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ludmilkaskok [199]
3 years ago
13

You have quarters and dimes that total $2.80. your friend says it is possible that the number of quarters is 8 more than the num

ber of dimes. is your friend correct? explain. let dd represent the number of dimes.an equation that models this situation is?
Mathematics
1 answer:
ss7ja [257]3 years ago
4 0
To answer this question, you need to know that a quarter is worth $0.25 and dimes worth $0.10. In this case, the total number of money that made from quarters and dimes is $2.80
With Q=quarter and D=dime, then the equation should be:
2.80= 0.25 Q + 0.10 D

The condition that needs to be fulfilled in this case is that the quarter should be 8 more than dimes. Then you need to find out the highest possible number of the quarter and lowest possible number of the dimes.  
If you divide 2.8 with 0.25 you will found it will be 11.2 but 11 quarter will result with 0.05 value which you cannot remove, so the highest possible of the quarter should be 10.
If the highest possible of the quarter is 10, then the lowest possible of dimes should be:
2.8= 0.25(10) + 0.10D
0.10D= 0.3
D=3

Since 10-3 is 7, then it is not possible for the quarter to be 8 more than the dimes.
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Answer:

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3 years ago
Let f(x,y,z) = ztan-1(y2) i + z3ln(x2 + 1) j + z k. find the flux of f across the part of the paraboloid x2 + y2 + z = 3 that li
Sophie [7]
Consider the closed region V bounded simultaneously by the paraboloid and plane, jointly denoted S. By the divergence theorem,

\displaystyle\iint_S\mathbf f(x,y,z)\cdot\mathrm dS=\iiint_V\nabla\cdot\mathbf f(x,y,z)\,\mathrm dV

And since we have

\nabla\cdot\mathbf f(x,y,z)=1

the volume integral will be much easier to compute. Converting to cylindrical coordinates, we have

\displaystyle\iiint_V\nabla\cdot\mathbf f(x,y,z)\,\mathrm dV=\iiint_V\mathrm dV
=\displaystyle\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=1}\int_{z=2}^{z=3-r^2}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta
=\displaystyle2\pi\int_{r=0}^{r=1}r(3-r^2-2)\,\mathrm dr
=\dfrac\pi2

Then the integral over the paraboloid would be the difference of the integral over the total surface and the integral over the disk. Denoting the disk by D, we have

\displaystyle\iint_{S-D}\mathbf f\cdot\mathrm dS=\frac\pi2-\iint_D\mathbf f\cdot\mathrm dS

Parameterize D by

\mathbf s(u,v)=u\cos v\,\mathbf i+u\sin v\,\mathbf j+2\,\mathbf k
\implies\mathbf s_u\times\mathbf s_v=u\,\mathbf k

which would give a unit normal vector of \mathbf k. However, the divergence theorem requires that the closed surface S be oriented with outward-pointing normal vectors, which means we should instead use \mathbf s_v\times\mathbf s_u=-u\,\mathbf k.

Now,

\displaystyle\iint_D\mathbf f\cdot\mathrm dS=\int_{u=0}^{u=1}\int_{v=0}^{v=2\pi}\mathbf f(x(u,v),y(u,v),z(u,v))\cdot(-u\,\mathbf k)\,\mathrm dv\,\mathrm du
=\displaystyle-4\pi\int_{u=0}^{u=1}u\,\mathrm du
=-2\pi

So, the flux over the paraboloid alone is

\displaystyle\iint_{S-D}\mathbf f\cdot\mathrm dS=\frac\pi2-(-2\pi)=\dfrac{5\pi}2
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4 years ago
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