Answer:
mkkkmnsine kgsk jtb jgjdb Geeta kl John ku
We know that:

is an equation of a circle.
When we substitute x and y (from the pairs we have), we'll get a system of equations:

and all we have to do is solve it for a, b and r.
There will be:

From equations (II) and (III) we have:

and from (I) and (II):

Now we can easly calculate a and b:

Finally we calculate

:

And the equation of the circle is:
Answer:
Final answer is
.
Step-by-step explanation:
given functions are
and
.
Now we need to find about what is the value of
.
simply means we need to multiply the value of
and
.


Hence final answer is
.
9514 1404 393
Answer:
24 cm²
Step-by-step explanation:
To find the area of the trapezium, we must know the length CD. That means we must know the length BC. Fortunately, the perimeter of ABCF is given, so we have ...
P = 2(AB +BC)
BC = (P/2) -AB = (20 cm)/2 - 6 cm = 4 cm
Then CD is ...
CD = BD -BC = 9 cm -4 cm = 5 cm
The area of the trapezium is given by the formula ...
A = (1/2)(b1 +b2)h
A = (1/2)(5 cm + 3 cm)(6 cm) = 24 cm²
The area of trapezium CDEF is 24 cm².
Answer:
- 10 liters of orange juice
- 5 liters of champagne
Step-by-step explanation:
Let c represent the number of liters of champagne Lauren uses. Then (15-c) will be the number of liters of orange juice. The total cost of the mix will be ...
12c +1.50(15-c) = 5.00(15)
10.5c = 52.50 . . . . . subtract 22.50, simplify
52.50/10.5 = c = 5 . . . . divide by the coefficient of c
Then the amount of orange juice is ...
15 -c = 15 -5 = 10 . . . . liters
Lauren should use 5 liters of champagne and 10 liters of orange juice.