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3 years ago

A store holiday sale has an item marked down by $10 with an additional discount of 25% off the

Mathematics

1 answer:

ValentinkaMS [17]3 years ago

5 0

Answer:

C : $58.36

Step-by-step explanation:

Given:

A store holiday sale has an item marked down by $10.

Discount on new price = 25%

Final price was $36.27.

Question asked:

what was the original ?

Solution:

Let original price = $x$

New price = $x - 10$

Original price - marked down amount - discount amount = $36.27

$x - 10 -25\% of(x - 10 ) = 36.27\\$

$x - 10 - \frac{25}{100} (x - 10) = 36.27\\x - 10 - 0.25(x - 10) = 36.27\\x - 10 -0.25x +2.5 = 36.27\\0.75x - 7.5 = 36.27\\$

Adding both side by 7.5

$0.75x = 43.77\\$

Dividing both side by 0.75

$x = 58.36$

Therefore, the original price was $58.36

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In Cartesian coordinates, the region (call it $R$ ) is the set

$R = \left\{(x,y,z) ~:~ x\ge0 \text{ and } y\ge0 \text{ and } 2 \le z \le 4-x^2-y^2\right\}$

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which is a circle with radius $\sqrt2$ . Then we can better describe the solid by

$R = \left\{(x,y,z) ~:~ 0 \le x \le \sqrt2 \text{ and } 0 \le y \le \sqrt{2 - x^2} \text{ and } 2 \le z \le 4 - x^2 - y^2 \right\}$

so that the volume is

$\displaystyle \iiint_R dV = \int_0^{\sqrt2} \int_0^{\sqrt{2-x^2}} \int_2^{4-x^2-y^2} dz \, dy \, dx$

While doable, it's easier to compute the volume in cylindrical coordinates.

$\begin{cases} x = r \cos(\theta) \\ y = r\sin(\theta) \\ z = \zeta \end{cases} \implies \begin{cases}x^2 + y^2 = r^2 \\ dV = r\,dr\,d\theta\,d\zeta\end{cases}$

Then we can describe $R$ in cylindrical coordinates by

$R = \left\{(r,\theta,\zeta) ~:~ 0 \le r \le \sqrt2 \text{ and } 0 \le \theta \le\dfrac\pi2 \text{ and } 2 \le \zeta \le 4 - r^2\right\}$

so that the volume is

$\displaystyle \iiint_R dV = \int_0^{\pi/2} \int_0^{\sqrt2} \int_2^{4-r^2} r \, d\zeta \, dr \, d\theta \\\\ ~~~~~~~~ = \frac\pi2 \int_0^{\sqrt2} \int_2^{4-r^2} r \, d\zeta\,dr \\\\ ~~~~~~~~ = \frac\pi2 \int_0^{\sqrt2} r((4 - r^2) - 2) \, dr \\\\ ~~~~~~~~ = \frac\pi2 \int_0^{\sqrt2} (2r-r^3) \, dr \\\\ ~~~~~~~~ = \frac\pi2 \left(\left(\sqrt2\right)^2 - \frac{\left(\sqrt2\right)^4}4\right) = \boxed{\frac\pi2}$