.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: 12.32 is the answer.
Step-by-step explanation: Using the pythagorean thereom we can solve this by using the formula A^2+B^2=C^2.
We substitute in the values,
17^2+B^2=21^2.
Now we simplify.
289+B^2=441
We subtract the 289 from both sides,
B^2=152
Now we use square root to find the answer.
12.32 is the answer.
Hope this helps!
Answer:
C. )
Step-by-step explanation:
Answer:
X=0
Step-by-step explanation:
It will be 1 of the 100block
5 of the tens block and 2 single
