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4 years ago

What is the standard equation for a circle with the center (2,1) and passing through (0, 1)

2 answers:

Volgvan4 years ago

4 0

The radius is the distance from the center to any point. The distance from (2,1) to (0,1) is of course 2. The general formula for a circle with radius r and center (a,b) is

$(x-a)^2 + (y-b)^2 = r^2$

In our case we get

$(x-2)^2 + (y-1)^2 = 4$

BigorU [14]4 years ago

3 0

So this is what I got (x-2)^2+(y-1)^2=4

You might be interested in

What is the value of x in the equation 2.5 (6x - 4) = 10 + 4 (1.5 + 0.5x)

ioda

Answer:

x = 2

Step-by-step explanation:

Given

2.5(6x-4) = 10 + 4 (1.5+0.5x)

First of all we will have to solve the parenthesis

(2.5)(6x)-(2.5)(4) = 10 + (4)(1.5)+(4)(0.5x)

15x-10 = 10+6+2x

We have to isolate x so that we can find its value

15x-10 = 16+2x

15x = 16+2x+10

15x-2x = 26

13x = 26

13x/13 = 26/13

x = 2 ..

8 0

3 years ago

Write an equation for the line parallel to the given line that contains C

Ksenya-84 [330]

Answer:

The equation would be y = -2x + 10

Step-by-step explanation:

To find the line parallel going through the given point, first note that parallel lines have the same slope. This means that we have a slope of -2 in the original line, therefore the parallel line will also have a slope of -2. Now we can use this and the given point to find the line using point-slope form.

y - y1 = m(x - x1)

y - 6 = -2(x - 2)

y - 6 = -2x + 4

y = -2x + 10

7 0

3 years ago

If any one could help me and explain how they would do this it would make my entire day it has to be a fraction please I can not

Mrrafil [7]

So slope intercept form is y=mx+b m is the slope and b is the y intercept. The equation of this line would be y=1/4x+2 because the rise over run (slope) is 2/8 or 1/4 and the line intercepts the y axis at 2.

8 0

2 years ago

Read 2 more answers

If a number is multiplied by 4 it gives the same result as 2/7 of 504 what is the number

NNADVOKAT [17]

507/7*2=144.86/4=36.21

5 0

3 years ago

What is the average rate of change of f over the interval -1 on a separate sheet of paper and upload the work as a photo. Show a

Anarel [89]

Answer:

Average rate of change for the function $f(x)= x^2-x-1$ over the interval -1<x<1 is -1

Step-by-step explanation:

We need to find average rate of change of f over the interval -1 < x < 1

The function given is: $f(x)= x^2-x-1$

The formula used to find average rate of change is:

$Average\:rate\:of\:change=\frac{f(b)-f(a)}{b-a}$

We have, a = -1 and b = 1

Finding f(b) when b=1

$f(x)=x^2-x-1\\f(1)=(1)^2-(1)-1\\f(1)=1-1-1\\f(1)=-1$

Now, finding f(a), when a= -1

$f(x)=x^2-x-1\\f(-1)=(-1)^2-(-1)-1\\f(1)=1+1-1\\f(-1)=2-1\\f(-1)=1$

Now, putting values and finding average rate of change

$Average\:rate\:of\:change=\frac{f(b)-f(a)}{b-a}\\Average\:rate\:of\:change=\frac{f(1)-f(-1)}{1-(-1)}\\Average\:rate\:of\:change=\frac{-1-(1)}{1-(-1)}\\Average\:rate\:of\:change=\frac{-1-1}{1+1}\\Average\:rate\:of\:change=\frac{-2}{2}\\Average\:rate\:of\:change=-1$

So, average rate of change for the function $f(x)= x^2-x-1$ over the interval -1<x<1 is -1

4 0

3 years ago