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3 years ago

Kristin decides to spend at most $50 for a birthday dinner at a restaurant, including a 15% tip.,

1 answer:

7 0

0.15x + x <(or equal to) 50
Where x is the meal's cost (no tip included)

50 - 15% = 42.5.
x >(or equal to) 42.50.

(nothing is copied content, I copy pasted a number into the right position, feel free to check it mods)

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PRECALCULUS QUESTION PLEASE HELP!!!!!!

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2 years ago

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2 years ago

A pair of skis was discounted 20%. How many percent will the discounted price have to be increased to return to the original pri

denis-greek [22]

Answer:

25%

Step-by-step explanation:

Let the original price of A pair of skis be $100

discount given on A pair of skis = 20% on original price

discount value of A pair of skis in $ = 20% * $100

= (20 * 100)/100 = $20

discounted price of A pair of skis = original price - discount = 100 -20 = $80

in order to increase discounted price to original price of $100

there must be increase of $20 to discounted price which is $80

percentage increase on discounted price to return to the original price =

$(increased \ amount / discounted \ price) * 100 \\=>( 20/80) * 100 = 25\\$

therefore 25% must be increase on discounted price to return to the original price

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2 years ago

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2 years ago

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,

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

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3 years ago