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shepuryov [24]
3 years ago
10

How does decreasing only the mass change an objects density

Mathematics
1 answer:
Sladkaya [172]3 years ago
3 0

?                                                                    

ufjkdjfzkvbz,dhfbvc,jc,vjbvjkxb,gbjkfnkgjfs,kbbcvcxccccc

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A line passes through the points (-3, 5) and (2, 3). What is the slope of this line
vfiekz [6]

Answer:

slope is -2/5

Step-by-step explanation:

m=y2-y1/x2-x1

m=3-5/2-(-3))

m=-2/2+3

m=-2/5

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The graph of y=-4x+7 is:
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G(x) = 4x + 4<br> F(x)=3/5x - 9/5
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Step-by-step explanation:

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4 years ago
Find the line integral with respect to arc length ∫C(9x+5y)ds, where C is the line segment in the xy-plane with endpoints P=(2,0
Gemiola [76]

(a) You can parameterize <em>C</em> by the vector function

<em>r</em><em>(t)</em> = (<em>x(t)</em>, <em>y(t)</em> ) = <em>P</em> (1 - <em>t </em>) + <em>Q</em> <em>t</em> = (2 - 2<em>t</em>, 7<em>t</em> )

where 0 ≤ <em>t</em> ≤ 1.

(b) From the above parameterization, we have

<em>r</em><em>'(t)</em> = (-2, 7)   ==>   ||<em>r</em><em>'(t)</em>|| = √((-2)² + 7²) = √53

Then

d<em>s</em> = √53 d<em>t</em>

and the line integral is

\displaystyle\int_C(9x(t)+5y(t))\,\mathrm ds = \boxed{\sqrt{53}\int_0^1(17t+18)\,\mathrm dt}

(c) The remaining integral is pretty simple,

\displaystyle\sqrt{53}\int_0^1(17t+18)\,\mathrm dt = \sqrt{53}\left(\frac{17}2t^2+18t\right)\bigg|_{t=0}^{t=1} = \boxed{\frac{53^{3/2}}2}

6 0
3 years ago
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