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2 years ago

I need this asap so ye

Mathematics

2 answers:

Sauron [17]2 years ago

6 0

Answer:

First one is D second is G

Step-by-step explanation:

Arturiano [62]2 years ago

5 0

First one is D and second is H

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Complete the factoring<br> 7y+42=7()

stiks02 [169]

7y + 42 = 7(y + 6)

Hope this helps.

3 0

3 years ago

Read 2 more answers

38.Randy and Amy left school at the same time and began walking in oppo-

RideAnS [48]

Answer:

DISTANCE=RATE*TIME. THEREFORE

X=3.6*1/6+4.2*1/6 (1/6 HOUR=10 MIN)

X=.6+.7

X=1.3 KM IS HOW FAR THEY WIL BE APART AFTER 10 MINUTES.

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

David ran for president of the chess club, and he received 76 votes. there were 80 members in the club. what percentage of the c

TEA [102]

76 out of 80 people for David.

Therefore, 76/80 people for David.

Further simplifying, every 19 out of 20 people voted for David.

What's 19 out of 20?

20 x 5 = 100
19 x 5 = 95

95% of the club members voted for David.

5 0

4 years ago

Help

sdas [7]

I think the answer for this question is letter C: SAS

8 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