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jeka94
3 years ago
9

Yesterday, Nancy's lunch cost $9.80, and she left a $1.47 tip. Today, Nancy's lunch cost $12.60. If she tips at the same rate th

at she did yesterday, how much of a tip should Nancy leave?
Mathematics
2 answers:
Semmy [17]3 years ago
8 0

Answer:

$1.47/ $9.80 = 0.15

15% rate

$12.60 * 15% = $1.89

Step-by-step explanation:

If her lunch <em>cost</em> $9.80 and she left a $1.47 <em>tip</em>, if we do <u>$1.47/ $9.80 = 0.15. </u> So 15% is our percent of tip.

(That is the <em>cost</em> of the first day and the <em>tip</em> of the second day divided.) If we want to find out how much her next<em> tip</em> should be, we have to multiply the percent of tip (15%) with the next days <em>cost </em>($12.60.) So, <u>$12.60 * 15%= $1.89 </u>

Sedbober [7]3 years ago
5 0
$1.47 = $9.80P
$1.47/$9.80 = $9.80P/$9.80
P = 0.15
The percent of tip is 15%
Today’s Lunch $12.60 x 15% = $12.60 x 0.15 = $1.89
She leaves $1.89 tip.
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37. Verify Green's theorem in the plane for f (3x2- 8y2) dx + (4y - 6xy) dy, where C is the boundary of the
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• (38) was addressed in 24438105

• (39) was addressed in 24434477

• (40) and (41) were both addressed in 24434541

In both parts, we're considering the line integral

\displaystyle \int_C (3x^2-8y^2)\,\mathrm dx + (4y-6xy)\,\mathrm dy

and I assume <em>C</em> has a positive orientation in both cases

(a) It looks like the region has the curves <em>y</em> = <em>x</em> and <em>y</em> = <em>x</em> ² as its boundary***, so that the interior of <em>C</em> is the set <em>D</em> given by

D = \left\{(x,y) \mid 0\le x\le1 \text{ and }x^2\le y\le x\right\}

• Compute the line integral directly by splitting up <em>C</em> into two component curves,

<em>C₁ </em>: <em>x</em> = <em>t</em> and <em>y</em> = <em>t</em> ² with 0 ≤ <em>t</em> ≤ 1

<em>C₂</em> : <em>x</em> = 1 - <em>t</em> and <em>y</em> = 1 - <em>t</em> with 0 ≤ <em>t</em> ≤ 1

Then

\displaystyle \int_C = \int_{C_1} + \int_{C_2} \\\\ = \int_0^1 \left((3t^2-8t^4)+(4t^2-6t^3)(2t))\right)\,\mathrm dt \\+ \int_0^1 \left((-5(1-t)^2)(-1)+(4(1-t)-6(1-t)^2)(-1)\right)\,\mathrm dt \\\\ = \int_0^1 (7-18t+14t^2+8t^3-20t^4)\,\mathrm dt = \boxed{\frac23}

*** Obviously this interpretation is incorrect if the solution is supposed to be 3/2, so make the appropriate adjustment when you work this out for yourself.

• Compute the same integral using Green's theorem:

\displaystyle \int_C (3x^2-8y^2)\,\mathrm dx + (4y-6xy)\,\mathrm dy = \iint_D \frac{\partial(4y-6xy)}{\partial x} - \frac{\partial(3x^2-8y^2)}{\partial y}\,\mathrm dx\,\mathrm dy \\\\ = \int_0^1\int_{x^2}^x 10y\,\mathrm dy\,\mathrm dx = \boxed{\frac23}

(b) <em>C</em> is the boundary of the region

D = \left\{(x,y) \mid 0\le x\le 1\text{ and }0\le y\le1-x\right\}

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• Using Green's theorem:

\displaystyle \int_C (3x^2-8y^2)\,\mathrm dx + (4y-6xy)\,\mathrm dx = \int_0^1\int_0^{1-x}10y\,\mathrm dy\,\mathrm dx = \boxed{\frac53}

4 0
3 years ago
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