12. ∆ABD is similar to ∆PQD, so
QD/z = BD/x
Likewise ∆CDB is similar to ∆PQB, so
QB/z = BD/y
Since QD + QB = BD, you have
QD/z + QB/z = BD/x + BD/y
BD/z = BD/x + BD/y
Dividing by BD gives the desired result:
1/x + 1/y = 1/z
13. Triangles ABD, BCD, and ACB are all similar. This means
AB/AC = AD/AB
AB² = AC×AD = (4 cm)×(9 cm)
AB² = 36 cm²
AB = 6 cm
and
BD/CD = AD/BD
BD² = CD×AD = (5 cm)×(4 cm)
BD² = 20 cm²
BD = 2√5 cm
Multiply both numerator and denominator with 8+5i. Then you get
Let's follow the transformations that happen to A, to get to A' and A''.
Point A is at (-5, -2)
It moves to (-5, 2) which is where A' is located. Note the x coordinate stays the same while the y coordinate flips from negative to positive. This must mean we applied a reflection over the x axis.
That rule in general is 
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Now compare A'(-5,2) and A''(1,4). We can shift A' 6 units to the right and then 2 units up so we move from A' to A''.
Algebraically this is stated as 
Whatever the x coordinate is, add 6 to it. For the y coordinate, we add on 2.
Applying that rule to B'(-1,2) gets us to

which is the proper location of B''
The same applies to moving C' to C''

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In summary, we started off by reflecting over the x axis. Then we applied the translation rule of "shift to the right 6 units, shift up 2 units".
In terms of algebra, combining the rules
and
will have us end up with 
Answer:J
Jack
Step-by-step explanation:
Jack: 275/5.5 = 50mph
Gavin: 288/6 = 48mph