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Ede4ka [16]
2 years ago
15

30 POINTS AVAILABLE

Mathematics
1 answer:
Solnce55 [7]2 years ago
5 0

Find the center, which is the midpoint of the two endpoints given.

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

(4-2)/2 , (-3+5)/2

2/2 , 2/2

Midpoint = (1,1)

Find the radius, which is the distance from the center to an end point.

√(x2-x1)^2 +(y2-y1)^2

√(4-1)^2 + (-3-1)^2

√(3^2 + -4^2)

√9+16

√25

r = 5

The equation for a circle is written as (x-h)^2 + (y-k)^2 = r^2

h and k are the center points.

Replace the letters with their values to get:

(x-1)^2 + (y-1)^2 =25

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HELP ME ASAP PLEASE!!
Daniel [21]

Answer:

set both of them equal to each other:

8x - 8° = 5x + 25°

subtract 5x by both sides of the equations and it becomes:

3x-8° = 25°

now add 8° both sides:

3x = 33°

divide three by both sides:

x = 11°

now thats done to find B:

plug your x value in the B equation:

5x + 25°

5(11) +25°

55 +25°

not sure if you can add here sorry :(. I hope this helped so far :)

5 0
3 years ago
Blank divided by blank equals 6 what is the two blanks
Travka [436]
36 by 6 blah blah blah
4 0
2 years ago
Read 2 more answers
Please solve and show working for 13-((4/5)+(6/8))=
UNO [17]
<span>13-((4/5)+(6/8))
Make your fractions have common denominators 
</span>13-((32/40)+(30/40))
Add your fractions and simplify
13-(62/40)
or
13-(31/20)
or
13-(1 11/20)
Then turn 13 into a fraction with a common denominator! Im going to use the second fraction method (31/20)
13 written as a fraction is 13/1, its LCD with 31/20 is 20. I now multiply the top and bottom by 20
260/20
Now I rewrite the problem again
(260/20)-(31/20)
Which equals
229/20!
This is your unsimplified answer
Finally you simplify and get
11 9/20

6 0
3 years ago
A square has an area of 289 cm2. What is the length of one side?
Delvig [45]

Answer:

144.5cm

Step-by-step explanation:

Since it's a square, all sides should be equal so you just need to divide 289 by 2(are of a square is side times side)

3 0
3 years ago
Can someone help please?
zalisa [80]

r =  \frac{15}{7}

8 0
2 years ago
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