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Law Incorporation [45]
3 years ago
13

Suppose a circle with center (14,9) passes through point (16, 12). Which equation represents the circle?

Mathematics
1 answer:
kifflom [539]3 years ago
8 0

Answer:

( x - 14)^2 + ( y - 9)^2 = 13

Step-by-step explanation:

The equation of the circle with a center and a point

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

( 14 , 9) - center-( a, b)

a = 14

b = 9

( 16 , 12) - point - ( x, y)

x = 16

y = 12

Step 1: substitute the center into the equation

( x - 14)^2 + ( y - 9)^2 = r^2

Step 2: substitute the point into the equation

( x - 14)^2 +( y - 9)^2 = r^2

( 16 - 14)^2 + ( 12 - 9)^2 = r^2

2^2 + 3^2 = r^2

4 + 9 = r^2

13 = r^2

Step 3: subs the radius into the equation

( x - 14)^2 + ( y - 9)^2 = r^2

( x - 14)^2 + (y - 9)^2 = 13

Therefore, the equation of the circle is

( x - 14)^2 + ( y - 9)^2 = 13

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Which of the following definitions describe functions from the domain to the codomain given? Which functions are one-to-one? Whi
LenKa [72]

Answer:

  • a) f is a function. It is not 1-1, it is not onto.
  • b) g is not a function.
  • c) h is a function. It is not 1-1, it is onto.
  • f) h is a function. It is a bijection, and h^-1(x,y)=(y-1,x-1)

Step-by-step explanation:

a)  For all x∈ℤ, the number f(x)=x²+1 exists and is unique because f(x) is defined using the operations addition (+) and multiplication (·) on ℤ. Then f is a function. f is not one-to-one: consider -1,1∈ℤ. -1≠1 but f(-1)=f(1)=2- Because two different elements in the domain have the same image under f, f is not 1-1. f is not onto: x²≥0 for all x∈ℤ then f(x)=x²+1≥1>0 for all x∈ℤ. Then 0∈ℕ but for all x∈ℤ f(x)≠0, which means that one element of the codomain doen't have a preimage, so f is not onto.

b) 0∈ℕ, so 0 is an element of the domain of g, but g(0)=1/0 is undefined, therefore g is not a function.

c) Let (z,n)∈ℤ x ℕ. The number h(z,n)=z·1/(n+1) is unique and it's always defined because n+1>0, then h is a function. h is not 1-1: consider the ordered pairs (1,2), (2,5). They are different elements of the domain, but h(2,5)=2/6=1/3=h(1,2). h is onto: any rational number q∈ℚ can be written as q=a/b for some integer a and positive integer b. Then (a,b-1)∈ ℤ x ℕ and h(a,b-1)=a/b=q.

f) For all (x,y)∈ℝ², the pair h(x,y)=(y+1,x+1) is defined and is unique, because the definition of y+1 and x+1 uses the addition operation on ℝ. f is 1-1; suppose that (x,y),(u,v)∈ℝ² are elements of the domain such that h(x,y)=h(u,v). Then (y+1,x+1)=(v+1,u+1), so by equality of ordered pairs y+1=v+1 and x+1=u+1. Thus x=u and y=x, therefore (x,y)=(u,v). f is onto; let (a,b)∈ℝ² be an element of the codomain. Then (b-1,a-1)∈ℝ² is an element of the domain an h(b-1,a-1)=(a-1+1,b-1+1)=(a,b). Because h is 1-1 and onto, then h is a bijection so h has a inverse h^-1 such that for all (x,y)∈ℝ² h(h^-1(x,y))=(x,y) and h^-1(h^(x,y))=(x,y). The previous proof of the surjectivity of h (h onto) suggests that we define h^-1(x,y)=(y-1,x-1). This is the inverse, because h(h^-1(x,y))=h(y-1,x-1)=(x,y) and h^-1(h^(x,y))=h^-1(y+1,x+1)=(x,y).

5 0
3 years ago
13. Find the length of the side of a square with an area of 16.
4vir4ik [10]

Answer:

length of the side of the square   = 4    

Step-by-step explanation:

All the sides of a square are equal in length.

So, let's get the unknown side be x.

Area = length × width

  16  = x ×  x

   16 = x²

 √16 = x

    4  = x

Length of the side of the square   = 4      

Hope this helps you.

6 0
2 years ago
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