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:
Answered by Norbert J.M..
Hope this helps.
Hello there!
n - 5 ≤ 5n - 1
Solve for n
Let's start by subtracting 5n from both sides
n - 5 - 5n ≤ 5n - 5n - 1
n - 5 - 5n ≤ -1
We need to transfer -5 on the other side, we can do that by adding 5 on both sides
n - 5 - 5n + 5 ≤ -1 + 5
n - 5n ≤ -4
-4n ≤ -4
Finally divide both sides by -4
-4n/-4 ≤ -4/-4
n ≥ -1 (This is the answer)!
Do you know why I changed the sign?
A lot of students failed to remember that rules, which is:
When you are solving an inequality, if you divide both sides by a negative sign, you MUST change the symbol as well. If it was this <, it will change to this >. Got it? Cool!
I hope the steps are clear to understand. If you have questions, feel free to let me know...
As always, I am here to help!
Answer:
54.431 kilograms
Step-by-step explanation:
don't have step by step but 1kg is 2.205lbs
Answer:
y=3x-38
Step-by-step explanation:
We know that Tracy spent $38 in less than 3 times than what Daniel spent. Given this information we can put our eqaution together.
Y=3x-38
Hope this helps! :)