is a half.

2 hours and 30 minutes.

Hope this helps & good luck!

Feel free to message me if you need more help! :)

3 0

4 years ago

Read 2 more answers

Write as an algebraic expression

Mazyrski [523]

Answer:

13) x=n-11

15) x=3*10

Step-by-step explanation:

I used x as the variable. The questions are pretty straightforward. They are asking you to put the terms in the description on the right side of the equation. The questions = x. That was probably confusing, but it is correct. I hope this helped!

8 0

3 years ago

Read 2 more answers

Pls help me with this math problem 90 points for you in return.

Flura [38]

Answer:

K = -2J + 28

Step-by-step explanation:

To find the slope just look at rise over run.

The rise is -4 as it goes down and run is 2 to the right.

slope = rise/run = -4/2 = -2

The second blank is 28, you just have to see where the line touches the y axis.

7 0

2 years ago

What is 30 tens in hundreds

Dima020 [189]

Answer:

300

Step-by-step explanation:

each ten is worth 10 so 10x30+ 300

plz mark brainliest

4 0

3 years ago

If c is the line segment connecting (x1,y1) to (x2,y2), show that the line integral of xdy-ydx=x1y2-x2y1......this is in a chapt

MatroZZZ [7]

Green's theorem doesn't really apply here. GT relates the line integral over some *closed* connected contour that bounds some region (like a circular path that serves as the boundary to a disk). A line segment doesn't form a region since it's completely one-dimensional.

At any rate, we can still compute the line integral just fine. It's just that GT is irrelevant.

We parameterize the line segment by

$\mathbf r(t)=\langle x_1,y_1\rangle(1-t)+\langle x_2,y_2\rangle t$
$\implies\mathbf r(t)=\langle x_1+(x_2-x_1)t,y_1+(y_2-y_1)t\rangle$

with $0\le t\le1$ . Then we find the differential:

$\mathbf r(t)\equiv\langle x,y\rangle=\langle x_1+(x_2-x_1)t,y_1+(y_2-y_1)t\rangle$
$\implies\mathrm d\mathbf r\equiv\langle\mathrm dx,\mathrm dy\rangle=\langle x_2-x_1,y_2-y_1\rangle\,\mathrm dt$

with $0\le t\le1$ .

Here, the line integral is

$\displaystyle\int_{\mathcal C}x\,\mathrm dy-y\,\mathrm dx=\int_{\mathcal C}\langle-y,x\rangle\cdot\langle\mathrm dx,\mathrm dy\rangle$
$=\displaystyle\int_{t=0}^{t=1}\langle-y_1-(y_2-y_1)t,x_1+(x_2-x_1)t\rangle\cdot\langle x_2-x_1,y_2-y_1\rangle\,\mathrm dt$
$=\displaystyle\int_{t=0}^{t=1}(x_1y_2-x_2y_1)\,\mathrm dt$
$=(x_1y_2-x_2y_1)\displaystyle\int_{t=0}^{t=1}\,\mathrm dt$
$=x_1y_2-x_2y_1$

as required.

4 0

4 years ago

Can someone help me?27 : (11-8)