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Sindrei [870]
3 years ago
7

In how many ways can you seat 5 women and 5 men in a row if women must seat next to each other,men must seat next to each other,

and 2 women are indistinguishable siamese twins (also dressed alike)?
Mathematics
1 answer:
makvit [3.9K]3 years ago
8 0

Answer: 288ways

Step-by-step explanation:

There are 5men and 5women to be arranged, since the men must seat together, they will be arranged in 5! ways. For the women, since they must also seat together but with siamese twins between them, they can be arranged in 4! ways instead of 5! ways and this is due to presence of the twins among them.

Note that Siamese twins cannot be separated as such both are taken as one making it 4!.

Since the women and men are always sitting together, they can be arranged in 2! ways i.e 2 sexes

The final seating arrangement can be done in 2!×(5!+4!) ways

= 2× (120+24)

= 2×144

= 288ways.

Note that the arrangement of the men and women are added because they can only be arranged differently to ensure different sex are not sitting together.

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Sergeeva-Olga [200]

Q2. By the chain rule,

\dfrac{dy}{dx} = \dfrac{dy}{dt} \cdot \dfrac{dt}{dx} = \dfrac{\frac{dy}{dt}}{\frac{dx}{dt}}

We have

x=2t \implies \dfrac{dx}{dt}=2

y=t^4+1 \implies \dfrac{dy}{dt}=4t^3

The slope of the tangent line to the curve at t=1 is then

\dfrac{dy}{dx} \bigg|_{t=1} = \dfrac{4t^3}{2} \bigg|_{t=1} = 2t^3\bigg|_{t=1} = 2

so the slope of the normal line is -\frac12. When t=1, we have

x\bigg|_{t=1} = 2t\bigg|_{t=1} = 2

y\bigg|_{t=1} = (t^4+1)\bigg|_{t=1} = 2

so the curve passes through (2, 2). Using the point-slope formula for a line, the equation of the normal line is

y - 2 = -\dfrac12 (x - 2) \implies y = -\dfrac12 x + 3

Q3. Differentiating with the product, power, and chain rules, we have

y = x(x+1)^{1/2} \implies \dfrac{dy}{dx} = \dfrac{3x+2}{2\sqrt{x+1}} \implies \dfrac{dy}{dx}\bigg|_{x=3} = \dfrac{11}4

The derivative vanishes when

\dfrac{3x+2}{2\sqrt{x+1}} = 0 \implies 3x+2=0 \implies x = -\dfrac23

Q4. Differentiating  with the product and chain rules, we have

y = (2x+1)e^{-2x} \implies \dfrac{dy}{dx} = -4xe^{-2x}

The stationary points occur where the derivative is zero.

-4xe^{-2x} = 0 \implies x = 0

at which point we have

y = (2x+1)e^{-2x} \bigg|_{x=0} = 1

so the stationary point has coordinates (0, 1). By its "nature", I assume the question is asking what kind of local extremum this point. Compute the second derivative and evaluate it at x=0.

\dfrac{d^2y}{dx^2}\bigg|_{x=0} = (8x-4)e^{-2x}\bigg|_{x=0} = -4 < 0

The negative sign tells us this stationary point is a local maximum.

Q5. Differentiating the volume equation implicitly with respect to t, we have

V = \dfrac{4\pi}3 r^3 \implies \dfrac{dV}{dt} = 4\pi r^2 \dfrac{dr}{dt}

When r=5\,\rm cm, and given it changes at a rate \frac{dr}{dt}=-1.5\frac{\rm cm}{\rm s}, we have

\dfrac{dV}{dt} = 4\pi (5\,\mathrm{cm})^2 \left(-1.5\dfrac{\rm cm}{\rm s}\right) = -150\pi \dfrac{\rm cm^3}{\rm s}

Q6. Given that V=400\pi\,\rm cm^3 is fixed, we have

V = \pi r^2h \implies h = \dfrac{400\pi}{\pi r^2} = \dfrac{400}{r^2}

Substitute this into the area equation to make it dependent only on r.

A = \pi r^2 + 2\pi r \left(\dfrac{400}{r^2}\right) = \pi r^2 + \dfrac{800\pi}r

Find the critical points of A.

\dfrac{dA}{dr} = 2\pi r - \dfrac{800\pi}{r^2} = 0 \implies r = \dfrac{400}{r^2} \implies r^3 = 400 \implies r = 2\sqrt[3]{50}

Check the sign of the second derivative at this radius to confirm it's a local minimum (sign should be positive).

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\dfrac{dA}{dt} = 4x\dfrac{dx}{dt} + (12\,\mathrm{cm})\dfrac{dx}{dt} = \left(4(\sqrt{114}-3\,\mathrm {cm}) + 12\,\mathrm{cm}\right)\left(0.05\dfrac{\rm cm}{\rm s}\right) = \dfrac{\sqrt{114}}5 \dfrac{\rm cm^2}{\rm s}

5 0
1 year ago
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