2⃣, -4⃣ = x; no radical necessary.
Answer:
∠RPQ = 27
Step-by-step explanation:
In ΔSRQ,
∠R = 90
∠SQR = 36°
∠R + ∠SQR + ∠RSQ = 180 {Angle sum property of triangle}
90 + 36 + ∠RSQ = 180
126 + ∠RSQ = 180
∠RSQ = 180 - 126
∠RSQ = 54°
∠PSQ +∠RSQ = 180 {Linear pair}
∠PSQ + 54 = 180
∠PSQ = 180 - 54
∠PSQ = 126
In ΔPSQ,
SQ = PS ,
So, ∠SQP = ∠SPQ {Angles opposite to equal sides are equal}
∠SQP = ∠SPQ =x
∠PSQ + x +x = 180 {Angle sum property of triangle}
126 + 2x = 180
2x = 180 - 126
2x = 54
x = 54/2
x = 27
∠RPQ = 27°
The converted to radians is 4(pi)
Answer: Hello mate!
Let's define the variable t as the time, and define t = 0 as the moment when the first skater starts to move:
We know that the speed of the first skater is 8 m/s, and we need to find the position as a function of time, then we need to integrate the velocity over time
if v1(t) = 8m/s
then p1(t) = (8m/s)*t
now we also know that the second skater has a velocity of 9m/s and enters in the frozen lake at t= 10s.
then the velocity of the second skater is: v2(t) = 9m/s, and the position is:
p2(t) = (9m/s)*(t - 10s)
now we want to know how many seconds after the second skater starts are needed for the second skater to overtake the first one.
this is equivalent to see when his positions will be equal.
so p1(t) = p2(t):
(8m/s)*t = (9m/s)(t - 10s) = (9m/s)*t - 90m
(8m/s)*t - (9m/s)*t = -90m
(-1m/s)*t = 90m
t = 90m/(1m/s) = 90s
Then in t = 90 seconds, the second skater will overtake the first one, and knowing that the second skater started at t = 10 seconds; there are 80 seconds after the second skater started needed to overtake the first skater.
