Let's define a few variables to start.
Let A = one digit of the three-digit number
Let B = one digit of the three-digit number
Let C = one digit of the three-digit number
For problems like these, it is generally helpful to find an expression that relates everything together. \(ABC, ACB, BAC, BCA, CAB, CBA\) represent all the six different 3-digit numbers. We can then represent the sum of all the unique 3-digit numbers with the expression \(222(A+B+C)\) because each digit is added twice in the units, tens, and hundreds digit.
Also, it is generally helpful to find a bound of some kind that restricts the possibilities. This way, it is not necessary to check every single possibility mindlessly. We can achieve this by recognizing that we can represent the sum in a different way. The sum must be at least \(3231+123=3354\) and must not exceed \(3231+987=4218\).
With this information, we can solve an inequality that restricts the possibilities for \(A+B+C\).
\(3354\leq 222(A+B+C)\leq4218\\ 15.108...\leq A+B+C \leq 19\)
Of course, in the context of this problem, \(A+B+C\) is restricted to the set of whole numbers. Therefore, \(A+B+C\) equals 16,17,18, or 19. After this point, we will just test all the possibilities since we only have 4 to check anyway.
We can represent the 6th number as \(222(A+B+C)-3231\). Below, I have included a table for each possibility for \(A+B+C\):
\(A+B+C\) | 6th Number |
\(16\) | \(321\) |
\(17\) | \(543\) |
\(18\) | \(765\) |
\(19\) | \(987\) |
From this table, only one of possibilities for the 6th number reigns supreme, 765, because 765 is the only number where its digits add to its corresponding \(A+B+C\) value.
\((x-h)^2+(y-k)^2=r^2\) is the center-radius form for a circle such that \((h,k)\) is the coordinates of the center of the circle on the coordinate plane and \(r\) is the radius.
1) Find the equation of the circle centered at (1,-4) with radius 3
Since \((h,k)\) of \((x-h)^2+(y-k)^2=r^2\) represents the coordinates of the center and \((1,-4)\) is already known to be the center, then we know that \(h=1\text{ and }k=-4\) . \(r\) is the radius, so r=3.
Substitute these values into the center-radius form of a circle to generate the equation of this particular circle.
\(h=1,k=-4,r=3\\ (x-h)^2+(y-k)^2=r^2\\ (x-1)^2+(y-(-4))^2=3^2\\ (x-1)^2+(y+4)^2=9\)
2) I encourage you to use the information I gave you above to try and work out the second problem.
3) Find the distance between the points \((0,8)\text{ and }(2,-5)\).
We can use the distance formula, \(d=\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}\) . Substitute the points that we want to find the distance from.
\(d=\sqrt{(8-(-5))^2+(0-2)^2}\\ d=\sqrt{13^2+(-2)^2}\\ d=\sqrt{169+4}\\ d=\sqrt{173}\) | Substitute in the known coordinates and simplify completely. 173 has no perfect square factors, so the radical cannot be simplified any further. |