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Anni [7]
2 years ago
9

Traces of surface x2 + y2 − z2 = 1. Determine the equation for the family of traces in x = n.

Mathematics
1 answer:
pickupchik [31]2 years ago
6 0

Answer:

\mathbf{y^2 -z^2 =1- n^2}

Step-by-step explanation:

Give that:

the surface equation is  x^2 +y^2 -z^2 =1

from the family of traces in x = n given that x^2 +y^2 -z^2 =1 , the equation can be represented as :

n^2 +y^2 -z^2 =1

\mathbf{y^2 -z^2 =1- n^2}

This represents a family of hyperbola for all values of n expects that n = ± 1

So, if n = ± 1,

Then

y^2 - z^2 = 0

(y-z) (y+z) = 0

y = ± z

So, for n = ± 1, it is a pair of line for y = z, y = -z

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Area = 139.27 m²

Step-by-step explanation:

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Plug in the values

Area = (10*10) + ½(π*5²)

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Amanda went to the store to purchase ink pens. She found three kinds of pens. The first cost $4 each; the price of the second ki
KonstantinChe [14]

Answer:

Amanda buy first kind of pen = 3

Amanda buy second kind of pen = 2

Amanda buy third kind of pen = 15

Step-by-step explanation:

Given - Amanda went to the store to purchase ink pens. She found three    

            kinds of pens. The first cost $4 each; the price of the second kind

            was 4 for $1; and the cost for the third kind was 2 for $1. She bought

            20 pens and she bought at least one of each kind. (It is possible to

            buy only 1 of the pens that are "4 for $1" or "2 for $1".) The cost was

            $20.

To find - When she got back to her office, Amanda decided to turn this into

              a math problem for me. She asked: how many of each kind did I

              buy?

Proof -

Let Amanda buy first kind of pen = x

                      second kind of pen = y

                           third kind of pen = z

As given,

She bought total pen = 20

⇒x + y + z = 20          ...............(1)

Now,

As given,

cost for first kind pen = $4 for 1 pen

As she bought x pens of first kind , so

Cost of x pens of first kind = $4x

Now,

The price of the second kind was 4 for $1

⇒Cost of second kind = $\frac{1}{4} for 1 pen

As she bought y pens of send kind , so

Cost of y pens of second kind = $\frac{1}{4}y

Now,

The price of the third kind was 2 for $1

⇒Cost of third kind = $\frac{1}{2} for 1 pen

As she bought z pens of send kind , so

Cost of z pens of third kind = $\frac{1}{2}z

Now,

As given, The cost was $20

⇒4x + \frac{1}{4}y + \frac{1}{2}z = 20

⇒16x + y + 2z = 80             .....................(2)

∴ we get 2 equations

x + y + z = 20                   .....................(1)

16x + y + 2z = 80             .....................(2)

Now,

Subtract equation (1) from equation (2) , we get

16x + y + 2z  - ( x+ y + z )= 80 - 20

⇒16x + y + 2z - x - y - z = 60

⇒15x + z = 60

⇒z = 60 - 15x

Now,

Put the value of z in equation (1) , we get

x + y + 60 - 15 x = 20

⇒ y - 14x = 20 - 60

⇒y - 14x = -40

⇒14x - y = 40

⇒y = 14x - 40

Now,

we get

z = 60 - 15x

y = 14x - 40

As given

she bought at least one of each kind

it means x > 1, y > 1, z > 1

Now,

If x = 1, then y = 14 - 40 = -26

Not possible

If x = 2 , then y = 14(2) - 40 = -12

Not possible

If x = 3, then y = 14(3) - 40 = 2 and z = 60 - 15(3) = 15

Possible.

If x = 4,  then y = 14(4) - 40 = 16 and z = 60 - 15(4) = 0

Not Possible.

If x = 5, then y = 14(5) - 40 = 30 and z = 60 - 15(5) = -15

Not Possible.

∴ we get

x = 3, y = 2, z = 15

Amanda buy first kind of pen = x = 3

Amanda buy second kind of pen = y = 2

Amanda buy third kind of pen = z = 15

7 0
2 years ago
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