.hello :
an equation of the circle <span>Center at the w(a,b) and ridus : r is :
(x-a)² +(y-b)² = r²
in this exercice : r = 1
</span><span>The points (-18,15) and (-20,15) lie on a circle with a radius of 1:
</span>(-18-a)²+(15-b)² = 1 ....(1)
(-20-a)² +(15-b)² = 1 ....(2)
solve this system :
(1) -(2) : (-18-a)² - (-20-a)² =0
(-18-a)² =(-20-a)² =0
( -18-a = -20-a) or (-18-a = - (-20-a))
1 ) ( -18-a = -20-a) no solution confused : -18=-20
2 ) -18-a =20+a
-2a =38
a = -19
subst in (1) :(-18+19)²+(15-b)² =1
(15-b)² = 0.... 15-b = 0 .... b = 15
the center is :w(-19,15)
Answer:
Step-by-step explanation:
Finding the domain:
The domain of a function is the set of possible input values for which the function is real and defined.
Given the function
The function has no undefined points nor domain constraints. Hence, the domain is
i.e.
Finding the range:
The range of a function is the set of possible output values (dependent variable y values) for which a function is defined.
The range of polynomials with odd degree is all the real numbers.
Hence, the domain is
i.e.
Answer:
1200
Step-by-step explanation:
ghgvhjhv g fty ftvyuibn
Answer:
c. 14
Step-by-step explanation:
Let the missing y value be y2.
If the data represent a linear function, then:
Multiply both sides by 2
Add 6 to both sides
The missing value is 14