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kolbaska11 [484]
3 years ago
9

Cecily's teacher held a raffle. To win the raffle, a student has to pick a paper scroll with an integer written on it. The chart

shows the scrolls picked by Cecily and her friends.
Mathematics
1 answer:
AfilCa [17]3 years ago
7 0

Complete question :

Cecily’s teacher held a raffle. To win the raffle, a student has to pick a paper scroll with an integer written on it. The chart shows the scrolls picked by Cecily and her friends.

Name

Paper Scroll

Cecily

A scroll labeled 6.7.

Marty

A scroll labeled negative 37.

Jon

A scroll labeled StartFraction 7 Over 9 EndFraction

Fiona

A scroll labeled 3 and one-third.

Who picked the winning scroll?

Cecily picked the winning scroll.

Marty picked the winning scroll.

Jon picked the winning scroll.

Fiona picked the winning scroll.

Answer:

Marty picked the winning scroll.

Step-by-step explanation:

Based on the rule given in other to win, the winning scroll is that which has integer value written on it.

An integer simply means a number devoid of fraction. Hence, it could be a positive or negative while number.

Given that the students pick below :

Cecily = 6.7 (not an integer)

Marty = - 37 (integer)

Jon = 7/9 (not an integer)

The only scroll which meets the condition of being an integer is Marty's scroll ; Hence, she picked the winning scroll.

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A painting is purchased for $250. if the value of the painting doubles every 5 years, then its value is given by the function v(
nignag [31]
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Method 2, using function given. 
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3 0
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3 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.

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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.
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Step-by-step explanation:

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it represents a parabola with center (4,0)

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