Answer:
r = (ab)/(a+b)
Step-by-step explanation:
Consider the attached sketch. The diagram shows base b at the bottom and base a at the top. The height of the trapezoid must be twice the radius. The point where the slant side of the trapezoid is tangent to the inscribed circle divides that slant side into two parts: lengths (a-r) and (b-r). The sum of these lengths is the length of the slant side, which is the hypotenuse of a right triangle with one leg equal to 2r and the other leg equal to (b-a).
Using the Pythagorean theorem, we can write the relation ...
((a-r) +(b-r))^2 = (2r)^2 +(b -a)^2
a^2 +2ab +b^2 -4r(a+b) +4r^2 = 4r^2 +b^2 -2ab +a^2
-4r(a+b) = -4ab . . . . . . . . subtract common terms from both sides, also -2ab
r = ab/(a+b) . . . . . . . . . divide by the coefficient of r
The radius of the inscribed circle in a right trapezoid is r = ab/(a+b).
_____
The graph in the second attachment shows a trapezoid with the radius calculated as above.
3/6 because there are six sides to a dice.
(3782 - 192) : 2 = 1795 (first number)
1795 + 192 = 1987 ( second number)
--------------------------
1987 - 1795 = 192
1987 + 1795 = 3782
Diagonals of a rectangle are congruent.
18x - 28 = x + 380
Solving for x, I get 17.
So, x = 17.
PR = 18x - 28
PR = 18(17) - 28
PR = 278
QS = x + 380
QS = 17 + 380
QS = 397
The answers are
f
o
g
(
x
)
=
−
2
x
+
23
and
g
o
f
(
x
)
=
−
2
x
+
5
Explanation:
f
(
x
)
=
−
2
x
+
11
g
(
x
)
=
x
−
6
f
o
g
(
x
)
=
f
(
g
(
x
)
)
=
f
(
x
−
6
)
=
−
2
(
x
−
6
)
+
11
=
−
2
x
+
12
+
11
=
−
2
x
+
23
g
o
f
(
x
)
=
g
(
f
(
x
)
)
=
g
(
−
2
x
+
11
)
=
−
2
x
+
11
−
6
=
−
2
x
+
5
I think that the equations speak by themselves.
Of course,
f
o
g
(
x
)
≠
g
o
f
(
x
)