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klio [65]
3 years ago
9

The cost C of moving rubble is proportional to the product of the mass M of the rubble and the distance D the rubble must be mov

ed. Use K as your proportionality constant (Note: we don't use C since we are already using it for cost).
(a) Find C(M,D).
C(M,D)=___________________
(b) If it costs 50 dollars to move 1 ton of rubble 1 mile, how much does it cost to move 16 tons 15 miles?.
$_____________
Mathematics
1 answer:
Mandarinka [93]3 years ago
3 0

Answer:

a) C(M,D) = KMD

b) $12,000                  

Step-by-step explanation:

We are given the following information in the question:

The cost C of moving rubble is proportional to the product of the mass M of the rubble and the distance D the rubble must be moved.

This can be written as:

a)

C \propto M\times D\\C = KMD,\\\text{where K is the constant of proportionality.}

C(M,D) = KMD

b) It costs 50 dollars to move 1 ton of rubble 1 mile

50 = K\times 1\times 1\\\Rightarrow K= 50

We have to find  cost to move 16 tons 15 miles.

Putting the values in the function:

C(16,15) = 50\times 16\times 15 = 12000

Thus, $12,00 is the cost to move 16 tons 15 miles.

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63.38

Step-by-step explanation:

(49.90)(20%)(6%)

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(59.80)(6%)

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8 0
2 years ago
Find the volume of the solid.
dmitriy555 [2]

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\}

In the plane z=2, we have

2 = 4 - x^2 - y^2 \implies x^2 + y^2 = 2 = \left(\sqrt2\right)^2

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}

3 0
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Answer:

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1,024 is 4^5 hope this helps :)
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