I did another one just like this about an hour ago. I did it the elegant, tedious way,
and it took about 2 feet of screen. Then I realized there was a totally bourgeois but
easy way to do it. For 5 points, I'll give you the method for the masses:
List all the possible pairs of digits that add up to 9, along with their flips:
18 . . . . . 81
27 . . . . . 72
36 . . . . . 63
45 . . . . . 54
Do you see any pair that are different by 27 ?
How about 36 and 63 ?
So the original number is 63, and when you flip it, it becomes 36, which is 27 less.
(AB)^2 + (2 m)^2 = (8 m)^2, or
(AB)^2 + 4 m^2 = 64 m^2, or (AB)^2 = 60 m^2.
Taking the square root of both sides, (length of AB) = + 2sqrt(15) (answer)
It is the first one you see on the top of the page
Answer:
54
Step-by-step explanation:
To solve problems like this, always recall the "Two-Tangent theorem", which states that two tangents of a circle are congruent if they meet at an external point outside the circle.
The perimeter of the given triangle = IK + KM + MI
IK = IJ + JK = 13
KM = KL + LM = ?
MI = MN + NI ?
Let's find the length of each tangents.
NI = IJ = 5 (tangents from external point I)
JK = IK - IJ = 13 - 5 = 8
JK = KL = 8 (Tangents from external point K)
LM = MN = 14 (Tangents from external point M)
Thus,
IK = IJ + JK = 5 + 8 = 13
KM = KL + LM = 8 + 14 = 22
MI = MN + NI = 14 + 5 = 19
Perimeter = IK + KM + MI = 13 + 22 + 19 = 54
F(-2) = 7
f(0) = 6
f(6) = 3
f represents the function and the number in the parentheses is the input for the function, y on the graph is the output