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2 years ago

Suppose B is a proper subset of C,

Mathematics

1 answer:

Tomtit [17]2 years ago

3 0

Answer:

Maximum: 7

Minimum: 0

Step-by-step explanation:

A proper subset B of a set C, denoted $B\subset C$ , is a subset that is strictly contained in C and so necessarily excludes at least one member of C.

This means that the number of elements in B must be at least 1 less than the number of elements in C. If the number of elements in C is 8, then the maximum number of elements in B can be 7.

The empty set is a proper subset of any nonempty set. Hence, the minimum number of elements in B can be 0.

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195% of what number is 28?

Paladinen [302]

Answer:

14.35

Step-by-step explanation:

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2 years ago

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Here is a retangle ABCD the length of the rectangle is incresed by 10% the width increased by 5%

gladu [14]

Answer:

the ans:

Step-by-step explanation:

The percentage increase in the perimeter of the rectangle is 6.67 %.

Percentage :

A percentage is a number or ratio that can be expressed as a fraction of 100.

The length of the rectangle is increased by 10% ,

10%

If initial length of rectangle is 10 then new length of rectangle will be 11 unit.

5 %

If initial width of rectangle is 20 then new width of rectangle will be 21 unit.

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2 years ago

Charlie has been in an automobile accident. No one was injured, but his car is pretty banged up. The damage to his car is estima

IceJOKER [234]

Answer:

Step-by-step explanation:

Damage to car = $2,500

Deductible to be paid = $500

As he has to pay the deductible of $500 When he will pay this deductible then insurance company will cover up everything. So the insurance company will cover up $2500 damage to his car he just has to pay the deductible.

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2 years ago

The perimeter of a playing field for a certain sport is 178 ft. The field is a rectangle, and the length is 43 ft longer than th

timofeeve [1]

Answer:

23 and 66 are the dimensions

Step-by-step explanation:

178/2=89

89+43=132

178-132=46

132/2=66

46/2=23

(I think this is the answer I'm so sorry if I am wrong)

5 0

2 years ago

Evaluate the surface integral. ∫∫s (x2 + y2 + z2) ds s is the part of the cylinder x2 + y2 = 16 that lies between the planes z =

KonstantinChe [14]

Decompose the surface into three components, $\mathbf r_1,\mathbf r_2,\mathbf r_3$ , corresponding respectively to the cylindrical region and the top and bottom disks:

$\mathbf r_1(u,v)=\begin{cases}x(u,v)=4\cos u\\y(u,v)=4\sin u\\z(u,v)=v\end{cases}$
where $0\le u\le2\pi$ and $0\le v\le5$ ,

$\mathbf r_2(u,v)=\begin{cases}x(u,v)=u\cos v\\y(u,v)=u\sin v\\z(u,v)=0\end{cases}$
where $0\le u\le4$ and $0\le v\le2\pi$ , and

$\mathbf r_3(u,v)=\begin{cases}x(u,v)=u\cos v\\y(u,v)=u\sin v\\z(u,v)=5\end{cases}$
where $0\le u\le4$ and $0\le v\le2\pi$ .

For the cylinder, we have

$\dfrac{\partial\mathbf r_1}{\partial u}\times\dfrac{\partial\mathbf r_1}{\partial v}=\langle4\cos u,4\sin u,0\rangle\implies\left\|\dfrac{\partial\mathbf r_1}{\partial u}\times\dfrac{\partial\mathbf r_1}{\partial v}\right\|=4$

and the integral over this surface is

$\displaystyle\iint_{\text{cyl}}(x^2+y^2+z^2)\,\mathrm dS=4\int_{v=0}^{v=5}\int_{u=0}^{u=2\pi}((4\cos u)^2+(4\sin u)^2+v^2)\,\mathrm du\,\mathrm dv$
$=\displaystyle320\int_{u=0}^{u=2\pi}\mathrm du+8\pi\int_{v=0}^{v=5}v^2\,\mathrm dv$
$=640\pi+\dfrac83\pi(125)$
$=\dfrac{2920\pi}3$

Bottom disk:

$\dfrac{\partial\mathbf r_2}{\partial u}\times\dfrac{\partial\mathbf
 r_2}{\partial v}=\langle0,0,u\rangle\implies\left\|\dfrac{\partial\mathbf r_2}{\partial 
u}\times\dfrac{\partial\mathbf r_2}{\partial v}\right\|=u$

and the integral over the bottom disk is

$\displaystyle\iint_{z=0}(x^2+y^2+z^2)\,\mathrm dS=\int_{v=0}^{v=2\pi}\int_{u=0}^{u=4}u((u\cos v)^2+(u\sin v)^2)\,\mathrm du\,\mathrm dv$
$=\displaystyle2\pi\int_{u=0}^{u=4}u^3\,\mathrm du$
$=128\pi$

The setup for the integral along the top disk is similar to that for the bottom disk, except that $z=5$ :

$\displaystyle\iint_{z=5}(x^2+y^2+z^2)\,\mathrm 
dS=\int_{v=0}^{v=2\pi}\int_{u=0}^{u=4}u((u\cos v)^2+(u\sin 
v)^2+5^2)\,\mathrm du\,\mathrm dv$
$=\displaystyle2\pi\int_{u=0}^{u=4}(u^3+25u)\,\mathrm du$
$=528\pi$

Finally, the value of the integral over the entire surface is the sum of the integrals over the component surfaces:

$\displaystyle\iint_S(x^2+y^2+z^2)\,\mathrm dS=\frac{2920\pi}3+128\pi+528\pi=\dfrac{4888\pi}3$

7 0

2 years ago