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VLD [36.1K]
3 years ago
5

which do you think would seat more people a 4ft by 6 ft rectangular table or a circular table with a diameter of 6 ft? How many

people would sit at each table? Explain your reasoning.
Mathematics
1 answer:
sesenic [268]3 years ago
5 0
To determine which seating arrangement would be able to seat more people you will find the distance around each table.

For the rectangle, you will find the perimeter:

P = 2l + 2w
   2 x 6 + 2 x 4
P = 20 feet

For the circle you will find the circumference:

C = pi x d
      3.14 x 6
C = 18.84 feet

The rectangle would seat more people because the distance around the table is longer.

You would be able to fit about 7-8 people around the rectangular table and 6-7 people around the circular table.
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Answer:

y =1/2

Step-by-step explanation:

solve for ; y

10(1 – 2y) = -5(2Y - 1)

-10(2y-1) +5(2Y - 1) =0

(2y-1)(-10+5) = 0

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2y-1=0

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7 0
3 years ago
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Hi, help with question 18 please. thanks​
Vadim26 [7]

Answer:

See Below.

Step-by-step explanation:

We are given the equation:

\displaystyle y^2 = 1 + \sin x

And we want to prove that:

\displaystyle 2y\frac{d^2y}{dx^2} + 2\left(\frac{dy}{dx}\right) ^2 + y^2 = 1

Find the first derivative by taking the derivative of both sides with respect to <em>x: </em>

<em />\displaystyle 2y \frac{dy}{dx}  = \cos x<em />

Divide both sides by 2<em>y: </em>

<h3><em />\displaystyle \frac{dy}{dx} = \frac{\cos x}{2y}<em /></h3>

<em />

Find the second derivative using the quotient rule:

\displaystyle \begin{aligned} \frac{d^2y}{dx^2} &= \frac{(\cos x)'(2y) - (\cos x)(2y)'}{(2y)^2}\\ \\  &= \frac{-2y\sin x-2\cos x \dfrac{dy}{dx}}{4y^2} \\ \\ &= -\frac{y\sin x + \cos x\left(\dfrac{\cos x}{2y}\right)}{2y^2} \\ \\ &= -\frac{2y^2\sin x+\cos ^2 x}{4y^3}\end{aligned}

Substitute:

\displaystyle 2y\left(-\frac{2y^2\sin x+\cos ^2 x}{4y^3}\right)  + 2\left(\frac{\cos x}{2y}\right)^2 +y^2 = 1

Simplify:

\displaystyle \frac{-2y^2\sin x-\cos ^2x}{2y^2} + \frac{\cos ^2 x}{2y^2} + y^2 = 1

Combine fractions:

\displaystyle \frac{\left(-2y^2\sin x -\cos^2 x\right)+\left(\cos ^2 x\right)}{2y^2} + y^2 = 1

Simplify:

\displaystyle \frac{-2y^2\sin x }{2y^2} + y^2 = 1

Cancel:

\displaystyle -\sin x + y^2 = 1

Substitute:

-\sin x + \left( 1 + \sin x\right) =1

Simplify. Hence:

1\stackrel{\checkmark}{=}1

Q.E.D.

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