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3 years ago

12R If V= ---- (r+R) Make R the subject of the equation .

1 answer:

6 0

V= $\frac{12R}{r+R}$
note, r and R are differnt
so multipl both sides by (r+R)
V(r+R)=12R
distribute
Vr+VR=12R
minus VR from both sides
Vr=12R-VR
undistribute R
Vr=R(12-V)
divide both sides by (12-V)
$\frac{Vr}{12-V}=R$ or
$R= \frac{Vr}{12-V}$

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Evaluate the line integral, where c is the given curve. (x + 9y) dx + x2 dy, c c consists of line segments from (0, 0) to (9, 1)

viktelen [127]

$\displaystyle\int_C(x+9y)\,\mathrm dx+x^2\,\mathrm dy=\int_C\langle x+9y,x^2\rangle\cdot\underbrace{\langle\mathrm dx,\mathrm dy\rangle}_{\mathrm d\mathbf r}$

The first line segment can be parameterized by $\mathbf r_1(t)=\langle0,0\rangle(1-t)+\langle9,1\rangle t=\langle9t,t\rangle$ with $0\le t\le1$ . Denote this first segment by $C_1$ . Then

$\displaystyle\int_{C_1}\langle x+9y,x^2\rangle\cdot\mathbf dr_1=\int_{t=0}^{t=1}\langle9t+9t,81t^2\rangle\cdot\langle9,1\rangle\,\mathrm dt$
$=\displaystyle\int_0^1(162t+81t^2)\,\mathrm dt$
$=108$

The second line segment ( $C_2$ ) can be described by $\mathbf r_2(t)=\langle9,1\rangle(1-t)+\langle10,0\rangle t=\langle9+t,1-t\rangle$ , again with $0\le t\le1$ . Then

$\displaystyle\int_{C_2}\langle x+9y,x^2\rangle\cdot\mathrm d\mathbf r_2=\int_{t=0}^{t=1}\langle9+t+9-9t,(9+t)^2\rangle\cdot\langle1,-1\rangle\,\mathrm dt$
$=\displaystyle\int_0^1(18-8t-(9+t)^2)\,\mathrm dt$
$=-\dfrac{229}3$

Finally,

$\displaystyle\int_C(x+9y)\,\mathrm dx+x^2\,\mathrm dy=108-\dfrac{229}3=\dfrac{95}3$

5 0

3 years ago

Land in downtown Columbia is valued at $10 a square foot. What is the value of a triangular lot with sides of lengths 119, 147,

stiks02 [169]

Answer:

$87,461

Step-by-step explanation:

Given that the dimensions or sides of lengths of the triangle are 119, 147, and 190 ft

where S is the semi perimeter of the triangle, that is, s = (a + b + c)/2.

S = (119 + 147 + 190) / 2 = 456/ 2 = 228

Using Heron's formula which gives the area in terms of the three sides of the triangle

= √s(s – a)(s – b)(s – c)

Therefore we have = √228 (228 - 119)(228 - 147)(228 - 190)

=> √228 (109)(81)(38)

= √228(335502)

=√76494456

= 8746.1109071 * $10

= 87461.109071

≈$87,461

Hence, the value of a triangular lot with sides of lengths 119, 147, and 190 ft is $87,461.

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3 years ago

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vovikov84 [41]

Answer:

Your answer is 23/24

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3 years ago

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babunello [35]

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Katyanochek1 [597]

Answer:

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