Answer and solution below:
Answer:
Andrew must run 1 5/12 Laps to complete the race.
Step-by-step explanation:
Total lap run by Michael, Nate, Ben, and Andrew = 4 laps
laps run by Michael = 1 1/3 laps =(3+1)/3 = 4/3 laps
laps run by Nate =1/2 laps
laps run by ben =3/4 laps
Let laps run by Andrew = x
therefore
laps run by Michael + laps run by Nate + laps run by ben + laps run by Andrew = 4 laps
4/3 + 1/2 + 3/4 + x = 4
LCM of 3,2 and 4 is 12
=> (4*4 + 6*1 + 3*3+12x)/12 = 4
=> (16+6+9+12x)/12 = 4
=> 31+12x = 12*4 = 48
=> 12x = 48-31= 17
=> x = 17/12 = 1 5/12
Thus, Andrew must run 1 5/12 Laps to complete the race.
Answer:
y
Step-by-step explanation:
The value of y is dependent on the value of x+3 because y represents leilas age.
9514 1404 393
Answer:
(i) x° = 70°, y° = 20°
(ii) ∠BAC ≈ 50.2°
(iii) 120
(iv) 300
Step-by-step explanation:
(i) Angle x° is congruent with the one marked 70°, as they are "alternate interior angles" with respect to the parallel north-south lines and transversal AB.
x = 70
The angle marked y° is the supplement to the one marked 160°.
y = 20
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(ii) The triangle interior angle at B is x° +y° = 70° +20° = 90°, so triangle ABC is a right triangle. With respect to angle BAC, side BA is adjacent, and side BC is opposite. Then ...
tan(∠BAC) = BC/BA = 120/100 = 1.2
∠BAC = arctan(1.2) ≈ 50.2°
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(iii) The bearing of C from A is the sum of the bearing of B from A and angle BAC.
bearing of C = 70° +50.2° = 120.2°
The three-digit bearing of C from A is 120.
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(iv) The bearing of A from C is 180 added to the bearing of C from A:
120 +180 = 300
The three-digit bearing of A from C is 300.