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2 years ago

What is the equation of a hyperbola with a = 1 and c = 9? Assume that the transverse axis is horizontal.

Mathematics

1 answer:

belka [17]2 years ago

8 0

Answer:

Step-by-step explanation:

The general formula for a horizontal hyperbola :

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

Given: a = 1 we know a² = 1 ,and that c = 9 so we know c² = 9² = 81

We also know that the relation between a, b, c for a hyperbola is c²= a²+b²

c²= a²+b², substitute what we know

81 = 1 +b², subtract 1 from both sides of the equation

80 = b²

The equation of our hyperbola is:

$\frac{x^{2} }{1 } -\frac{y^{2} }{80 } =1$ or $x^{2} -\frac{y^{2} }{80} =1$

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A circular swimming pool has a diameter of 40 meters. The depth of the pool is constant along west-east lines and increases line

Nataliya [291]

Answer:

Volume is $2000\pi\ m^{3}$

Solution:

As per the question:

Diameter, d = 40 m

Radius, r = 20 m

Now,

From north to south, we consider this vertical distance as 'y' and height, h varies linearly as a function of y:

iff

h(y) = cy + d

Then

when y = 1 m

h(- 20) = 1 m

1 = c.(- 20) + d = - 20c + d (1)

when y = 9 m

h(20) = 9 m

9 = c.20 + d = 20c + d (2)

Adding eqn (1) and (2)

d = 5 m

Using d = 5 in eqn (2), we get:

$c = \frac{1}{5}$

Therefore,

$h(y) = \frac{1}{5}y + 5$

Now, the Volume of the pool is given by:

$V = \int h(y)dA$

where

A = $r\theta$

$A = rdr\ d\theta$

Thus

$V = \int (\frac{1}{5}y + 5)dA$

$V = \int_{0}^{2\pi}\int_{0}^{20} (\frac{1}{5}rsin\theta + 5) rdr\ d\theta$

$V = \int_{0}^{2\pi}\int_{0}^{20} (\frac{1}{5}r^{2}sin\theta + 5r}) dr\ d\theta$

$V = \int_{0}^{2\pi} (\frac{1}{15}20^{3}sin\theta + 1000) d\theta$

$V = [- 533.33cos\theta + 1000\theta]_{0}^{2\pi}$

$V = 0 + 2\pi \times 1000 = 2000\pi\ m^{3}$

7 0

3 years ago

You don’t know how important this is Please help me!!

Natali5045456 [20]

Answer:

f(x)= $\frac{7}{5}$ x- $\frac{6}{5}$

Step-by-step explanation:

First put 7x-5y=6 into slope intercept form

y= $\frac{7}{5}$ x- $\frac{6}{5}$

Then put it into f(x) form

7 0

2 years ago

A line has an y-intercept of 4 and a x-intercept of 1. Use this information to write a function in slope intercept form (y = mx

Ad libitum [116K]

Answer: D. y = -4x + 4

Step-by-step explanation:

Notice that the y-intercept is when x is zero and the x intercept is when y is 0.

and to write in the equation in slope intercept form , you will need the slope and the intercept.

We will use the y and x intercepts to find the slope, by find the difference in their y coordinates and dividing it by the difference in their x coordinates.

y intercept: ( 0,4)

x intercept: (1,0)

y coordinates difference: 4 - 0 = 4

x coordinates difference: 0 - 1 = - 1

Slope: 4/-1 = -4

Since the slope is -4 and the the y intercept is 4 , the equation will be

y = -4x + 4

7 0

3 years ago

JUST THE FIRST PART NOT THE DRAWING <br> IM BEGGING AT THIS POINT

Butoxors [25]

Answer:

75 (.2)= 15

75-15=60

The small bag cost $60.

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Find the perimeter of pentagon STUVW WITH VERTICES S(0,0) T(3,-2) U(2,-5) V(-2,-5) W(-3,-2)

andrew11 [14]

Use the distance formula.

$\sqrt{( x_{2} - x_{1} )^2 + (y_{2} - y_{1})^2}$

Points S and W.
$\sqrt{(3)^2 + (2)^2}$

$\sqrt{9+4}$

$\sqrt{13}$

~3.6

Points S and T
$\sqrt{(3 - 0)^2 + (-2 - 0)^2}$

$\sqrt{(3)^2 + (-2)^2}$

$\sqrt{9+4}$

$\sqrt{13}$

~3.6

Points T and U
$\sqrt{(3 - 2)^2 + (-2 + 5)^2}$

$\sqrt{(1)^2 + (3)^2}$

$\sqrt{1+9}$

$\sqrt{10}$

~3.1

Points U and V
$\sqrt{(2+2)^2 + (-5 + 5)^2}$

$\sqrt{(4)^2 + (0)^2}$

$\sqrt{16}$

~4

Points V and W
$\sqrt{(-2+3)^2 + (-5 + 2)^2}$

$\sqrt{(1)^2 + (-3)^2}$

$\sqrt{2+9}$

$\sqrt{11}$

~3.3

Add all these together.

3.3 + 3.1 + 4 + 3.1 + 3.6
≈17

4 0

3 years ago