All categories

3 years ago

The end points of a diameter of a circle are (6,2) and (-4,7).

Mathematics

1 answer:

nata0808 [166]3 years ago

8 0

Answer:

Step-by-step explanation:

The standard form equation of a circle with radius r is expressed as

( x − h )^2 + ( y − k )^2 =r ^2 ,

where r represents the radius

h and k are the coordinates of the center of the circle C( h , k )

To determine the coordinates at the center of the circle, the midpoint formula would be used. It is expressed as

[(x1 + x2)/2 , (y1 + y2)/2]

Midpoint of the circle =

(6 - 4)/2 , (2 + 7)/2 = (1, 4.5)

h coordinate of the center = 1

k coordinate of the center = 4.5

r^2 = (x - h)^2 + (2 - k)^2

r^2 = (6 - 1)^2 + (2 - 4.5)^2

r^2 = 5^2 + (- 2.5)^2 = 25 + 6.25

r^2 = 31.25

Substituting r^2 = 31.25, h = 1 and k = 4.5 into (x − h )^2 + ( y − k )^2 = r^2, the standard equation of the circle becomes

(x − 1 )^2 + ( y − 4.5 )^2 = 31.25

You might be interested in

Can anyone help me please I’ll mark as brainliest.

sergejj [24]

Answer:

First tables blanks are 160, 195, 230, 300

The rate of change would be 35/1 cos the cost is going up by 35 and the months are going up by one

The second tables first row is 1 2 3 4 5, the second row is 5 7 9 11 13

5 0

3 years ago

When solving a word problem, what step should you always do after reading the problem, but prior to writing the equation?

Hoochie [10]

Properly identifying and assigning the proper variables to each part of the problem before attempting to solve it

4 0

3 years ago

Suppose the clean water of a stream flows into Lake Alpha, then into Lake Beta, and then further downstream. The in and out flow

Gala2k [10]

Answer:

a) dx / dt = - x / 800

b) x = 500*e^(-0.00125*t)

c) dy/dt = x / 800 - y / 200

d) y(t) = 0.625*e^(-0.00125*t)*( 1 - e^(-4*t) )

Step-by-step explanation:

Given:

- Out-flow water after crash from Lake Alpha = 500 liters/h

- Inflow water after crash into lake beta = 500 liters/h

- Initial amount of Kool-Aid in lake Alpha is = 500 kg

- Initial amount of water in Lake Alpha is = 400,000 L

- Initial amount of water in Lake Beta is = 100,000 L

Find:

a) let x be the amount of Kool-Aid, in kilograms, in Lake Alpha t hours after the crash. find a formula for the rate of change in the amount of Kool-Aid, dx/dt, in terms of the amount of Kool-Aid in the lake x:

b) find a formula for the amount of Kook-Aid in kilograms, in Lake Alpha t hours after the crash

c) Let y be the amount of Kool-Aid, in kilograms, in Lake Beta t hours after the crash. Find a formula for the rate of change in the amount of Kool-Aid, dy/dt, in terms of the amounts x,y.

d) Find a formula for the amount of Kool-Aid in Lake Beta t hours after the crash.

Solution:

- We will investigate Lake Alpha first. The rate of flow in after crash in lake alpha is zero. The flow out can be determined:

dx / dt = concentration*flow

dx / dt = - ( x / 400,000)*( 500 L / hr )

dx / dt = - x / 800

- Now we will solve the differential Eq formed:

Separate variables:

dx / x = -dt / 800

Integrate:

Ln | x | = - t / 800 + C

- We know that at t = 0, truck crashed hence, x(0) = 500.

Ln | 500 | = - 0 / 800 + C

C = Ln | 500 |

- The solution to the differential equation is:

Ln | x | = -t/800 + Ln | 500 |

x = 500*e^(-0.00125*t)

- Now for Lake Beta. We will consider the rate of flow in which is equivalent to rate of flow out of Lake Alpha. We can set up the ODE as:

conc. Flow in = x / 800

conc. Flow out = (y / 100,000)*( 500 L / hr ) = y / 200

dy/dt = con.Flow_in - conc.Flow_out

dy/dt = x / 800 - y / 200

- Now replace x with the solution of ODE for Lake Alpha:

dy/dt = 500*e^(-0.00125*t)/ 800 - y / 200

dy/dt = 0.625*e^(-0.00125*t)- y / 200

- Express the form:

y' + P(t)*y = Q(t)

y' + 0.005*y = 0.625*e^(-0.00125*t)

- Find the integrating factor:

u(t) = e^(P(t)) = e^(0.005*t)

- Use the form:

( u(t) . y(t) )' = u(t) . Q(t)

- Plug in the terms:

e^(0.005*t) * y(t) = 0.625*e^(0.00375*t) + C

y(t) = 0.625*e^(-0.00125*t) + C*e^(-0.005*t)

- Initial conditions are: t = 0, y = 0:

0 = 0.625 + C

C = - 0.625

Hence,

y(t) = 0.625*( e^(-0.00125*t) - e^(-0.005*t) )

y(t) = 0.625*e^(-0.00125*t)*( 1 - e^(-4*t) )

6 0

3 years ago

Marina had 24,500 to invest. She divided the money into three different accounts. At the end of the year, she had made RM1,300 i

galben [10]

Answer:

See below

Bold parts are important parts. They are the equations.

Marina had RM24,500 to invest.

If the amount of money in the 4% account was four times the amount of money in the 5.5% account.

" At the end of the year, she had made RM1,300 in interest. The annual yield on each of the three accounts was 4%, 5.5%, and 6%."

"If the amount of money in the 4% account was four times the amount of money in the 5.5% account,"

a = 4b

Down is the equations.

let a = amt in the 4% acct

let b = amt in the 5.5% acct

let c = amt in the 6%

"Marina had RM 24,500 to invest."

a + b + c = 24500

Replace a with 4b in both equations, simplify

b = $2000 in the 5.5% investment

a = $8000 in the 4% acct

Hope this helps.

8 0

3 years ago

Identify the common difference of the sequence: –8, –1, 6, 13

allochka39001 [22]

Answer

Step-by-step explanation: They are skipping -7, 5, -7

7 0

3 years ago

Read 2 more answers