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adell [148]
3 years ago
13

What are the dimensions of the pond? Please help.

Mathematics
1 answer:
balu736 [363]3 years ago
8 0

Let's assume

length of box is l

width of box is w

we know that

area of rectangle = length*width

21=l*w...................(1)

now, we are given

fencing length is 20 foot

fencing length will be equal to perimeter

and we know that

perimeter=2(l+w)

so, we get

2(l+w)=20

now, we can solve for l

l=10-w

now, we can plug that in first equation

21=(10-w)*w

now, we can solve for w

10w-w^2=21

w^2-10w+21=0

(w-7)(w-3)=0

w=7 , w=3

now, we can find length

l=10-3=7

or

l=10-7=3

so, option-B..................Answer

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Answer:

\dfrac{x^{2}}{1210000 } - \dfrac{y^{2}}{5759600} = 1

Step-by-step explanation:

A hyperbola is a curve for which the difference of the distances |d₂ - d₁| of any point P from the foci is constant.

The definition leads to the equation

\dfrac{x^{2}}{a^{2}} - \dfrac{y^{2}}{b^{2}} = 1

1. Calculate the value of a²

Assume the neighbours are at the foci A and B.

The explosion is farther from B than from A, so it occurred on the right branch of the hyperbola.

Then AB = 1 mi = 5280 ft

and the distance from the focus to the y-axis is

c = 2640

The vertices are (±a,0).

The speed of sound is 1100 ft/s, so B is 2200 ft further from the explosion.  

The distance from A to the vertex V₁ is c - a. Then

AB = 2200 + 2(c - a) = 2200 -2(2640 - 2a) = 2200 +5280- 2a = 5280

2a = 2200

a = 1100

a² = 1 210 000

2. Calculate the value of b²

\begin{array}{rcl}a^{2} + b^{2} & = & c^{2}\\1100^{2} + b^{2} & = & 2640^{2}\\b^{2} & = & 2640^{2} - 1100^{2}\\& = & 6969600 - 1210000\\& = & \mathbf{5759600}\\\end{array}

3. Write the equation for the hyperbola

\mathbf{\dfrac{x^{2}}{1210000 } - \dfrac{y^{2}}{5759600}} = \mathbf{1}

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